Bernie Sanders

From: JoeBiden.com
Date: Tue, Mar 3, 2020 at 11:38 PM
Subject: Bernie Sanders
To: ajay mishra

This week, Bernie Sanders’ campaign reported that they raised a staggering $46 million in February.

‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌
ajay — we know we keep emailing you and asking you for $5.

Here’s why: this week, Bernie Sanders’ campaign reported that they raised a staggering $46 million in February.

And while that might seem like bad news, we actually have a real chance to compete against this latest haul.

Look, given our surge in small-dollar fundraising following our win in South Carolina, we are in a better position than ever before to compete head-to-head against the Sanders campaign’s money.

But, we have to keep closing the money gap before polls close tonight, and that’s where you come in, ajay.

If you pitch in $5 before the polls close tonight, we’ll be in a terrific position to win tonight in Super Tuesday states and beyond »

Countdown Until the First Polls Close Tonight
Deadline approaching
Donor Record for ajay mishra
ZIP: 78746
Donated to Joe Biden? No, not yet
Suggested Action? Pitch in $5 before polls close tonight »

If you’ve saved payment information with ActBlue Express, your donation to Joe Biden will go through immediately:

Chip in $5 ➞
Chip in $25 ➞
Chip in $50 ➞
Chip in $100 ➞
Chip in $250 ➞
Or donate another amount ➞

Let’s make sure that we have every resource at our disposal to win tonight and keep the momentum going, ajay!

Pitch in what you can today:

https://go.joebiden.com/Donate

Thank you,

JoeBiden.com

Any donor history information in this email reflects what we have on file for this specific email address. If you have donated with a different email, with a check, or with a spouse — thank you so much! We have that on file and cannot thank you enough for supporting this campaign.

This email was sent to ajayinsead03. If we got your name or any information wrong — we’re so sorry! To update and correct your information click here. If you would like to receive fewer emails, click here. If you would like to unsubscribe, click here.

ajay, thank you so much for supporting Joe Biden’s Presidential campaign.

Change occurs because the conscience of a country begins to rise up and demand — demand change.

This isn’t the time to be complacent. If you are ready to fight for the soul of this nation, you can start by donating to elect Joe Biden by clicking the button below.

DONATE TO ELECT JOE BIDEN »

We know we send a lot of emails, and we are sorry about that. The reason? We are relying on grassroots supporters like you (we’re serious!).

But ajay, we don’t want to bother you. If you’d like to only receive our most important emails, click here. If you’d only like to receive volunteer emails, click here. If you’d like to unsubscribe from our emails, you can click here.

To make a contribution by mail, click here for instructions.

We sincerely thank you for your help and support.
– The entire Joe Biden for President team

Paid for by Biden for President

open.gif

FROM OLGA LEDNICHENKO TO BILL CLINTON I THINK ABOUT MAKE – MODEL AND MATERIAL OF MY TABLE AND COMPARE IT WITTH THAT OF OTHER PEOPELS; KITCHEN TBALES – IS THIS A GOOD SUMMARY FOR MALES AND U ? : I MEAN ITS JUST TO BE MAPPED WITH THE AMERICAN DREAM RAHMBO – SEE BELOW AND HOPE THIS IS GOOD ENOUGH FOR NOT JUST THE WELL OFF JOHN DOE BUT ALSO FOR THE AVERGAE JOE: PS: MAYBE RUSSIANS ARE MORE OR LESS ARTUCULATE THAN GOOGLE, BUT HERE I PRESENT MY SWANG SONG TO U PRESDIENT CLINTON… DIDNT U SAY TO AL GORE ITS CALED BRIDGE TO THE 21ST CETRUTY IS YANDEX GOOGLE AND BAIDU ? WELL, I GT LIVE JOIURNAL FROM U AND LIVE DORO FROM JAPAN PRESIDNET CLINTON AND YES.. THANKS FOR COMING TO RUSSIA AND WE LOVE U – EVEN IN RUSSIANS NOT JUST IN HINDOOS AND NO – THEY DONT LIKE CLINTON WORD TOO MUCH IN ARBA WORLD

 

PRIVET BUBBA KAK DILA TIBYA - REGARDS OLGA

https://www.google.com/search?q=FROM+OLGA+LEDNICHENKO+TO+BILL+CLINTON+I+THINK+ABOUT+MAKE+-+MODEL+AND+MATERIAL+OF+MY+TABLE+AND+COMPARE+IT+WITTH+THAT+OF+OTHER+PEOPELS%3B+KITCHEN+TBALES+-+IS+THIS+A+GOOD+SUMMARY+FOR+MALES+AND+U+%3F+&tbm=isch&ved=2ahUKEwjyjOWo8f7nAhUU4DgGHaezDxgQ2-cCegQIABAA&oq=FROM+OLGA+LEDNICHENKO+TO+BILL+CLINTON+I+THINK+ABOUT+MAKE+-+MODEL+AND+MATERIAL+OF+MY+TABLE+AND+COMPARE+IT+WITTH+THAT+OF+OTHER+PEOPELS%3B+KITCHEN+TBALES+-+IS+THIS+A+GOOD+SUMMARY+FOR+MALES+AND+U+%3F+&gs_l=img.3…23168.24492..25038…0.0..0.0.0…….0….1..gws-wiz-img.NIaeuiss01A&ei=S5teXrK7EJTA4-EPp-e-wAE&bih=758&biw=1519&hl=en#imgrc=Q3hN9l7QB9RWiM

2 of 18,802 Someone just viewed: Fwd: Someone just viewed: Fwd: Someone just viewed: Fwd: Failed attachments for FAREEINDS SEE THISE – OLGA IS NOR EXAPRT ON JOE KING AND NOR SERIOSIEASECTS AEVAN THAU AJAY TRIES SO HARD TO DISTARTEDC HER – SEETIS BUT ME AND WILIAME CLINTONE IS SUCH COOL NOR AJAY AND OBAMA AND BB NETANYAHOO AND THY ARE EVAN TRYONG TO GET PUTIN ONTHEIR ANGLES – WHEECH IS SUCH BAD CONASWEQIERES AGAINST ME WHEECH I CDMEET IN PRIVATE I WONT ADEEMT BUT WTHIN SMALL MOMENST OLGA ANDN ME WITH FEIKO AND MY SCRETE FAREIEDN NAEMD AJAY ROOMATE – WHOC IS ALSO FOOL, HE HAS SISTER – WHO MARRIED SOEONEW WHO INTO SPACE AND ISRO – AND THAR PERSONES ISIN CHICAGO WHEECH IAJAY AND OBAMA SAYS THAT IS CHICAGO MAFIA BUT THAR IS NOTHING TO ME BECAUSE CHICGAO MAFIA NR KNWOWS THAT HAILEIR CLINTON IS ALSO FROM CHICAGO – REGARDS – HOTMAN AND OLGA AND HILLAEIRE CLINTON Inbox x BIG SHOT FROM CHICAGO x Streak 11:05 PM (18 minutes ago) to me Someone just viewed: “Fwd: Someone just viewed: Fwd: Someone just viewed: Fwd: Failed attachments for FAREEINDS SEE THISE – OLGA IS NOR EXAPRT ON JOE KING AND NOR SERIOSIEASECTS AEVAN THAU AJAY TRIES SO HARD TO DISTARTEDC HER – SEETIS BUT ME AND WILIAME CLINTONE IS SUCH COOL NOR AJAY AND OBAMA AND BB NETANYAHOO AND THY ARE EVAN TRYONG TO GET PUTIN ONTHEIR ANGLES – WHEECH IS SUCH BAD CONASWEQIERES AGAINST ME WHEECH I CDMEET IN PRIVATE I WONT ADEEMT BUT WTHIN SMALL MOMENST OLGA ANDN ME WITH FEIKO AND MY SCRETE FAREIEDN NAEMD AJAY ROOMATE – WHOC IS ALSO FOOL, HE HAS SISTER – WHO MARRIED SOEONEW WHO INTO SPACE AND ISRO – AND THAR PERSONES ISIN CHICAGO WHEECH IAJAY AND OBAMA SAYS THAT IS CHICAGO MAFIA BUT THAR IS NOTHING TO ME BECAUSE CHICGAO MAFIA NR KNWOWS THAT HAILEIR CLINTON IS ALSO FROM CHICAGO – REGARDS – HOTMAN AND OLGA AND HILLAEIRE CLINTON” People on thread: 208 Westhaven Drive 78746 Blog Post By Email Device: PC Location: Chicago, IL

[1]

HOTMAN WHARS THSIE SUBJECTS ON OLGA TELL ME

[2]

HOTMAN WHARS THSIE SUBJECTS ON OLGA TELL ME

[3]

HOTMAN WHARS THSIE SUBJECTS ON OLGA TELL ME

643361467 FOR 324368521 OLGA SHULMAN LEDNCIHENJO SANKAY DUTT KABALA NUMBERS KABALAH REGARDS 324368521 10 ASPECTS OF GOD - U CAN AND U CANT KNOW WHAT DOES DOES ONLY WAHT GOD DOESNT DO - SAID M FOR WHO - OLGA SHULMAN LEDNICHENKO BOOOK - CHAPTER 1.2 AJAY MISHRA--- KABALAH REGARDS 324368521 10 ASPECTS OF GOD - U CAN AND U CANT KNOW WHAT DOES DOES ONLY WAHT GOD DOESNT DO - SAID M FOR WHO - OLGA SHULMAN LEDNICHENKO BOOOK - CHAPTER 1.2 AJAY MISHRA AJAY OBAAM KABALA NUMBERS YES ROBERT THAT IS TRUE - TO ROBERT SPANO AND KABALA NUMBERS FROM OLGA AND AJAY LEDNICHENKO AND SANJAY AND WHO KABALA NUMBERS - 643361467 AND 6 AND AJAY MISHRA AND YONI NETANAYHU UNKNOWN NUMBERS 324368521 OGA SHULMAN LEDNICHENKO WORDS MAPS DIAGRAMS, AND NUMBERS - K FOR KABALA , K FOR KABAL K FOR WHAT -- OLGA LCINTON AND MISHRA MODI MAPS 324368521 OGA SHULMAN LEDNICHENKO WORDS MAPS DIAGRAMS, AND NUMBERS - K FOR KABALA , K FOR KABAL K FOR WHAT

DOWN ARROW

CLICKS THIS

Super Tuesday: Official Texas Straw Poll

I LIKE JOE BIDEN AND KAMALA HARRIS

From: for Ajay’s eyes only
Date: Tue, Mar 3, 2020 at 11:01 PM
Subject: Super Tuesday: Official Texas Straw Poll
To: Ajay Mishra

We need [23] responses by 11:59pm.

‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌
End Citizens United is dedicated to electing Democrats, transforming our broken campaign finance system, and ultimately ending Citizens United. If you really want to receive only our most urgent emails or unsubscribe, click here.

➤➤ RESPOND NOW: Begin Official Super Tuesday Straw Poll >>
END CITIZENS UNITED OFFICIAL 2020 FOCUS GROUP
Ajay: You have been selected to take our Official Super Tuesday Straw Poll >>
Super Tuesday is TODAY. The candidate who wins the most states today is likely to be the Democratic nominee.
So we are polling Austin, Texas Democrats on their preferences, but our poll closes tonight at 11:59 PM.
8 Democratic candidates are running to defeat Trump, but it remains unclear who has the most support among End Citizens United members.
Begin your official Straw Poll to let us know your 2020 choice:
BEGIN OFFICIAL 2020 STRAW POLL >>
NAME: AJAY MISHRA
STATUS: TOP DEMOCRAT
It’s essential we get your response to know who top Democrats from Texas are supporting in the Presidential Primary in order to beat Trump in 2020.
And Ajay, you were selected as one of our top Texas Democrats to take our Official 2020 Straw Poll right now. So don’t delay:
BEGIN OFFICIAL 2020 STRAW POLL >>
Thank you for your valuable input,
-End Citizens United Research Team

This message was sent to: ajayinsead03@gmail.com
Click here to receive fewer emails. | Click here to unsubscribe.
End Citizens United
Thank you for supporting End Citizens United PAC. Our entire team is working to the bone to pass SWEEPING Campaign Finance Reform — the future of our Democracy depends on it:
DONATE TO END CITIZENS UNITED ➞
When we pass the Constitutional Amendment to overturn Citizens United, our movement will go down in history as HEROES. And Ajay, when the next generation asks what you did to stop Donald Trump, you’ll be able to say you were a Proud Member of End Citizens United.
We know we send a lot of emails. And you might be wondering why: We aren’t a SuperPAC. We’re not funded by billionaires. We rely on Democrats giving an average gift of $13 to emails like this to fuel our entire operation. We spend EVERY DIME so efficiently by running proven effective ads to win the most competitive races, pushing HARD to pass Sweeping Campaign Finance Reform, and building our movement.
However, if you’d like to receive fewer emails, you can click here. If you’d like to unsubscribe from our emails, click here. If you’d like to donate to help fund our efforts to fix America’s broken campaign finance system, please click here.
From the entire End Citizens United team, thanks for your support!

Tell Kamala which one of these issues is most important to you >>

I THINK ANYONE WHO SAYS KITCHEN TABLE ISNT IMORTNT IS DENYING THEIR OWN FAMILY

AND

THSOE WHO SAY MENDELEEV TBLE CAN BE COMLETELY UNEARTHEED HAVE TO LOOK AT SOME OTHER BACK YARDS

AND ALSO, ITS ALSO IMORTANT THE SIZE OF THE KITCHEN TABLE – MENAING – ARE WE FEEL COMFORTABLY NUMB THET

[A] SOME PEOPLE ARE HAVING FOOD FIGHTS AT THE JITCHEN TABLE

OR

[B] SOME PEOPLE ARE JIST AFRAIF OF INCREAISING THE SIZE OF THEIR EXISTING TABLE – BECAUSE – KIDS MEANS DIAPRES AND NAPKINS – AND THEIR COST HAS GONE THRU THE ROOF

OR

[C] SOME PEOPLE CANT AFFROD BEYOND THE SYTLE OF THEIR KICTHEN TABLE, BECAUS ETHE UDBUDEGT LINE IS MAKING DEMANDS ON THE OTHER SIDE

OR

[D] SOME PEOPLE ARE WONDERING IF OTHERS ON THE OTHER SIDE OF FLORIDA’S KEY WEST AND HAWAII’S WEST – ARE HAVING ENOUGH ON THERI TABLE, SO AS TO NOT REACH A TAGE OF DEPSRETATON – SO MUCH SO THAT OUR KISTCHJENS MAY BE DEMOMOLISHED BY AN ANGRY MOB, OUT SIDE OUR FRONT OR BACK YARD

OR

[E] SOME PEOPLE ARE WORRIED THAT – WHILE WE SIT IN OUR ARMS CHAIRS AND SVIVEL AROUND LIKE A MAFIA DON – ARE WE COGNIZANT ABOUT THE TABLES AND CHAIRS –

IN OTEHR PEOPELS HOMES HOMES THEY CALL TEHRI WON AND HAVE PROBLEMS IF YOU ARE INSENSITIVE OR NOT INCENTIVIZED BCEUA EOF OTEITEHR MOON

OR THE VOTE BACK OF YOUR RBOWN AND LIBERAL LEFT OR – FEARFUL THAT YOUR MACHO – MARLBOROO, MEN LIKE TO PALY RUSSION ROULELTTE IN MIDEDLE EASTA DN AFGHAN AND ALL YOU CARE OF IS YOUR TBLES AND YOUR KITCHNE NOT THEIRS

OR

[6] ALL OF THE ABOVE AND MORE ITEMS SONGS – I CAN ADD BUT THE ABOVE 6 – IS ENOUGH FOR EME TO CHEW ON..

AS A GIRL, IHAVE FOUND AMERICAN MEN TO BE HIGH OF TESTOSTRRSTERONE AND ON BOMBSTIC WORDS,

BUT MAY I DARE SAY WE HAVE A WORD CALED M FOR MENSCH ALSO – NOT JSJST M FOR MARLBORO MAN –

MAY BE DOUG CAN EXPLAIN WHY JEWISH MEN ARE SOFT, AND SWEET AND COTTON AND CANDY AND WHILE ST TRUE THAT THEY ARENT AS MACHO – AS IS THE BROWN MEDIVIAL MAN – BUT ITS A TARDE OFF – ANY GIRL CAN MAKE :

I HOPE U TAKE THAT AS A COMPLIMENT – AND YES, U HAVE A SOULFUL FACE – BUT THE JEW – CAME ON THE STAGE TO PROETCT YOU FROM ROWDY LIBERAL LEFT OF THE LEFT – ANARCH MAN

:)

HOPE THIS IS ENOUGH FOR AN EVENING : MATCH :)

REGARDS

OLGA – THE RUSSHIAN – THE ONE WHO HAS ENOUGH TESTOSTERONE SUPPLIED BY A MARLBRO MAN TO LAST A LIFE TIME FOR THE JEW AND THE SISSY TRIBE -:)

PPS: I THINK THIS TIME YOU DINT OOZE GARANDISSONEESS AND CONFINED YOURSEELDF TO BEING A PROSECTOR –

AND DONT THEY SAY – IN USA – ALL FOREIGNES SAY – THEY DREAM BIIIIIIIIIIIG LIKE EVERYTHING IS BIG IN TEXAS –

KAMALA HARRIS - REGADSING KAMALA HARRIS MY OPINION - THANKS, AJAY

HELLO KAMALA SO WHAT S YOUR OPINION ON THIS SUBAJCTS KAMALA

KAMALA HAARIIS ON EASTMAN AND KODAK - KODIAK IS MISSING BUT THATS OK FOR NOW - NOW READ - ONE LINER - HARRIS ON TARDE TRUMP AND TARRIF kamala harris - freedom vector FROM OLGA AND CLINTON TO JOE AND KAMALA how 6 kamala TO KAMALA FROM OLGA LEDNCIHENKO

LIKE WITH BARACK – IT WA S GREEK COLOLUMN WITH OPRAH CYRING WHEN HE SAID MOON AND THE TIDES AND LAWRENCE OF ARABAI , GANDHI, GODFATEHR, AND CABALANCA IN THE SAME LINE :)

 

TO KAMALA HARRIS - YES.. IN MY HINDU OPINION - THATS - THE RIGHT WAY IE THE RIGHT STARTEGY FOR U AMDIST THE 200O PLUS - KAMALA DID U SAY FOOD ON TABLE

404

NA - HOTMAN WHARS ALL THSIE TELL ME HPTMN WHO THSIE IS THIS U OR OLGA RENT RELIEF ACT OF KAMALA HARRIS - SEE WHY ITS HARD TO LIVE ON INCOME THAT WE HAVE IN USA thanks, Olga

From: Team Kamala
Date: Tue, Mar 3, 2020 at 10:25 PM
Subject: Tell Kamala which one of these issues is most important to you >>
To: Olga Shulman Lednichenko

We are hoping to get a lot of responses today. Olga,

From time to time, we like to ask you what your #1 issue is, especially as we get closer and closer to election day.

We do this because it gives Kamala an idea about what her top supporters care about most.

So whether it’s climate change, gun violence, or healthcare we want to know what motivates you the most to get involved in the political process.

We are hoping to get a lot of responses today so we can show Kamala exactly what her supporters care about, and we would especially love to hear from you!

Please take a minute to click her and complete our brief survey and let us know what your #1 issue is heading into the 2020 Election.

TAKE THE SURVEY

Understanding what your top priorities are is essential to electing more Democrats so we can flip the Senate and make Donald Trump a one-term president.

We care about your opinions, and this is an opportunity for you to tell Kamala exactly where you stand.

Give your input today by taking our brief survey:

https://action.kamalaharris.org/signup/March-Issue-Survey

Thank you for making your voice heard,

Team Kamala

PAID FOR BY FRIENDS OF KAMALA HARRIS

This email was sent to LEDNICHENKOOLGA. You received this message because you are subscribed to Senator Kamala Harris’ mailing list. These emails are an important way to stay in touch with Senator Harris’ work, but if you would like you can unsubscribe from receiving further emails here. Finally, the vast majority of donations to Senator Harris’ re-election campaign come from lots of people making small donations. Can you add one today? Use this link.

From: Team Kamala Date: Tue, Mar 3, 2020 at 10:25 PM Subject: Tell Kamala which one of these issues is most important to you >> To: Olga Shulman Lednichenko   We are hoping to get a lot of responses today.Olga,From time to time, we like to ask you what your #1 issue is, especially as we get closer and closer to election day.We do this because it gives Kamala an idea about what her top supporters care about most.So whether it’s climate change, gun violence, or healthcare we want to know what motivates you the most to get involved in the political process.We are hoping to get a lot of responses today so we can show Kamala exactly what her supporters care about, and we would especially love to hear from you!Please take a minute to click her and complete our brief survey and let us know what your #1 issue is heading into the 2020 Election. TAKE THE SURVEYUnderstanding what your top priorities are is essential to electing more Democrats so we can flip the Senate and make Donald Trump a one-term president.We care about your opinions, and this is an opportunity for you to tell Kamala exactly where you stand.Give your input today by taking our brief survey:https://action.kamalaharris.org/signup/March-Issue-SurveyThank you for making your voice heard,Team Kamala PAID FOR BY FRIENDS OF KAMALA HARRIS This email was sent to LEDNICHENKOOLGA@GMAIL.COM. You received this message because you are subscribed to Senator Kamala Harris’ mailing list. These emails are an important way to stay in touch with Senator Harris’ work, but if you would like you can unsubscribe from receiving further emails here. Finally, the vast majority of donations to Senator Harris’ re-election campaign come from lots of people making small donations. Can you add one today? Use this link.            

From: Team Kamala <info@kamalaharris.org>
Date: Tue, Mar 3, 2020 at 10:25 PM
Subject: Tell Kamala which one of these issues is most important to you >>
To: Olga Shulman Lednichenko <LEDNICHENKOOLGA@gmail.com>

 

We are hoping to get a lot of responses today.Olga,

From time to time, we like to ask you what your #1 issue is, especially as we get closer and closer to election day.

We do this because it gives Kamala an idea about what her top supporters care about most.

So whether it’s climate change, gun violence, or healthcare we want to know what motivates you the most to get involved in the political process.

We are hoping to get a lot of responses today so we can show Kamala exactly what her supporters care about, and we would especially love to hear from you!

Please take a minute to click her and complete our brief survey and let us know what your #1 issue is heading into the 2020 Election.

TAKE THE SURVEY

Understanding what your top priorities are is essential to electing more Democrats so we can flip the Senate and make Donald Trump a one-term president.

We care about your opinions, and this is an opportunity for you to tell Kamala exactly where you stand.

Give your input today by taking our brief survey:

https://action.kamalaharris.org/signup/March-Issue-Survey

Thank you for making your voice heard,

Team Kamala

PAID FOR BY FRIENDS OF KAMALA HARRIS

This email was sent to LEDNICHENKOOLGA@GMAIL.COM. You received this message because you are subscribed to Senator Kamala Harris’ mailing list. These emails are an important way to stay in touch with Senator Harris’ work, but if you would like you can unsubscribe from receiving further emails here. Finally, the vast majority of donations to Senator Harris’ re-election campaign come from lots of people making small donations. Can you add one today? Use this link.

THIS ISNOT SATURDAY BUT ITS LIKE SERIOUS AND FUNNY: SO AMEIRICAN, IS ALL ABOUT WHAT – WHAT DID YONI SAY IN HIS BOOK” GIRLS AND CARS ? AND WHAT ELS EIS USA ABOUT ? OK SO I VE A QUETSIONS ITS A DICULT QUETION WHOIS BLOND? HINT COLOUR MATTERS, NT JUST COLOR

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

 

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

OLGA SHULMAN LEDNICHENKO PICAACO - VASSILY KANDISNKY - PARIS 8TH ARRONDISOMENT AND ALENXADER NEVSKY AND SEPTEMBER 11 AND RUSSHIAN CHURCH EQUAL TO

HOWZ 6 OLGA PRODUCTIONS

HERE IS SOME QUIZ HELP

HERE

AND SEE HERE

http://yoninetanyahu.com/wp-content/uploads/2020/03/WHO-IS-BLUE-AND-BLONDE.jpg

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

HEIDI KLUM AND ALINA MATSENKO CURTESEY OLGASHULMAN LEDNICHENKO AJAY KISHRA SALMNA KHAN KATRINA KAIF OSAMA BIN LADEN DAWOOD IBRAHIM CHTA SHAKEEL - BLLYWOOD STRUCTURES HELLO ALINA - SO, WHAT DID U ASK, SO BBODY LIKE ALINA - OR SHCISINA, AND - THEN U KOW PAIULINA - AND MERLY STREEP - AND - IF I HEAR EXCSUES - THENU KNOW IT - JUST TRUST ME U DONT KNOW WHAT CATEGROY PERSO I AM MRE PAINFUL THANU CAN BE OLGA AND ALINA - IS THIS PAIR alina - modeling OH AND THIS IS ALINA

Image result for FROM PRESIDENT WILLIAM JEFFERSON CLINTON - WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE - HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN - THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS - :)

HOWZ 6 OLGA PRODUCTIONS

 https://www.youtube.com/watch?v=VrsLXWGtRcs

 

HERE IS SOME QUIZ HELP

HERE

AND SEE HERE

http://yoninetanyahu.com/wp-content/uploads/2020/03/WHO-IS-BLUE-AND-BLONDE.jpg

Your Home Report for 1909 Payne Ave – Zestimate and Neighborhood Updates

404

 

404

From: Zillow
Date: Tue, Mar 3, 2020 at 9:31 PM
Subject: Your Home Report for 1909 Payne Ave – Zestimate and Neighborhood Updates
To:

RgRgQQRhPVcGemlsbG93QgoAI2F_Xl4UkvrPUh1hamF5Lm1pc2hyYUBhbHVtbmkuaW5zZWFkLmVkdVgEAAAAAA~~
The latest data on your home and real estate activity near you.
?tok=72274223-69ac-4290-9cdc-fbb41a72a318~X1-ZUz01ilitwqlu1_56tmi&service=emailtrackingservice

Zillow
Zestimate: $404,958 Zestimate Arrow Image
1909 ‌Payne ‌Ave, ‌Austin, ‌TX
2 bd, 1 ba, 720 sqft
prop-status-off-market.png Off Market
See how it changed
Check out your news feed
Feed Image
6309 Burns St # C-201 listed for rent for $1,299/mo
1mi away • 40m ago
Feed Image
6309 Burns St # D-307 listed for rent for $1,369/mo
1mi away • 41m ago
Feed Image
2211 W North Loop Blvd APT 223 listed for rent for $999/mo
1mi away • 11h ago
View More
Median Neighborhood Zestimate
Median Zestimate Image
+0.8% past 30 days
$476,800
Learn more
Just Listed
$380,000
3 bd | 2 ba | 1,204 sqft
Austin
$465,000
4 bd | 1 ba | 1,430 sqft
Austin
$899,000
4 bd | 3 ba | 2,300 sqft
Austin
View More
Sold Nearby
$389,000
3 bd | 2 ba | 1,102 sqft
Austin
$845,292
4 bd | 3 ba | 1,892 sqft
Austin
View More
Zillow, ‌Inc.
1301 ‌Second ‌Avenue, ‌Floor ‌31
Seattle, ‌WA ‌98101
© ‌2006‌-‌2020
Privacy policy | Unsubscribe from this email | Update your preferences

Streak 9:56 PM (11 minutes ago) to me Translate message Turn off for: Hindi Someone just viewed: ” BHULEKH Uttar Pradesh खाता विवरण (अप्रमाणित प्रति) ग्राम का नाम : बक्कास परगना : (मोहनलालगंज) तहसील : मोहनलालगंज जनपद : लखनऊ फसली वर्ष : 1423-1428 भाग : 1 खाता संख्या : 00074 खातेदार का नाम / पिता पति संरक्षक का नाम / निवास स्थान खसरा संख्या क्षेत्रफल (हे.) आदेश टिप्पणी श्रेणी : 1-क / भूमि जो संक्रमणीय भूमिधरों केअधिकार में हो। ओम प्रकाश श्रीवास्तव / स्व.एफ.पी.श्रीवास्तव / म.न.ए.706 से. सी.महानगर लखनऊ श्रीमती रेनू प्रकाश / ओ.पी. श्रीवास्तव / नि. A 706 सेक्टर सी महानग र लखनऊ प्रमोद कुमार सोनकर / विश्वनाथ सोनकर / 164/54अस्तबल मौलवीगंज चारबाग लखनऊ 750 0.8790 आदेश ना.तह.गोसाईगंज वाद स.टी-20161046390311136/5.12.16 खाता स.74 गाटा स.750/0.879हे. का 1/4 भाग रकबा 0.21975हे. विक्रेता के सम्‍पूर्ण अंश मा.गु.60ख से विक्रेता प्रमोद कुमार सोनकर पुत्र विश्‍वनाथ सोनकर नि.164/54 अस्‍तबल मौलवीगंज चारबाग लखनऊ का नाम खारिज करके क्रेता कमलेश कुमार पुत्र स्‍व.रामप्रसाद नि.दुल्‍लापुर पर. व तह. मोहनलालगंज लखनऊ का नाम जरिये बैनामा दि.15.10.16 दर्ज हो । ह.र.का./6.1.17 न्‍यायालय असिस्‍टेन्‍ट कलेक्‍टर प्रथम श्रेणी मोहनलालगंज लखनऊ वाद स.टी-2017104639038182 ता.फै.15.9.17 धारा 80 उ.प्र.राजस्‍व संहिता 2006 कमलेश बनाम सरकार ग्राम बक्‍कास वादीय भूमि गाटा स.750/0.879हे. का 1/4 भाग मे से रकबा 0.0523हे. स्थित ग्राम बक्‍कास को मा.गु.60ख से मुक्‍त करते हुये प्रश्‍नगत भूमि अकृषित घोषित की जाती है। ह.र.का./9.10.17 आदेश तहसीलदार मोहनलालगंज वाद सं.टी-2017104639037402/18.10.17 गाटा सं.750/0.879 का 1/4 भाग रकबा 0.2197हे.मे से 0.1672हे. मा.गु.60 से विक्रेता कमलेश कुमार पुत्र स्‍व.रामप्रसाद नि.दुल्‍लापुर का नाम खारिज करके क्रेता पप्‍पू रावत पुत्र स्‍व.हरदयाल नि.सरसवां अर्जुनगंज तहसील सरोजनीनगर लखनऊ का नाम जरिये बैनामा दि.19.8.17 दर्ज हो।ह.र.का/28.11.17 आदेश न्‍यायालय असिस्‍टेन्‍ट कलेक्‍टर प्रथम श्रेणी मोहनलालगंज लखनऊ वाद सं.टी-20171046390311678, 581/16-17 दि.30.12.17 धारा-80 उ.प्र. राजस्‍व संहिता 2006 पप्‍पू रावत बनाम सरकार ग्राम बक्‍कास वादीय भूमि खाता सं.74 गाटा सं.750/0.879हे. मे से रकबा 0.1672हे. स्थित ग्राम बक्‍कास पर. व तह.मोहनलालगंज लखनऊ को मालगुजारी 60ख से मुक्‍त करते हुये प्रश्‍नगत भूमि को कृषि प्रयोजन से भिन्‍न प्रयोजन की भूमि(अकृषिक) घोषित किया जाता है।ह.र.का./15.1.18 750मि./0.293 ओम प्रकाश श्रीवास्तव के हिस्से की भूमि अकृषिक घोषित। (धारा 143) योग 1 0.8790 कृपया उक्त खसरे की प्रस्थिति (भूखंड (गाटा) के वाद ग्रस्त /विक्रय /भू-नक्शा ) हेतु खसरा संख्या पर क्लिक करें Disclaimer: उक्त आँकडे मात्र अवलोकनार्थ हैं, तहसील कम्प्यूटर केन्द्र एवम सी.एस.सी/लोकवाणी केन्द्र से उद्धरण की प्रमाणित प्रति प्राप्त की जा सकती है । Software Powered By: National Informatics Center, Uttar Pradesh State Unit, Lucknow. ” People on thread: Yoni Netanyahu Olga Blog Post By Email Device: Android Tablet Location: Jakarta, JK

BHULEKA

 

HELLO KAJASRTA

WHERE IS KJAKARTA BUBBA

OLGA AND SANJAY AND ALINA AND CHISTINA, HWERE IS KAKARTA ?

[youtube http://youtube.com/w/?v=6D9vAItORgE]

HWREE IS JAKARTA?

[youtube http://youtube.com/w/?v=Zn8q0LsO4PA]

OH HERE IS JAKARTA ?

[youtube http://youtube.com/w/?v=qOSwBIstZUs]

 

HWREE IS JAKARTA?

OH HERE IS JAKARTA ?

https://www.google.com/search?q=BHULEKH+Uttar+Pradesh+%E0%A4%96%E0%A4%BE%E0%A4%A4%E0%A4%BE+%E0%A4%B5%E0%A4%BF%E0%A4%B5%E0%A4%B0%E0%A4%A3+(%E0%A4%85%E0%A4%AA%E0%A5%8D%E0%A4%B0%E0%A4%AE%E0%A4%BE%E0%A4%A3%E0%A4%BF%E0%A4%A4+%E0%A4%AA%E0%A5%8D%E0%A4%B0%E0%A4%A4%E0%A4%BF)+%E0%A4%97%E0%A5%8D%E0%A4%B0%E0%A4%BE%E0%A4%AE+%E0%A4%95%E0%A4%BE+%E0%A4%A8%E0%A4%BE%E0%A4%AE+:+%E0%A4%AC%E0%A4%95%E0%A5%8D%E0%A4%95%E0%A4%BE%E0%A4%B8+%E0%A4%AA%E0%A4%B0%E0%A4%97%E0%A4%A8%E0%A4%BE+:+(%E0%A4%AE%E0%A5%8B%E0%A4%B9%E0%A4%A8%E0%A4%B2%E0%A4%BE%E0%A4%B2%E0%A4%97%E0%A4%82%E0%A4%9C)+%E0%A4%A4%E0%A4%B9%E0%A4%B8%E0%A5%80%E0%A4%B2+:+%E0%A4%AE%E0%A5%8B%E0%A4%B9%E0%A4%A8%E0%A4%B2%E0%A4%BE%E0%A4%B2%E0%A4%97%E0%A4%82%E0%A4%9C+%E0%A4%9C%E0%A4%A8%E0%A4%AA%E0%A4%A6+:+%E0%A4%B2%E0%A4%96%E0%A4%A8%E0%A4%8A+%E0%A4%AB%E0%A4%B8%E0%A4%B2%E0%A5%80+%E0%A4%B5%E0%A4%B0%E0%A5%8D%E0%A4%B7+:+1423-1428+%E0%A4%AD%E0%A4%BE%E0%A4%97+:+1+%E0%A4%96%E0%A4%BE%E0%A4%A4%E0%A4%BE+%E0%A4%B8%E0%A4%82%E0%A4%96%E0%A5%8D%E0%A4%AF%E0%A4%BE+:+00074+%E0%A4%96%E0%A4%BE%E0%A4%A4%E0%A5%87%E0%A4%A6%E0%A4%BE%E0%A4%B0+%E0%A4%95%E0%A4%BE+%E0%A4%A8%E0%A4%BE%E0%A4%AE+/+%E0%A4%AA%E0%A4%BF%E0%A4%A4%E0%A4%BE+%E0%A4%AA%E0%A4%A4%E0%A4%BF+%E0%A4%B8%E0%A4%82%E0%A4%B0%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%95+%E0%A4%95%E0%A4%BE+%E0%A4%A8%E0%A4%BE%E0%A4%AE+/+%E0%A4%A8%E0%A4%BF%E0%A4%B5%E0%A4%BE%E0%A4%B8+%E0%A4%B8%E0%A5%8D%E0%A4%A5%E0%A4%BE%E0%A4%A8+%E0%A4%96%E0%A4%B8%E0%A4%B0%E0%A4%BE+%E0%A4%B8%E0%A4%82%E0%A4%96%E0%A5%8D%E0%A4%AF%E0%A4%BE+%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A5%87%E0%A4%A4%E0%A5%8D%E0%A4%B0%E0%A4%AB%E0%A4%B2+(%E0%A4%B9%E0%A5%87.)+%E0%A4%86%E0%A4%A6%E0%A5%87%E0%A4%B6+%E0%A4%9F%E0%A4%BF%E0%A4%AA%E0%A5%8D%E0%A4%AA%E0%A4%A3%E0%A5%80+%E0%A4%B6%E0%A5%8D%E0%A4%B0%E0%A5%87%E0%A4%A3%E0%A5%80+:+1-%E0%A4%95+/+%E0%A4%AD%E0%A5%82%E0%A4%AE%E0%A4%BF+%E0%A4%9C%E0%A5%8B+%E0%A4%B8%E0%A4%82%E0%A4%95%E0%A5%8D%E0%A4%B0%E0%A4%AE%E0%A4%A3%E0%A5%80%E0%A4%AF+%E0%A4%AD%E0%A5%82%E0%A4%AE%E0%A4%BF%E0%A4%A7%E0%A4%B0%E0%A5%8B%E0%A4%82+%E0%A4%95%E0%A5%87%E0%A4%85%E0%A4%A7%E0%A4%BF%E0%A4%95%E0%A4%BE%E0%A4%B0+%E0%A4%AE%E0%A5%87%E0%A4%82+%E0%A4%B9%E0%A5%8B%E0%A5%A4+%E0%A4%93%E0%A4%AE+%E0%A4%AA%E0%A5%8D%E0%A4%B0%E0%A4%95%E0%A4%BE%E0%A4%B6+%E0%A4%B6%E0%A5%8D%E0%A4%B0%E0%A5%80%E0%A4%B5%E0%A4%BE%E0%A4%B8%E0%A5%8D%E0%A4%A4%E0%A4%B5+/+%E0%A4%B8%E0%A5%8D%E0%A4%B5.%E0%A4%8F%E0%A4%AB.%E0%A4%AA%E0%A5%80.%E0%A4%B6%E0%A5%8D%E0%A4%B0%E0%A5%80%E0%A4%B5%E0%A4%BE%E0%A4%B8%E0%A5%8D%E0%A4%A4%E0%A4%B5+/+%E0%A4%AE.%E0%A4%A8.%E0%A4%8F.706+%E0%A4%B8%E0%A5%87.+%E0%A4%B8%E0%A5%80.%E0%A4%AE%E0%A4%B9%E0%A4%BE%E0%A4%A8%E0%A4%97%E0%A4%B0+%E0%A4%B2%E0%A4%96%E0%A4%A8%E0%A4%8A+%E0%A4%B6%E0%A5%8D%E0%A4%B0%E0%A5%80%E0%A4%AE%E0%A4%A4%E0%A5%80+%E0%A4%B0%E0%A5%87%E0%A4%A8%E0%A5%82&source=lnms&tbm=isch&sa=X&ved=2ahUKEwigqd643P7nAhVFT30KHSyvBx4Q_AUoBHoECAYQBg&biw=1536&bih=758

From: Streak
Date: Tue, Mar 3, 2020 at 9:56 PM
Subject: Someone just viewed: BHULEKH Uttar Pradesh खाता विवरण (अप्रमाणित प्रति) ग्राम का नाम : बक्कास परगना : (मोहनलालगंज) तहसील : मोहनलालगंज जनपद : लखनऊ फसली वर्ष : 1423-1428 भाग : 1 खाता संख्या : 00074 खातेदार का नाम / पिता पति संरक्षक का नाम / निवास स्थान खसरा संख्या क्षेत्रफल (हे.) आदेश टिप्पणी श्रेणी : 1-क / भूमि जो संक्रमणीय भूमिधरों केअधिकार में हो। ओम प्रकाश श्रीवास्तव / स्व.एफ.पी.श्रीवास्तव / म.न.ए.706 से. सी.महानगर लखनऊ श्रीमती रेनू �
To:

5c06a15b95c700eebb7b9823_headerLockup_2x.png

Someone just viewed: ” BHULEKH Uttar Pradesh खाता विवरण (अप्रमाणित प्रति) ग्राम का नाम : बक्कास परगना : (मोहनलालगंज) तहसील : मोहनलालगंज जनपद : लखनऊ फसली वर्ष : 1423-1428 भाग : 1 खाता संख्या : 00074 खातेदार का नाम / पिता पति संरक्षक का नाम / निवास स्थान खसरा संख्या क्षेत्रफल (हे.) आदेश टिप्पणी श्रेणी : 1-क / भूमि जो संक्रमणीय भूमिधरों केअधिकार में हो। ओम प्रकाश श्रीवास्तव / स्व.एफ.पी.श्रीवास्तव / म.न.ए.706 से. सी.महानगर लखनऊ श्रीमती रेनू प्रकाश / ओ.पी. श्रीवास्तव / नि. A 706 सेक्टर सी महानग र लखनऊ प्रमोद कुमार सोनकर / विश्वनाथ सोनकर / 164/54अस्तबल मौलवीगंज चारबाग लखनऊ 750 0.8790 आदेश ना.तह.गोसाईगंज वाद स.टी-20161046390311136/5.12.16 खाता स.74 गाटा स.750/0.879हे. का 1/4 भाग रकबा 0.21975हे. विक्रेता के सम्‍पूर्ण अंश मा.गु.60ख से विक्रेता प्रमोद कुमार सोनकर पुत्र विश्‍वनाथ सोनकर नि.164/54 अस्‍तबल मौलवीगंज चारबाग लखनऊ का नाम खारिज करके क्रेता कमलेश कुमार पुत्र स्‍व.रामप्रसाद नि.दुल्‍लापुर पर. व तह. मोहनलालगंज लखनऊ का नाम जरिये बैनामा दि.15.10.16 दर्ज हो । ह.र.का./6.1.17 न्‍यायालय असिस्‍टेन्‍ट कलेक्‍टर प्रथम श्रेणी मोहनलालगंज लखनऊ वाद स.टी-2017104639038182 ता.फै.15.9.17 धारा 80 उ.प्र.राजस्‍व संहिता 2006 कमलेश बनाम सरकार ग्राम बक्‍कास वादीय भूमि गाटा स.750/0.879हे. का 1/4 भाग मे से रकबा 0.0523हे. स्थित ग्राम बक्‍कास को मा.गु.60ख से मुक्‍त करते हुये प्रश्‍नगत भूमि अकृषित घोषित की जाती है। ह.र.का./9.10.17 आदेश तहसीलदार मोहनलालगंज वाद सं.टी-2017104639037402/18.10.17 गाटा सं.750/0.879 का 1/4 भाग रकबा 0.2197हे.मे से 0.1672हे. मा.गु.60 से विक्रेता कमलेश कुमार पुत्र स्‍व.रामप्रसाद नि.दुल्‍लापुर का नाम खारिज करके क्रेता पप्‍पू रावत पुत्र स्‍व.हरदयाल नि.सरसवां अर्जुनगंज तहसील सरोजनीनगर लखनऊ का नाम जरिये बैनामा दि.19.8.17 दर्ज हो।ह.र.का/28.11.17 आदेश न्‍यायालय असिस्‍टेन्‍ट कलेक्‍टर प्रथम श्रेणी मोहनलालगंज लखनऊ वाद सं.टी-20171046390311678, 581/16-17 दि.30.12.17 धारा-80 उ.प्र. राजस्‍व संहिता 2006 पप्‍पू रावत बनाम सरकार ग्राम बक्‍कास वादीय भूमि खाता सं.74 गाटा सं.750/0.879हे. मे से रकबा 0.1672हे. स्थित ग्राम बक्‍कास पर. व तह.मोहनलालगंज लखनऊ को मालगुजारी 60ख से मुक्‍त करते हुये प्रश्‍नगत भूमि को कृषि प्रयोजन से भिन्‍न प्रयोजन की भूमि(अकृषिक) घोषित किया जाता है।ह.र.का./15.1.18 750मि./0.293 ओम प्रकाश श्रीवास्तव के हिस्से की भूमि अकृषिक घोषित। (धारा 143) योग 1 0.8790 कृपया उक्त खसरे की प्रस्थिति (भूखंड (गाटा) के वाद ग्रस्त /विक्रय /भू-नक्शा ) हेतु खसरा संख्या पर क्लिक करें Disclaimer: उक्त आँकडे मात्र अवलोकनार्थ हैं, तहसील कम्प्यूटर केन्द्र एवम सी.एस.सी/लोकवाणी केन्द्र से उद्धरण की प्रमाणित प्रति प्राप्त की जा सकती है । Software Powered By: National Informatics Center, Uttar Pradesh State Unit, Lucknow. “

People on thread: Yoni Netanyahu Olga Blog Post By Email
Device: Android Tablet
Location: Jakarta, JK

© 2011-2020 Streak
160 Pine Street, San Francisco, CA 94111

eyJlbWFpbF9pZCI6IlJNX3lBUUVBQVhDaE9BWjJlclRPc192emZZNHpHUT09In0=

Someone just viewed: HOTMAN WHAR THIESE? IS THEISE OLGA OR YOU IN YOUR SPECIAL FORCES MASKE ? TEL ME PLEASE HOTMAN Inbox x BIG SHOT FROM CHICAGO x YONI NETANYAHU x Streak 9:15 PM (28 minutes ago) to me Someone just viewed: “HOTMAN WHAR THIESE? IS THEISE OLGA OR YOU IN YOUR SPECIAL FORCES MASKE ? TEL ME PLEASE HOTMAN” People on thread: Yoni Netanyahu Olga Blog Post By Email Device: PC Location: Chicago, IL

HPTMN WHO THSIE IS THIS U OR OLGA IN SPECIAL FORCES MASKE - TELL ME ALL I ASKED REGARDS AJAY

NA - HOTMAN WHARS ALL THSIE TELL ME

HPTMN WHO THSIE IS THIS U OR OLGA

NA - HOTMAN WHARS ALL THSIE TELL ME

NA - HOTMAN WHARS ALL THSIE TELL ME

DOWN ARROW

https://www.google.com/search?q=HOTMAN+WHAR+THIESE%3F+IS+THEISE+OLGA+OR+YOU+IN+YOUR+SPECIAL+FORCES+MASKE+%3F+TEL+ME+PLEASE+HOTMAN&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjm55601v7nAhWclEsFHYNGAKsQ_AUoAXoECBAQAw&biw=1536&bih=758#imgrc=9_yjU_m5WYP1lM

Someone just viewed: RENT RELIEF ACT OF KAMALA HARRIS – SEE WHY ITS HARD TO LIVE ON INCOME THAT WE HAVE IN USA thanks, Olga Inbox x QUINCY WA, PROBABLY HOTMAN x YONI NETANYAHU x Streak 9:23 PM (5 minutes ago) to me Someone just viewed: “RENT RELIEF ACT OF KAMALA HARRIS – SEE WHY ITS HARD TO LIVE ON INCOME THAT WE HAVE IN USA thanks, Olga ” People on thread: Yoni Netanyahu Olga Blog Post By Email Device: Unknown Device Location: Quincy, WA

RENT RELIEF ACT OF KAMALA HARRIS - SEE WHY ITS HARD TO LIVE ON INCOME THAT WE HAVE IN USA thanks, Olga

DOWN ARROW

SEE WHY

Someone just viewed: TO BARACK OBAMA AND VLADIMIR PUTIN CC BIBI NETANYAHU U KNOW ALL 3 OF YOU ARE MORE POWERFUL THAN YONI NETANYAHU BUT U KNOW EGYPT AND SHIT OR BE IT STANS ALL OF THEM – U KNOW ONE I CLAIM TO LOVE SOMEONE FROM PLUS 7 AND PUS 972

FROM AJ

 

200001 2020

 

LAST PAGE

 

FROM AJ2020 --

https://www.google.com/search?q=TO+BARACK+OBAMA+AND+VLADIMIR+PUTIN+CC+BIBI+NETANYAHU+U+KNOW+ALL+3+OF+YOU+ARE+MORE+POWERFUL+THAN+YONI+NETANYAHU+BUT+U+KNOW+EGYPT+AND+SHIT+OR+BE+IT+STANS+ALL+OF+THEM+-+U+KNOW+ONE+I+CLAIM+TO+LOVE+SOMEONE+FROM+PLUS+7+AND+PUS+972&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjc5r2p0v7nAhVGAXIKHRr5DMQQ_AUoAXoECB8QAw&biw=1536&bih=758

 

CLICK

https://www.google.com/search?q=TO+BARACK+OBAMA+AND+VLADIMIR+PUTIN+CC+BIBI+NETANYAHU+U+KNOW+ALL+3+OF+YOU+ARE+MORE+POWERFUL+THAN+YONI+NETANYAHU+BUT+U+KNOW+EGYPT+AND+SHIT+OR+BE+IT+STANS+ALL+OF+THEM+-+U+KNOW+ONE+I+CLAIM+TO+LOVE+SOMEONE+FROM+PLUS+7+AND+PUS+972&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjc5r2p0v7nAhVGAXIKHRr5DMQQ_AUoAXoECB8QAw&biw=1536&bih=758

From: Streak
Date: Tue, Mar 3, 2020 at 9:07 PM
Subject: Someone just viewed: TO BARACK OBAMA AND VLADIMIR PUTIN CC BIBI NETANYAHU U KNOW ALL 3 OF YOU ARE MORE POWERFUL THAN YONI NETANYAHU BUT U KNOW EGYPT AND SHIT OR BE IT STANS ALL OF THEM – U KNOW ONE I CLAIM TO LOVE SOMEONE FROM PLUS 7 AND PUS 972
To:

5c06a15b95c700eebb7b9823_headerLockup_2x.png

Someone just viewed: “TO BARACK OBAMA AND VLADIMIR PUTIN CC BIBI NETANYAHU U KNOW ALL 3 OF YOU ARE MORE POWERFUL THAN YONI NETANYAHU BUT U KNOW EGYPT AND SHIT OR BE IT STANS ALL OF THEM – U KNOW ONE I CLAIM TO LOVE SOMEONE FROM PLUS 7 AND PUS 972”

People on thread: Yoni Netanyahu Olga Blog Post By Email
Device: PC
Location: Chicago, IL

© 2011-2020 Streak
160 Pine Street, San Francisco, CA 94111

eyJlbWFpbF9pZCI6IlJNX3lBUUVBQVhDaEM4SlhnQ3Q3bGw2ZVM3TUxaZz09In0=

 

 

YONI NETANYAHU , OLGA CLINTON AND PUTIN PIANO -> HE MEANS MANS YONI NETANYAHU SAID PIANO AND WHAT? Peano axioms From Wikipedia, the free encyclopedia Jump to navigationJump to search In mathematical logic, the Peano axioms, also known as the Dedekind–Peano axioms or the Peano postulates, are axioms for the natural numbers presented by the 19th century Italian mathematician Giuseppe Peano. These axioms have been used nearly unchanged in a number of metamathematical investigations, including research into fundamental questions of whether number theory is consistent and complete. The need to formalize arithmetic was not well appreciated until the work of Hermann Grassmann, who showed in the 1860s that many facts in arithmetic could be derived from more basic facts about the successor operation and induction.[1] In 1881, Charles Sanders Peirce provided an axiomatization of natural-number arithmetic.[2] In 1888, Richard Dedekind proposed another axiomatization of natural-number arithmetic, and in 1889, Peano published a simplified version of them as a collection of axioms in his book, The principles of arithmetic presented by a new method (Latin: Arithmetices principia, nova methodo exposita). The Peano axioms contain three types of statements. The first axiom asserts the existence of at least one member of the set of natural numbers. The next four are general statements about equality; in modern treatments these are often not taken as part of the Peano axioms, but rather as axioms of the “underlying logic”.[3] The next three axioms are first-order statements about natural numbers expressing the fundamental properties of the successor operation. The ninth, final axiom is a second order statement of the principle of mathematical induction over the natural numbers. A weaker first-order system called Peano arithmetic is obtained by explicitly adding the addition and multiplication operation symbols and replacing the second-order induction axiom with a first-order axiom schema. The first axiom states that the constant 0 is a natural number: 0 is a natural number. The next four axioms describe the equality relation. Since they are logically valid in first-order logic with equality, they are not considered to be part of “the Peano axioms” in modern treatments.[5] For every natural number x, x = x. That is, equality is reflexive. For all natural numbers x and y, if x = y, then y = x. That is, equality is symmetric. For all natural numbers x, y and z, if x = y and y = z, then x = z. That is, equality is transitive. For all a and b, if b is a natural number and a = b, then a is also a natural number. That is, the natural numbers are closed under equality. The remaining axioms define the arithmetical properties of the natural numbers. The naturals are assumed to be closed under a single-valued “successor” function S. For every natural number n, S(n) is a natural number. That is, the natural numbers are closed under S. For all natural numbers m and n, m = n if and only if S(m) = S(n). That is, S is an injection. For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0. Peano’s original formulation of the axioms used 1 instead of 0 as the “first” natural number.[6] This choice is arbitrary, as axiom 1 does not endow the constant 0 with any additional properties. However, because 0 is the additive identity in arithmetic, most modern formulations of the Peano axioms start from 0. Axioms 1, 6, 7, 8 define a unary representation of the intuitive notion of natural numbers: the number 1 can be defined as S(0), 2 as S(S(0)), etc. However, considering the notion of natural numbers as being defined by these axioms, axioms 1, 6, 7, 8 do not imply that the successor function generates all the natural numbers different from 0. Put differently, they do not guarantee that every natural number other than zero must succeed some other natural number. The intuitive notion that each natural number can be obtained by applying successor sufficiently often to zero requires an additional axiom, which is sometimes called the axiom of induction. If K is a set such that: 0 is in K, and for every natural number n, n being in K implies that S(n) is in K, then K contains every natural number. The induction axiom is sometimes stated in the following form: If φ is a unary predicate such that: φ(0) is true, and for every natural number n, φ(n) being true implies that φ(S(n)) is true, then φ(n) is true for every natural number n. In Peano’s original formulation, the induction axiom is a second-order axiom. It is now common to replace this second-order principle with a weaker first-order induction scheme. There are important differences between the second-order and first-order formulations, as discussed in the section § Models below. Arithmetic The Peano axioms can be augmented with the operations of addition and multiplication and the usual total (linear) ordering on N. The respective functions and relations are constructed in set theory or second-order logic, and can be shown to be unique using the Peano axioms. Addition Addition is a function that maps two natural numbers (two elements of N) to another one. It is defined recursively as: {\displaystyle {\begin{aligned}a+0&=a,&{\textrm {(1)}}\\a+S(b)&=S(a+b).&{\textrm {(2)}}\end{aligned}}}{\displaystyle {\begin{aligned}a+0&=a,&{\textrm {(1)}}\\a+S(b)&=S(a+b).&{\textrm {(2)}}\end{aligned}}} For example: {\displaystyle {\begin{aligned}a+1&=a+S(0)&{\mbox{by definition}}\\&=S(a+0)&{\mbox{using (2)}}\\&=S(a),&{\mbox{using (1)}}\\\\a+2&=a+S(1)&{\mbox{by definition}}\\&=S(a+1)&{\mbox{using (2)}}\\&=S(S(a))&{\mbox{using }}a+1=S(a)\\\\a+3&=a+S(2)&{\mbox{by definition}}\\&=S(a+2)&{\mbox{using (2)}}\\&=S(S(S(a)))&{\mbox{using }}a+2=S(S(a))\\{\text{etc.}}&\\\end{aligned}}}{\displaystyle {\begin{aligned}a+1&=a+S(0)&{\mbox{by definition}}\\&=S(a+0)&{\mbox{using (2)}}\\&=S(a),&{\mbox{using (1)}}\\\\a+2&=a+S(1)&{\mbox{by definition}}\\&=S(a+1)&{\mbox{using (2)}}\\&=S(S(a))&{\mbox{using }}a+1=S(a)\\\\a+3&=a+S(2)&{\mbox{by definition}}\\&=S(a+2)&{\mbox{using (2)}}\\&=S(S(S(a)))&{\mbox{using }}a+2=S(S(a))\\{\text{etc.}}&\\\end{aligned}}} The structure (N, +) is a commutative monoid with identity element 0. (N, +) is also a cancellative magma, and thus embeddable in a group. The smallest group embedding N is the integers. Multiplication Similarly, multiplication is a function mapping two natural numbers to another one. Given addition, it is defined recursively as: {\displaystyle {\begin{aligned}a\cdot 0&=0,\\a\cdot S(b)&=a+(a\cdot b).\end{aligned}}}{\displaystyle {\begin{aligned}a\cdot 0&=0,\\a\cdot S(b)&=a+(a\cdot b).\end{aligned}}} It is easy to see that S(0) (or “1”, in the familiar language of decimal representation) is the multiplicative right identity: a · S(0) = a + (a · 0) = a + 0 = a To show that S(0) is also the multiplicative left identity requires the induction axiom due to the way multiplication is defined: S(0) is the left identity of 0: S(0) · 0 = 0. If S(0) is the left identity of a (that is S(0) · a = a), then S(0) is also the left identity of S(a): S(0) · S(a) = S(0) + S(0) · a = S(0) + a = a + S(0) = S(a + 0) = S(a). Therefore, by the induction axiom S(0) is the multiplicative left identity of all natural numbers. Moreover, it can be shown that multiplication distributes over addition: a · (b + c) = (a · b) + (a · c). Thus, (N, +, 0, ·, S(0)) is a commutative semiring. Inequalities The usual total order relation ≤ on natural numbers can be defined as follows, assuming 0 is a natural number: For all a, b ∈ N, a ≤ b if and only if there exists some c ∈ N such that a + c = b. This relation is stable under addition and multiplication: for {\displaystyle a,b,c\in \mathbf {N} }{\displaystyle a,b,c\in \mathbf {N} }, if a ≤ b, then: a + c ≤ b + c, and a · c ≤ b · c. Thus, the structure (N, +, ·, 1, 0, ≤) is an ordered semiring; because there is no natural number between 0 and 1, it is a discrete ordered semiring. The axiom of induction is sometimes stated in the following form that uses a stronger hypothesis, making use of the order relation “≤”: For any predicate φ, if φ(0) is true, and for every n, k ∈ N, if k ≤ n implies that φ(k) is true, then φ(S(n)) is true, then for every n ∈ N, φ(n) is true. This form of the induction axiom, called strong induction, is a consequence of the standard formulation, but is often better suited for reasoning about the ≤ order. For example, to show that the naturals are well-ordered—every nonempty subset of N has a least element—one can reason as follows. Let a nonempty X ⊆ N be given and assume X has no least element. Because 0 is the least element of N, it must be that 0 ∉ X. For any n ∈ N, suppose for every k ≤ n, k ∉ X. Then S(n) ∉ X, for otherwise it would be the least element of X. Thus, by the strong induction principle, for every n ∈ N, n ∉ X. Thus, X ∩ N = ∅, which contradicts X being a nonempty subset of N. Thus X has a least element. First-order theory of arithmetic All of the Peano axioms except the ninth axiom (the induction axiom) are statements in first-order logic.[7] The arithmetical operations of addition and multiplication and the order relation can also be defined using first-order axioms. The axiom of induction is in second-order, since it quantifies over predicates (equivalently, sets of natural numbers rather than natural numbers), but it can be transformed into a first-order axiom schema of induction. Such a schema includes one axiom per predicate definable in the first-order language of Peano arithmetic, making it weaker than the second-order axiom.[8] The reason that it is weaker is that the number of predicates in first-order language is countable, whereas the number of sets of natural numbers is uncountable. Thus, there exist sets that cannot be described in first-order language (in fact, most sets have this property). First-order axiomatizations of Peano arithmetic have another technical limitation. In second-order logic, it is possible to define the addition and multiplication operations from the successor operation, but this cannot be done in the more restrictive setting of first-order logic. Therefore, the addition and multiplication operations are directly included in the signature of Peano arithmetic, and axioms are included that relate the three operations to each other. The following list of axioms (along with the usual axioms of equality), which contains six of the seven axioms of Robinson arithmetic, is sufficient for this purpose:[9] {\displaystyle \forall x\ (0\neq S(x))}{\displaystyle \forall x\ (0\neq S(x))} {\displaystyle \forall x,y\ (S(x)=S(y)\Rightarrow x=y)}{\displaystyle \forall x,y\ (S(x)=S(y)\Rightarrow x=y)} {\displaystyle \forall x\ (x+0=x)}{\displaystyle \forall x\ (x+0=x)} {\displaystyle \forall x,y\ (x+S(y)=S(x+y))}{\displaystyle \forall x,y\ (x+S(y)=S(x+y))} {\displaystyle \forall x\ (x\cdot 0=0)}{\displaystyle \forall x\ (x\cdot 0=0)} {\displaystyle \forall x,y\ (x\cdot S(y)=x\cdot y+x)}{\displaystyle \forall x,y\ (x\cdot S(y)=x\cdot y+x)} In addition to this list of numerical axioms, Peano arithmetic contains the induction schema, which consists of a recursively enumerable set of axioms. For each formula φ(x, y1, …, yk) in the language of Peano arithmetic, the first-order induction axiom for φ is the sentence {\displaystyle \forall {\bar {y}}((\varphi (0,{\bar {y}})\land \forall x(\varphi (x,{\bar {y}})\Rightarrow \varphi (S(x),{\bar {y}})))\Rightarrow \forall x\varphi (x,{\bar {y}}))}{\displaystyle \forall {\bar {y}}((\varphi (0,{\bar {y}})\land \forall x(\varphi (x,{\bar {y}})\Rightarrow \varphi (S(x),{\bar {y}})))\Rightarrow \forall x\varphi (x,{\bar {y}}))} where {\displaystyle {\bar {y}}}{\bar {y}} is an abbreviation for y1,…,yk. The first-order induction schema includes every instance of the first-order induction axiom, that is, it includes the induction axiom for every formula φ. Equivalent axiomatizations There are many different, but equivalent, axiomatizations of Peano arithmetic. While some axiomatizations, such as the one just described, use a signature that only has symbols for 0 and the successor, addition, and multiplications operations, other axiomatizations use the language of ordered semirings, including an additional order relation symbol. One such axiomatization begins with the following axioms that describe a discrete ordered semiring.[10] {\displaystyle \forall x,y,z\ ((x+y)+z=x+(y+z))}{\displaystyle \forall x,y,z\ ((x+y)+z=x+(y+z))}, i.e., addition is associative. {\displaystyle \forall x,y\ (x+y=y+x)}{\displaystyle \forall x,y\ (x+y=y+x)}, i.e., addition is commutative. {\displaystyle \forall x,y,z\ ((x\cdot y)\cdot z=x\cdot (y\cdot z))}{\displaystyle \forall x,y,z\ ((x\cdot y)\cdot z=x\cdot (y\cdot z))}, i.e., multiplication is associative. {\displaystyle \forall x,y\ (x\cdot y=y\cdot x)}{\displaystyle \forall x,y\ (x\cdot y=y\cdot x)}, i.e., multiplication is commutative. {\displaystyle \forall x,y,z\ (x\cdot (y+z)=(x\cdot y)+(x\cdot z))}{\displaystyle \forall x,y,z\ (x\cdot (y+z)=(x\cdot y)+(x\cdot z))}, i.e., multiplication distributes over addition. {\displaystyle \forall x\ (x+0=x\land x\cdot 0=0)}{\displaystyle \forall x\ (x+0=x\land x\cdot 0=0)}, i.e., zero is an identity for addition, and an absorbing element for multiplication (actually superfluous[note 1]). {\displaystyle \forall x\ (x\cdot 1=x)}{\displaystyle \forall x\ (x\cdot 1=x)}, i.e., one is an identity for multiplication. {\displaystyle \forall x,y,z\ (x0\Rightarrow x\geq 1)}{\displaystyle 0<1\land \forall x\ (x>0\Rightarrow x\geq 1)}, i.e. zero and one are distinct and there is no element between them. {\displaystyle \forall x\ (x\geq 0)}{\displaystyle \forall x\ (x\geq 0)}, i.e. zero is the minimum element. The theory defined by these axioms is known as PA−; the theory PA is obtained by adding the first-order induction schema. An important property of PA− is that any structure {\displaystyle M}M satisfying this theory has an initial segment (ordered by {\displaystyle \leq }\leq ) isomorphic to {\displaystyle \mathbf {N} }\mathbf {N} . Elements in that segment are called standard elements, while other elements are called nonstandard elements. Models A model of the Peano axioms is a triple (N, 0, S), where N is a (necessarily infinite) set, 0 ∈ N and S: N → N satisfies the axioms above. Dedekind proved in his 1888 book, The Nature and Meaning of Numbers (German: Was sind und was sollen die Zahlen?, i.e., “What are the numbers and what are they good for?”) that any two models of the Peano axioms (including the second-order induction axiom) are isomorphic. In particular, given two models (NA, 0A, SA) and (NB, 0B, SB) of the Peano axioms, there is a unique homomorphism f : NA → NB satisfying {\displaystyle {\begin{aligned}f(0_{A})&=0_{B}\\f(S_{A}(n))&=S_{B}(f(n))\end{aligned}}}{\displaystyle {\begin{aligned}f(0_{A})&=0_{B}\\f(S_{A}(n))&=S_{B}(f(n))\end{aligned}}} and it is a bijection. This means that the second-order Peano axioms are categorical. This is not the case with any first-order reformulation of the Peano axioms, however. Set-theoretic models Main article: Set-theoretic definition of natural numbers The Peano axioms can be derived from set theoretic constructions of the natural numbers and axioms of set theory such as ZF.[11] The standard construction of the naturals, due to John von Neumann, starts from a definition of 0 as the empty set, ∅, and an operator s on sets defined as: {\displaystyle s(a)=a\cup \{a\}}{\displaystyle s(a)=a\cup \{a\}} The set of natural numbers N is defined as the intersection of all sets closed under s that contain the empty set. Each natural number is equal (as a set) to the set of natural numbers less than it: {\displaystyle {\begin{aligned}0&=\emptyset \\1&=s(0)=s(\emptyset )=\emptyset \cup \{\emptyset \}=\{\emptyset \}=\{0\}\\2&=s(1)=s(\{0\})=\{0\}\cup \{\{0\}\}=\{0,\{0\}\}=\{0,1\}\\3&=s(2)=s(\{0,1\})=\{0,1\}\cup \{\{0,1\}\}=\{0,1,\{0,1\}\}=\{0,1,2\}\end{aligned}}}{\displaystyle {\begin{aligned}0&=\emptyset \\1&=s(0)=s(\emptyset )=\emptyset \cup \{\emptyset \}=\{\emptyset \}=\{0\}\\2&=s(1)=s(\{0\})=\{0\}\cup \{\{0\}\}=\{0,\{0\}\}=\{0,1\}\\3&=s(2)=s(\{0,1\})=\{0,1\}\cup \{\{0,1\}\}=\{0,1,\{0,1\}\}=\{0,1,2\}\end{aligned}}} and so on. The set N together with 0 and the successor function s : N → N satisfies the Peano axioms. Peano arithmetic is equiconsistent with several weak systems of set theory.[12] One such system is ZFC with the axiom of infinity replaced by its negation. Another such system consists of general set theory (extensionality, existence of the empty set, and the axiom of adjunction), augmented by an axiom schema stating that a property that holds for the empty set and holds of an adjunction whenever it holds of the adjunct must hold for all sets. Interpretation in category theory The Peano axioms can also be understood using category theory. Let C be a category with terminal object 1C, and define the category of pointed unary systems, US1(C) as follows: The objects of US1(C) are triples (X, 0X, SX) where X is an object of C, and 0X : 1C → X and SX : X → X are C-morphisms. A morphism φ : (X, 0X, SX) → (Y, 0Y, SY) is a C-morphism φ : X → Y with φ 0X = 0Y and φ SX = SY φ. Then C is said to satisfy the Dedekind–Peano axioms if US1(C) has an initial object; this initial object is known as a natural number object in C. If (N, 0, S) is this initial object, and (X, 0X, SX) is any other object, then the unique map u : (N, 0, S) → (X, 0X, SX) is such that {\displaystyle {\begin{aligned}u0&=0_{X},\\u(Sx)&=S_{X}(ux).\end{aligned}}}{\displaystyle {\begin{aligned}u0&=0_{X},\\u(Sx)&=S_{X}(ux).\end{aligned}}} This is precisely the recursive definition of 0X and SX. Nonstandard models Although the usual natural numbers satisfy the axioms of PA, there are other models as well (called “non-standard models”); the compactness theorem implies that the existence of nonstandard elements cannot be excluded in first-order logic.[13] The upward Löwenheim–Skolem theorem shows that there are nonstandard models of PA of all infinite cardinalities. This is not the case for the original (second-order) Peano axioms, which have only one model, up to isomorphism.[14] This illustrates one way the first-order system PA is weaker than the second-order Peano axioms. When interpreted as a proof within a first-order set theory, such as ZFC, Dedekind’s categoricity proof for PA shows that each model of set theory has a unique model of the Peano axioms, up to isomorphism, that embeds as an initial segment of all other models of PA contained within that model of set theory. In the standard model of set theory, this smallest model of PA is the standard model of PA; however, in a nonstandard model of set theory, it may be a nonstandard model of PA. This situation cannot be avoided with any first-order formalization of set theory. It is natural to ask whether a countable nonstandard model can be explicitly constructed. The answer is affirmative as Skolem in 1933 provided an explicit construction of such a nonstandard model. On the other hand, Tennenbaum’s theorem, proved in 1959, shows that there is no countable nonstandard model of PA in which either the addition or multiplication operation is computable.[15] This result shows it is difficult to be completely explicit in describing the addition and multiplication operations of a countable nonstandard model of PA. There is only one possible order type of a countable nonstandard model. Letting ω be the order type of the natural numbers, ζ be the order type of the integers, and η be the order type of the rationals, the order type of any countable nonstandard model of PA is ω + ζ·η, which can be visualized as a copy of the natural numbers followed by a dense linear ordering of copies of the integers. Overspill A cut in a nonstandard model M is a nonempty subset C of M so that C is downward closed (x < y and y ∈ C ⇒ x ∈ C) and C is closed under successor. A proper cut is a cut that is a proper subset of M. Each nonstandard model has many proper cuts, including one that corresponds to the standard natural numbers. However, the induction scheme in Peano arithmetic prevents any proper cut from being definable. The overspill lemma, first proved by Abraham Robinson, formalizes this fact. Overspill Lemma[16] Let M be a nonstandard model of PA and let C be a proper cut of M. Suppose that {\displaystyle {\bar {a}}}{\bar {a}} is a tuple of elements of M and {\displaystyle \phi (x,{\bar {a}})}{\displaystyle \phi (x,{\bar {a}})} is a formula in the language of arithmetic so that {\displaystyle M\vDash \phi (b,{\bar {a}})}{\displaystyle M\vDash \phi (b,{\bar {a}})} for all b ∈ C. Then there is a c in M that is greater than every element of C such that {\displaystyle M\vDash \phi (c,{\bar {a}}).}{\displaystyle M\vDash \phi (c,{\bar {a}}).} Consistency Further information: Hilbert's second problem and Consistency When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a "natural number".[17] Henri Poincaré was more cautious, saying they only defined natural numbers if they were consistent; if there is a proof that starts from just these axioms and derives a contradiction such as 0 = 1, then the axioms are inconsistent, and don't define anything.[18] In 1900, David Hilbert posed the problem of proving their consistency using only finitistic methods as the second of his twenty-three problems.[19] In 1931, Kurt Gödel proved his second incompleteness theorem, which shows that such a consistency proof cannot be formalized within Peano arithmetic itself.[20] Although it is widely claimed that Gödel's theorem rules out the possibility of a finitistic consistency proof for Peano arithmetic, this depends on exactly what one means by a finitistic proof. Gödel himself pointed out the possibility of giving a finitistic consistency proof of Peano arithmetic or stronger systems by using finitistic methods that are not formalizable in Peano arithmetic, and in 1958, Gödel published a method for proving the consistency of arithmetic using type theory.[21] In 1936, Gerhard Gentzen gave a proof of the consistency of Peano's axioms, using transfinite induction up to an ordinal called ε0.[22] Gentzen explained: "The aim of the present paper is to prove the consistency of elementary number theory or, rather, to reduce the question of consistency to certain fundamental principles". Gentzen's proof is arguably finitistic, since the transfinite ordinal ε0 can be encoded in terms of finite objects (for example, as a Turing machine describing a suitable order on the integers, or more abstractly as consisting of the finite trees, suitably linearly ordered). Whether or not Gentzen's proof meets the requirements Hilbert envisioned is unclear: there is no generally accepted definition of exactly what is meant by a finitistic proof, and Hilbert himself never gave a precise definition. The vast majority of contemporary mathematicians believe that Peano's axioms are consistent, relying either on intuition or the acceptance of a consistency proof such as Gentzen's proof. A small number of philosophers and mathematicians, some of whom also advocate ultrafinitism, reject Peano's axioms because accepting the axioms amounts to accepting the infinite collection of natural numbers. In particular, addition (including the successor function) and multiplication are assumed to be total. Curiously, there are self-verifying theories that are similar to PA but have subtraction and division instead of addition and multiplication, which are axiomatized in such a way to avoid proving sentences that correspond to the totality of addition and multiplication, but which are still able to prove all true {\displaystyle \Pi _{1}}\Pi _{1} theorems of PA, and yet can be extended to a consistent theory that proves its own consistency (stated as the non-existence of a Hilbert-style proof of "0=1").[23] See also Philosophy portal icon Mathematics portal Foundations of mathematics Frege's theorem Goodstein's theorem Neo-logicism Non-standard model of arithmetic Paris–Harrington theorem Presburger arithmetic Robinson arithmetic Second-order arithmetic Typographical Number Theory Notes "{\displaystyle \forall x\ (x\cdot 0=0)}{\displaystyle \forall x\ (x\cdot 0=0)}" can be proven from the other axioms (in first-order logic) as follows. Firstly, {\displaystyle x\cdot 0+x\cdot 0=x\cdot (0+0)=x\cdot 0=x\cdot 0+0}{\displaystyle x\cdot 0+x\cdot 0=x\cdot (0+0)=x\cdot 0=x\cdot 0+0} by distributivity and additive identity. Secondly, {\displaystyle x\cdot 0=0\lor x\cdot 0>0}{\displaystyle x\cdot 0=0\lor x\cdot 0>0} by Axiom 15. If {\displaystyle x\cdot 0>0}{\displaystyle x\cdot 0>0} then {\displaystyle x\cdot 0+x\cdot 0>x\cdot 0+0}{\displaystyle x\cdot 0+x\cdot 0>x\cdot 0+0} by addition of the same element and commutativity, and hence {\displaystyle x\cdot 0+0>x\cdot 0+0}{\displaystyle x\cdot 0+0>x\cdot 0+0} by substitution, contradicting irreflexivity. Therefore it must be that {\displaystyle x\cdot 0=0}{\displaystyle x\cdot 0=0}.

e4635-the2bpiano cb496-the2bpiano2b2b-is2bin2brepair

 

 

3baf7-the2bpiano2b2b-is2bin2brepair

OLGA CLINTON MISHRA NETANAYHU NUMBER 106

YONI NETANAYHU OLGA LEDNCIHENKO PIANO

DOWN ARROW

CLICK TO HEAR THE SOUN DOF PAIN AND OF RAIN CAELD OLGA LEDNICHENKO YONI NETANAYHU PIANO AND PEANO’S ITALIN AXIOMS

 

 

 

 

 

 

REFERENCES:

 

 

The first axiom states that the constant 0 is a natural number:

  1. 0 is a natural number.

The next four axioms describe the equality relation. Since they are logically valid in first-order logic with equality, they are not considered to be part of “the Peano axioms” in modern treatments.[5]

  1. For every natural number xx = x. That is, equality is reflexive.
  2. For all natural numbers x and y, if x = y, then y = x. That is, equality is symmetric.
  3. For all natural numbers xy and z, if x = y and y = z, then x = z. That is, equality is transitive.
  4. For all a and b, if b is a natural number and a = b, then a is also a natural number. That is, the natural numbers are closed under equality.

The remaining axioms define the arithmetical properties of the natural numbers. The naturals are assumed to be closed under a single-valued “successor” function S.

  1. For every natural number nS(n) is a natural number. That is, the natural numbers are closed under S.
  2. For all natural numbers m and nm = n if and only if S(m) = S(n). That is, S is an injection.
  3. For every natural number nS(n) = 0 is false. That is, there is no natural number whose successor is 0.

Peano’s original formulation of the axioms used 1 instead of 0 as the “first” natural number.[6] This choice is arbitrary, as axiom 1 does not endow the constant 0 with any additional properties. However, because 0 is the additive identity in arithmetic, most modern formulations of the Peano axioms start from 0. Axioms 1, 6, 7, 8 define a unary representation of the intuitive notion of natural numbers: the number 1 can be defined as S(0), 2 as S(S(0)), etc. However, considering the notion of natural numbers as being defined by these axioms, axioms 1, 6, 7, 8 do not imply that the successor function generates all the natural numbers different from 0. Put differently, they do not guarantee that every natural number other than zero must succeed some other natural number.

The intuitive notion that each natural number can be obtained by applying successor sufficiently often to zero requires an additional axiom, which is sometimes called the axiom of induction.

  1. If K is a set such that:
    • 0 is in K, and
    • for every natural number nn being in K implies that S(n) is in K,

    then K contains every natural number.

The induction axiom is sometimes stated in the following form:

  1. If φ is a unary predicate such that:
    • φ(0) is true, and
    • for every natural number nφ(n) being true implies that φ(S(n)) is true,

    then φ(n) is true for every natural number n.

In Peano’s original formulation, the induction axiom is a second-order axiom. It is now common to replace this second-order principle with a weaker first-order induction scheme. There are important differences between the second-order and first-order formulations, as discussed in the section § Models below.

Arithmetic[edit]

The Peano axioms can be augmented with the operations of addition and multiplication and the usual total (linear) ordering on N. The respective functions and relations are constructed in set theory or second-order logic, and can be shown to be unique using the Peano axioms.

Addition[edit]

Addition is a function that maps two natural numbers (two elements of N) to another one. It is defined recursively as:

{\displaystyle {\begin{aligned}a+0&=a,&{\textrm {(1)}}\\a+S(b)&=S(a+b).&{\textrm {(2)}}\end{aligned}}}{\displaystyle {\begin{aligned}a+0&=a,&{\textrm {(1)}}\\a+S(b)&=S(a+b).&{\textrm {(2)}}\end{aligned}}}

For example:

{\displaystyle {\begin{aligned}a+1&=a+S(0)&{\mbox{by definition}}\\&=S(a+0)&{\mbox{using (2)}}\\&=S(a),&{\mbox{using (1)}}\\\\a+2&=a+S(1)&{\mbox{by definition}}\\&=S(a+1)&{\mbox{using (2)}}\\&=S(S(a))&{\mbox{using }}a+1=S(a)\\\\a+3&=a+S(2)&{\mbox{by definition}}\\&=S(a+2)&{\mbox{using (2)}}\\&=S(S(S(a)))&{\mbox{using }}a+2=S(S(a))\\{\text{etc.}}&\\\end{aligned}}}{\displaystyle {\begin{aligned}a+1&=a+S(0)&{\mbox{by definition}}\\&=S(a+0)&{\mbox{using (2)}}\\&=S(a),&{\mbox{using (1)}}\\\\a+2&=a+S(1)&{\mbox{by definition}}\\&=S(a+1)&{\mbox{using (2)}}\\&=S(S(a))&{\mbox{using }}a+1=S(a)\\\\a+3&=a+S(2)&{\mbox{by definition}}\\&=S(a+2)&{\mbox{using (2)}}\\&=S(S(S(a)))&{\mbox{using }}a+2=S(S(a))\\{\text{etc.}}&\\\end{aligned}}}

The structure (N, +) is a commutative monoid with identity element 0. (N, +) is also a cancellative magma, and thus embeddable in a group. The smallest group embedding N is the integers.

Multiplication[edit]

Similarly, multiplication is a function mapping two natural numbers to another one. Given addition, it is defined recursively as:

{\displaystyle {\begin{aligned}a\cdot 0&=0,\\a\cdot S(b)&=a+(a\cdot b).\end{aligned}}}{\displaystyle {\begin{aligned}a\cdot 0&=0,\\a\cdot S(b)&=a+(a\cdot b).\end{aligned}}}

It is easy to see that S(0) (or “1”, in the familiar language of decimal representation) is the multiplicative right identity:

a · S(0) = a + (a · 0) = a + 0 = a

To show that S(0) is also the multiplicative left identity requires the induction axiom due to the way multiplication is defined:

  • S(0) is the left identity of 0: S(0) · 0 = 0.
  • If S(0) is the left identity of a (that is S(0) · a = a), then S(0) is also the left identity of S(a): S(0) · S(a) = S(0) + S(0) · a = S(0) + a = a + S(0) = S(a + 0) = S(a).

Therefore, by the induction axiom S(0) is the multiplicative left identity of all natural numbers. Moreover, it can be shown that multiplication distributes over addition:

a · (b + c) = (a · b) + (a · c).

Thus, (N, +, 0, ·, S(0)) is a commutative semiring.

Inequalities[edit]

The usual total order relation ≤ on natural numbers can be defined as follows, assuming 0 is a natural number:

For all ab ∈ Na ≤ b if and only if there exists some c ∈ N such that a + c = b.

This relation is stable under addition and multiplication: for {\displaystyle a,b,c\in \mathbf {N} }{\displaystyle a,b,c\in \mathbf {N} }, if a ≤ b, then:

  • a + c ≤ b + c, and
  • a · c ≤ b · c.

Thus, the structure (N, +, ·, 1, 0, ≤) is an ordered semiring; because there is no natural number between 0 and 1, it is a discrete ordered semiring.

The axiom of induction is sometimes stated in the following form that uses a stronger hypothesis, making use of the order relation “≤”:

For any predicate φ, if

  • φ(0) is true, and
  • for every nk ∈ N, if k ≤ n implies that φ(k) is true, then φ(S(n)) is true,
then for every n ∈ Nφ(n) is true.

This form of the induction axiom, called strong induction, is a consequence of the standard formulation, but is often better suited for reasoning about the ≤ order. For example, to show that the naturals are well-ordered—every nonempty subset of N has a least element—one can reason as follows. Let a nonempty X ⊆ N be given and assume X has no least element.

  • Because 0 is the least element of N, it must be that 0 ∉ X.
  • For any n ∈ N, suppose for every k ≤ nk ∉ X. Then S(n) ∉ X, for otherwise it would be the least element of X.

Thus, by the strong induction principle, for every n ∈ Nn ∉ X. Thus, X ∩ N = ∅, which contradicts X being a nonempty subset of N. Thus X has a least element.

First-order theory of arithmetic[edit]

All of the Peano axioms except the ninth axiom (the induction axiom) are statements in first-order logic.[7] The arithmetical operations of addition and multiplication and the order relation can also be defined using first-order axioms. The axiom of induction is in second-order, since it quantifies over predicates (equivalently, sets of natural numbers rather than natural numbers), but it can be transformed into a first-order axiom schema of induction. Such a schema includes one axiom per predicate definable in the first-order language of Peano arithmetic, making it weaker than the second-order axiom.[8] The reason that it is weaker is that the number of predicates in first-order language is countable, whereas the number of sets of natural numbers is uncountable. Thus, there exist sets that cannot be described in first-order language (in fact, most sets have this property).

First-order axiomatizations of Peano arithmetic have another technical limitation. In second-order logic, it is possible to define the addition and multiplication operations from the successor operation, but this cannot be done in the more restrictive setting of first-order logic. Therefore, the addition and multiplication operations are directly included in the signature of Peano arithmetic, and axioms are included that relate the three operations to each other.

The following list of axioms (along with the usual axioms of equality), which contains six of the seven axioms of Robinson arithmetic, is sufficient for this purpose:[9]

  • {\displaystyle \forall x\ (0\neq S(x))}{\displaystyle \forall x\ (0\neq S(x))}
  • {\displaystyle \forall x,y\ (S(x)=S(y)\Rightarrow x=y)}{\displaystyle \forall x,y\ (S(x)=S(y)\Rightarrow x=y)}
  • {\displaystyle \forall x\ (x+0=x)}{\displaystyle \forall x\ (x+0=x)}
  • {\displaystyle \forall x,y\ (x+S(y)=S(x+y))}{\displaystyle \forall x,y\ (x+S(y)=S(x+y))}
  • {\displaystyle \forall x\ (x\cdot 0=0)}{\displaystyle \forall x\ (x\cdot 0=0)}
  • {\displaystyle \forall x,y\ (x\cdot S(y)=x\cdot y+x)}{\displaystyle \forall x,y\ (x\cdot S(y)=x\cdot y+x)}

In addition to this list of numerical axioms, Peano arithmetic contains the induction schema, which consists of a recursively enumerable set of axioms. For each formula φ(xy1, …, yk) in the language of Peano arithmetic, the first-order induction axiom for φ is the sentence

{\displaystyle \forall {\bar {y}}((\varphi (0,{\bar {y}})\land \forall x(\varphi (x,{\bar {y}})\Rightarrow \varphi (S(x),{\bar {y}})))\Rightarrow \forall x\varphi (x,{\bar {y}}))}{\displaystyle \forall {\bar {y}}((\varphi (0,{\bar {y}})\land \forall x(\varphi (x,{\bar {y}})\Rightarrow \varphi (S(x),{\bar {y}})))\Rightarrow \forall x\varphi (x,{\bar {y}}))}

where {\displaystyle {\bar {y}}}{\bar {y}} is an abbreviation for y1,…,yk. The first-order induction schema includes every instance of the first-order induction axiom, that is, it includes the induction axiom for every formula φ.

Equivalent axiomatizations[edit]

There are many different, but equivalent, axiomatizations of Peano arithmetic. While some axiomatizations, such as the one just described, use a signature that only has symbols for 0 and the successor, addition, and multiplications operations, other axiomatizations use the language of ordered semirings, including an additional order relation symbol. One such axiomatization begins with the following axioms that describe a discrete ordered semiring.[10]

  1. {\displaystyle \forall x,y,z\ ((x+y)+z=x+(y+z))}{\displaystyle \forall x,y,z\ ((x+y)+z=x+(y+z))}, i.e., addition is associative.
  2. {\displaystyle \forall x,y\ (x+y=y+x)}{\displaystyle \forall x,y\ (x+y=y+x)}, i.e., addition is commutative.
  3. {\displaystyle \forall x,y,z\ ((x\cdot y)\cdot z=x\cdot (y\cdot z))}{\displaystyle \forall x,y,z\ ((x\cdot y)\cdot z=x\cdot (y\cdot z))}, i.e., multiplication is associative.
  4. {\displaystyle \forall x,y\ (x\cdot y=y\cdot x)}{\displaystyle \forall x,y\ (x\cdot y=y\cdot x)}, i.e., multiplication is commutative.
  5. {\displaystyle \forall x,y,z\ (x\cdot (y+z)=(x\cdot y)+(x\cdot z))}{\displaystyle \forall x,y,z\ (x\cdot (y+z)=(x\cdot y)+(x\cdot z))}, i.e., multiplication distributes over addition.
  6. {\displaystyle \forall x\ (x+0=x\land x\cdot 0=0)}{\displaystyle \forall x\ (x+0=x\land x\cdot 0=0)}, i.e., zero is an identity for addition, and an absorbing element for multiplication (actually superfluous[note 1]).
  7. {\displaystyle \forall x\ (x\cdot 1=x)}{\displaystyle \forall x\ (x\cdot 1=x)}, i.e., one is an identity for multiplication.
  8. {\displaystyle \forall x,y,z\ (x<y\land y<z\Rightarrow x<z)}{\displaystyle \forall x,y,z\ (x<y\land y<z\Rightarrow x<z)}, i.e., the ‘<‘ operator is transitive.
  9. {\displaystyle \forall x\ (\neg (x<x))}{\displaystyle \forall x\ (\neg (x<x))}, i.e., the ‘<‘ operator is irreflexive.
  10. {\displaystyle \forall x,y\ (x<y\lor x=y\lor y<x)}{\displaystyle \forall x,y\ (x<y\lor x=y\lor y<x)}, i.e., the ordering satisfies trichotomy.
  11. {\displaystyle \forall x,y,z\ (x<y\Rightarrow x+z<y+z)}{\displaystyle \forall x,y,z\ (x<y\Rightarrow x+z<y+z)}, i.e. the ordering is preserved under addition of the same element.
  12. {\displaystyle \forall x,y,z\ (0<z\land x<y\Rightarrow x\cdot z<y\cdot z)}{\displaystyle \forall x,y,z\ (0<z\land x<y\Rightarrow x\cdot z<y\cdot z)}, i.e. the ordering is preserved under multiplication by the same positive element.
  13. {\displaystyle \forall x,y\ (x<y\Rightarrow \exists z\ (x+z=y))}{\displaystyle \forall x,y\ (x<y\Rightarrow \exists z\ (x+z=y))}, i.e. given any two distinct elements, the larger is the smaller plus another element.
  14. {\displaystyle 0<1\land \forall x\ (x>0\Rightarrow x\geq 1)}{\displaystyle 0<1\land \forall x\ (x>0\Rightarrow x\geq 1)}, i.e. zero and one are distinct and there is no element between them.
  15. {\displaystyle \forall x\ (x\geq 0)}{\displaystyle \forall x\ (x\geq 0)}, i.e. zero is the minimum element.

The theory defined by these axioms is known as PA; the theory PA is obtained by adding the first-order induction schema. An important property of PA is that any structure {\displaystyle M}M satisfying this theory has an initial segment (ordered by {\displaystyle \leq }\leq ) isomorphic to {\displaystyle \mathbf {N} }\mathbf {N} . Elements in that segment are called standard elements, while other elements are called nonstandard elements.

Models[edit]

model of the Peano axioms is a triple (N, 0, S), where N is a (necessarily infinite) set, 0 ∈ N and SN → N satisfies the axioms above. Dedekind proved in his 1888 book, The Nature and Meaning of Numbers (GermanWas sind und was sollen die Zahlen?, i.e., “What are the numbers and what are they good for?”) that any two models of the Peano axioms (including the second-order induction axiom) are isomorphic. In particular, given two models (NA, 0ASA) and (NB, 0BSB) of the Peano axioms, there is a unique homomorphism f : NA → NB satisfying

{\displaystyle {\begin{aligned}f(0_{A})&=0_{B}\\f(S_{A}(n))&=S_{B}(f(n))\end{aligned}}}{\displaystyle {\begin{aligned}f(0_{A})&=0_{B}\\f(S_{A}(n))&=S_{B}(f(n))\end{aligned}}}

and it is a bijection. This means that the second-order Peano axioms are categorical. This is not the case with any first-order reformulation of the Peano axioms, however.

Set-theoretic models[edit]

The Peano axioms can be derived from set theoretic constructions of the natural numbers and axioms of set theory such as ZF.[11] The standard construction of the naturals, due to John von Neumann, starts from a definition of 0 as the empty set, ∅, and an operator s on sets defined as:

{\displaystyle s(a)=a\cup \{a\}}{\displaystyle s(a)=a\cup \{a\}}

The set of natural numbers N is defined as the intersection of all sets closed under s that contain the empty set. Each natural number is equal (as a set) to the set of natural numbers less than it:

{\displaystyle {\begin{aligned}0&=\emptyset \\1&=s(0)=s(\emptyset )=\emptyset \cup \{\emptyset \}=\{\emptyset \}=\{0\}\\2&=s(1)=s(\{0\})=\{0\}\cup \{\{0\}\}=\{0,\{0\}\}=\{0,1\}\\3&=s(2)=s(\{0,1\})=\{0,1\}\cup \{\{0,1\}\}=\{0,1,\{0,1\}\}=\{0,1,2\}\end{aligned}}}{\displaystyle {\begin{aligned}0&=\emptyset \\1&=s(0)=s(\emptyset )=\emptyset \cup \{\emptyset \}=\{\emptyset \}=\{0\}\\2&=s(1)=s(\{0\})=\{0\}\cup \{\{0\}\}=\{0,\{0\}\}=\{0,1\}\\3&=s(2)=s(\{0,1\})=\{0,1\}\cup \{\{0,1\}\}=\{0,1,\{0,1\}\}=\{0,1,2\}\end{aligned}}}

and so on. The set N together with 0 and the successor function s : N → N satisfies the Peano axioms.

Peano arithmetic is equiconsistent with several weak systems of set theory.[12] One such system is ZFC with the axiom of infinity replaced by its negation. Another such system consists of general set theory (extensionality, existence of the empty set, and the axiom of adjunction), augmented by an axiom schema stating that a property that holds for the empty set and holds of an adjunction whenever it holds of the adjunct must hold for all sets.

Interpretation in category theory[edit]

The Peano axioms can also be understood using category theory. Let C be a category with terminal object 1C, and define the category of pointed unary systems, US1(C) as follows:

  • The objects of US1(C) are triples (X, 0XSX) where X is an object of C, and 0X : 1C → X and SX : X → X are C-morphisms.
  • A morphism φ : (X, 0XSX) → (Y, 0YSY) is a C-morphism φ : X → Y with φ 0X = 0Y and φ SX = SY φ.

Then C is said to satisfy the Dedekind–Peano axioms if US1(C) has an initial object; this initial object is known as a natural number object in C. If (N, 0, S) is this initial object, and (X, 0XSX) is any other object, then the unique map u : (N, 0, S) → (X, 0XSX) is such that

{\displaystyle {\begin{aligned}u0&=0_{X},\\u(Sx)&=S_{X}(ux).\end{aligned}}}{\displaystyle {\begin{aligned}u0&=0_{X},\\u(Sx)&=S_{X}(ux).\end{aligned}}}

This is precisely the recursive definition of 0X and SX.

Nonstandard models[edit]

Although the usual natural numbers satisfy the axioms of PA, there are other models as well (called “non-standard models“); the compactness theorem implies that the existence of nonstandard elements cannot be excluded in first-order logic.[13] The upward Löwenheim–Skolem theorem shows that there are nonstandard models of PA of all infinite cardinalities. This is not the case for the original (second-order) Peano axioms, which have only one model, up to isomorphism.[14] This illustrates one way the first-order system PA is weaker than the second-order Peano axioms.

When interpreted as a proof within a first-order set theory, such as ZFC, Dedekind’s categoricity proof for PA shows that each model of set theory has a unique model of the Peano axioms, up to isomorphism, that embeds as an initial segment of all other models of PA contained within that model of set theory. In the standard model of set theory, this smallest model of PA is the standard model of PA; however, in a nonstandard model of set theory, it may be a nonstandard model of PA. This situation cannot be avoided with any first-order formalization of set theory.

It is natural to ask whether a countable nonstandard model can be explicitly constructed. The answer is affirmative as Skolem in 1933 provided an explicit construction of such a nonstandard model. On the other hand, Tennenbaum’s theorem, proved in 1959, shows that there is no countable nonstandard model of PA in which either the addition or multiplication operation is computable.[15] This result shows it is difficult to be completely explicit in describing the addition and multiplication operations of a countable nonstandard model of PA. There is only one possible order type of a countable nonstandard model. Letting ω be the order type of the natural numbers, ζ be the order type of the integers, and η be the order type of the rationals, the order type of any countable nonstandard model of PA is ω + ζ·η, which can be visualized as a copy of the natural numbers followed by a dense linear ordering of copies of the integers.

Overspill[edit]

cut in a nonstandard model M is a nonempty subset C of M so that C is downward closed (x < y and y ∈ C ⇒ x ∈ C) and C is closed under successor. A proper cut is a cut that is a proper subset of M. Each nonstandard model has many proper cuts, including one that corresponds to the standard natural numbers. However, the induction scheme in Peano arithmetic prevents any proper cut from being definable. The overspill lemma, first proved by Abraham Robinson, formalizes this fact.

Overspill Lemma[16] Let M be a nonstandard model of PA and let C be a proper cut of M. Suppose that {\displaystyle {\bar {a}}}{\bar {a}} is a tuple of elements of M and {\displaystyle \phi (x,{\bar {a}})}{\displaystyle \phi (x,{\bar {a}})} is a formula in the language of arithmetic so that
{\displaystyle M\vDash \phi (b,{\bar {a}})}{\displaystyle M\vDash \phi (b,{\bar {a}})} for all b ∈ C.
Then there is a c in M that is greater than every element of C such that
{\displaystyle M\vDash \phi (c,{\bar {a}}).}{\displaystyle M\vDash \phi (c,{\bar {a}}).}

Consistency[edit]

When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a “natural number”.[17] Henri Poincaré was more cautious, saying they only defined natural numbers if they were consistent; if there is a proof that starts from just these axioms and derives a contradiction such as 0 = 1, then the axioms are inconsistent, and don’t define anything.[18] In 1900, David Hilbert posed the problem of proving their consistency using only finitistic methods as the second of his twenty-three problems.[19] In 1931, Kurt Gödel proved his second incompleteness theorem, which shows that such a consistency proof cannot be formalized within Peano arithmetic itself.[20]

Although it is widely claimed that Gödel’s theorem rules out the possibility of a finitistic consistency proof for Peano arithmetic, this depends on exactly what one means by a finitistic proof. Gödel himself pointed out the possibility of giving a finitistic consistency proof of Peano arithmetic or stronger systems by using finitistic methods that are not formalizable in Peano arithmetic, and in 1958, Gödel published a method for proving the consistency of arithmetic using type theory.[21] In 1936, Gerhard Gentzen gave a proof of the consistency of Peano’s axioms, using transfinite induction up to an ordinal called ε0.[22] Gentzen explained: “The aim of the present paper is to prove the consistency of elementary number theory or, rather, to reduce the question of consistency to certain fundamental principles”. Gentzen’s proof is arguably finitistic, since the transfinite ordinal ε0 can be encoded in terms of finite objects (for example, as a Turing machine describing a suitable order on the integers, or more abstractly as consisting of the finite trees, suitably linearly ordered). Whether or not Gentzen’s proof meets the requirements Hilbert envisioned is unclear: there is no generally accepted definition of exactly what is meant by a finitistic proof, and Hilbert himself never gave a precise definition.

The vast majority of contemporary mathematicians believe that Peano’s axioms are consistent, relying either on intuition or the acceptance of a consistency proof such as Gentzen’s proof. A small number of philosophers and mathematicians, some of whom also advocate ultrafinitism, reject Peano’s axioms because accepting the axioms amounts to accepting the infinite collection of natural numbers. In particular, addition (including the successor function) and multiplication are assumed to be total. Curiously, there are self-verifying theories that are similar to PA but have subtraction and division instead of addition and multiplication, which are axiomatized in such a way to avoid proving sentences that correspond to the totality of addition and multiplication, but which are still able to prove all true {\displaystyle \Pi _{1}}\Pi _{1} theorems of PA, and yet can be extended to a consistent theory that proves its own consistency (stated as the non-existence of a Hilbert-style proof of “0=1”).[23]

See also[edit]

Notes[edit]

  1. ^ {\displaystyle \forall x\ (x\cdot 0=0)}{\displaystyle \forall x\ (x\cdot 0=0)}” can be proven from the other axioms (in first-order logic) as follows. Firstly, {\displaystyle x\cdot 0+x\cdot 0=x\cdot (0+0)=x\cdot 0=x\cdot 0+0}{\displaystyle x\cdot 0+x\cdot 0=x\cdot (0+0)=x\cdot 0=x\cdot 0+0} by distributivity and additive identity. Secondly, {\displaystyle x\cdot 0=0\lor x\cdot 0>0}{\displaystyle x\cdot 0=0\lor x\cdot 0>0} by Axiom 15. If {\displaystyle x\cdot 0>0}{\displaystyle x\cdot 0>0} then {\displaystyle x\cdot 0+x\cdot 0>x\cdot 0+0}{\displaystyle x\cdot 0+x\cdot 0>x\cdot 0+0} by addition of the same element and commutativity, and hence {\displaystyle x\cdot 0+0>x\cdot 0+0}{\displaystyle x\cdot 0+0>x\cdot 0+0} by substitution, contradicting irreflexivity. Therefore it must be that {\displaystyle x\cdot 0=0}{\displaystyle x\cdot 0=0}.

 

Someone just viewed: FROM PRESIDENT WILLIAM JEFFERSON CLINTON – WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE – HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN – THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS – :) Inbox x BIG SHOT FROM CHICAGO x Streak 7:00 PM (21 minutes ago) to me Someone just viewed: “FROM PRESIDENT WILLIAM JEFFERSON CLINTON – WHO WON THE MAXIMUM NUMBER OF COUNTIES UP AND DWON THE MISSISPI RIVER WITH OR WITHOT LOISUINA PURCHASE PROGRAMA ND DIXIE CHICSKS OR MASON DIXIES AND FROM CHELSEA AND OLGA CLINTON -> HELLO VERYONE – HOW IS 2 O 0 6 SOUND TO U IN EAR AND IN IN YEAR FORMAT ? AND HILALRY RODHMAN CLINTONREFSUES TO TALK TO LOW LEVEL WHITES NEGROS ABORGINALS OR BROWN OR JEWS EVEN – THANK U FOR YOUR ABIITY AND FOR YOUR COMPREHSNION AND FOR YOUR GEMRORTY AND ALBEGRA AND AIRTHMETCI AND ORTRY ANDFOR YOUR PROSE AND FOR YOUR EXCELLENCE IN LISTEINGA ND FOR YOUR ORATORY AND SPECHS AND FOR ? FOR YOUR DINENR TABLE CONVERSSTIONS – :)” People on thread: 208 Westhaven Drive 78746 Blog Post By Email Device: PC Location: Chicago, IL

[A]

FROM PRESDIENT BILL CLINTON TO ALL OF U

https://www.youtube.com/watch?v=-Dx_STgFLLo

 

[B]

Image result for BEE

FROM PRESDIENT BILL CLINTON TO ALL OF U --- BEE

 

SEA

FROM PRESDIENT BILL CLINTON TO ALL OF U --- BEA

DOWN ARROW

SSSSEEEEEEEEE SEA CCCCCCCC SEEEEE

Someone just viewed: ONLY IF U DONT CRY CAN I ELL 100 % TRUTH- DID ME SAY MY SOUL BURNS WHEN KATRINA QND SALMAN NEWS COMES THAT THEY ARE GAUREDE BY MIDDLE EASTERN ARMY FOR BHARTA DID ME SAY BAARCK – THEN I EXPECT U KNOW WORD EXPECTAION BARACK THEN I EXPETC OLGA THEN GUARDED BY SUPER POWER THIS DOENT MENA 3RD WORLD INDIAN AMRY CHEAP SHIT BUT – EITHER – OBMA OR CLINTON CIA – OR – BY ? -> HENCE – KREMLN ME IS PISED AND ? AND I WANT – LODON HARRY ND WILIA TO BE TAUGHT WHAT MEANSNCE UPON A RAJATHSN HOW MANY MOTHERFUCKING RAJPUR MEANS WARRIRO CALSS NOT PEN PAPER PENCIL SHIT – BRAHIM BUT WARRIOR CLASS DIED FOR THE HONOR ? AND -> MENS ALSO THAT ONCE UPON A KHANATE – HOW MANY MOTHERFUCKING MUSLIM MADARCHOD WERE TOLD WHAT MEANS RUSSIAN EMPIRE – OK ? – DO U NOW UNDERATND NARENDRA DAMODAR MODI – OR SHOULD I KNOWCK THE FUCK OUT OF YOUR WHATEVER IJAJAT YOU AHVE LEFT IN YOUR SOUL FRO YOURSELF – ? OK? BHARAT MADARCHOD IS A BRAHMIN NAME – MUSLIM MAFIA COK C=SUKERS AND MUSLIM DOMINATED BOLLYWOOD AND M Inbox x BIG SHOT FROM CHICAGO x Streak 7:00 PM (7 minutes ago) to me Someone just viewed: “ONLY IF U DONT CRY CAN I ELL 100 % TRUTH- DID ME SAY MY SOUL BURNS WHEN KATRINA QND SALMAN NEWS COMES THAT THEY ARE GAUREDE BY MIDDLE EASTERN ARMY FOR BHARTA DID ME SAY BAARCK – THEN I EXPECT U KNOW WORD EXPECTAION BARACK THEN I EXPETC OLGA THEN GUARDED BY SUPER POWER THIS DOENT MENA 3RD WORLD INDIAN AMRY CHEAP SHIT BUT – EITHER – OBMA OR CLINTON CIA – OR – BY ? -> HENCE – KREMLN ME IS PISED AND ? AND I WANT – LODON HARRY ND WILIA TO BE TAUGHT WHAT MEANSNCE UPON A RAJATHSN HOW MANY MOTHERFUCKING RAJPUR MEANS WARRIRO CALSS NOT PEN PAPER PENCIL SHIT – BRAHIM BUT WARRIOR CLASS DIED FOR THE HONOR ? AND -> MENS ALSO THAT ONCE UPON A KHANATE – HOW MANY MOTHERFUCKING MUSLIM MADARCHOD WERE TOLD WHAT MEANS RUSSIAN EMPIRE – OK ? – DO U NOW UNDERATND NARENDRA DAMODAR MODI – OR SHOULD I KNOWCK THE FUCK OUT OF YOUR WHATEVER IJAJAT YOU AHVE LEFT IN YOUR SOUL FRO YOURSELF – ? OK? BHARAT MADARCHOD IS A BRAHMIN NAME – MUSLIM MAFIA COK C=SUKERS AND MUSLIM DOMINATED BOLLYWOOD AND MUSLIM MIDLE FUCKING EAST > SANJAY -> MADARCHOD ENSGLSIH APTIETNTA DN CHANEFELIER – MAIAY LJE LEAUD EMEANS NUMBER 4 AND 44 – IS LONON TIME ZONE AND THATS CALED BARCK HSUEIN MUSLIM DNA – B OBAMA IS WEAK WHEN IT COEMS TO MUSLIM TEROR – BUT HERD WORF IF HE BECOMES YONI NOT GANDHI MAADRCJOD HINDU MUSLIM SHIT BEAHCOND – 3RD WORLD CHEAP WHOETS SELLING SKIN SRESM AND SHIT ND GETTING ACKNWOELDGEMENT FROM BARCK OBAMA OR – BRON SHIT NATONS TO SAY ME MODI ME WILTH AMABNI -> NARENDRA -> KEARN HUINDU LANAGUEG MAADRCHOD BOLYWOOD IF U WANTS ME TO PROETCT U – OR ELSE RUSSIAN KGB WILL KILL US AND EVE PUNISH ME SACRCAISTICALLY – ABY FIRST INVITING ME AND THEN AKSING SOME VERY OAUNFUL JARD TO IGNRIE QUESTION – OK ?” People on thread: 208 Westhaven Drive 78746 Blog Post By Email Device: PC Location: Chicago, IL

[A]

UN ITEEABLE AND ON TP T SAYS OLGA CLINTON NUMBER 65 AND THE RESULTS MEANS 1969 WINE AND WHAT THE OTHER NUMBER 1979 MINUS 1969 OLGA

]2]

SO PAY ATTENTON NOW NARENDRA MODI AND U ALL

DOWN ARROWDOWN ARROW

CLICK

a lot on the line today

From: Biden for President
Date: Tue, Mar 3, 2020 at 6:36 PM
Subject: a lot on the line today
To: ajay mishra

There are more than one thousand delegates being awarded across the country and we need to win as many as possible.

‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌
ajay,

There’s a lot on the line today.

There are more than one thousand delegates being awarded across the country, and we need to win as many as possible to seal the deal on making Joe the Democratic nominee. Can we count on you for $5 today?

If you’ve saved payment information with ActBlue Express, your donation will go through immediately:
$5 ➞
$25 ➞
$50 ➞
$100 ➞
$250 ➞
Other ➞

Let us break it down, ajay:

IF we can bring in a surge of donations today, we’ll be in great shape to win the votes we need in strategic districts across the country.

IF we can win in these districts, we’ll earn the delegate totals we need to secure the Democratic nomination and send Joe Biden to defeat Donald Trump.

BUT we can only make all of this happen if all our supporters rally around our campaign today and make it happen!

Your donation today can make a huge difference, ajay. Will you chip in $5 to make sure Joe Biden is our Democratic nominee come July?

If you’ve saved payment information with ActBlue Express, your donation will go through immediately:
$5 ➞
$25 ➞
$50 ➞
$100 ➞
$250 ➞
Other ➞

Thanks for standing with us,

Biden for President

Any donor history information in this email reflects what we have on file for this specific email address. If you have donated with a different email, with a check, or with a spouse — thank you so much! We have that on file and cannot thank you enough for supporting this campaign.

This email was sent to ajayinsead03. If we got your name or any information wrong — we’re so sorry! To update and correct your information click here. If you would like to receive fewer emails, click here. If you would like to unsubscribe, click here.

ajay, thank you so much for supporting Joe Biden’s Presidential campaign.

Change occurs because the conscience of a country begins to rise up and demand — demand change.

This isn’t the time to be complacent. If you are ready to fight for the soul of this nation, you can start by donating to elect Joe Biden by clicking the button below.

DONATE TO ELECT JOE BIDEN »

We know we send a lot of emails, and we are sorry about that. The reason? We are relying on grassroots supporters like you (we’re serious!).

But ajay, we don’t want to bother you. If you’d like to only receive our most important emails, click here. If you’d only like to receive volunteer emails, click here. If you’d like to unsubscribe from our emails, you can click here.

To make a contribution by mail, click here for instructions.

We sincerely thank you for your help and support.
– The entire Joe Biden for President team

Paid for by Biden for President

open.gif

Someone just viewed: Re: About 8 results (0.22 seconds) Search Results Web results HSL BUILDING SERVICES Ltd | ZoomInfo.com https://www.zoominfo.com/c/hsl-building-services/353669554 View HSL BUILDING SERVICES Ltd location, revenue, industry and description. Find related and similar companies as well as employees by title and much … Images for 353669554 Image result for 353669554 Image result for 353669554 Image result for 353669554 Image result for 353669554 Image result for 353669554 More images for 353669554 Report images Web results 保湿的面膜什么牌子的好?_百度知道 zhidao.baidu.com › 生活 › 美容/塑身 › 化妆 – Translate this page Apr 3, 2013 – 补水保湿效果好的面膜首选“玉妍春8倍水海藻”面膜,“玉妍春8倍水海藻”是目前补水效果最强的有机天然面膜,其海藻胶原的析出量高达8倍。 Каталог пользователей ВКонтакте » 353 669 501 — 353 669 600 https://vkfaces.com/vk/u Inbox x BIG SHOT FROM CHICAGO x YONI NETANYAHU x Streak 6:23 PM (5 minutes ago) to me Someone

[1]

OH HSL IS HERE WOW

[2]

OH HSL IS HERE WOW

DOWN ARROW

CLICK

 

 

BAIDI WOW ------------ BAIDI WOW --- BAIDI WOW

 

REFERENCES

 

Каталог пользователей ВКонтакте

353 669 501 — 353 669 600
Александр Чуприков
Александр Чуприков

Екатеринбург, Россия
Нурлан Ерланов
Нурлан Ерланов

36 подписчиков · Алматы, Казахстан
Димаш Каиров
Димаш Каиров

1 подписчик
Ray World
Ray World
страница заблокирована администрацией ВКонтакте
Dezman Ciko

2 подписчика
Руслан Самошкин
Руслан Самошкин

2 подписчика · Тула, Россия
Nursulu Zhansultankizi
Nursulu Zhansultankizi

45 подписчиков · 24 года, Кызылорда, Казахстан
Толганай Нурбек
Толганай Нурбек
страница заблокирована администрацией ВКонтакте
Елизавета Витальевна
Елизавета Витальевна

6 подписчиков · 22 года, Темиртау, Казахстан
Яна Молот
Яна Молот

9 подписчиков
Елена Леонтьева
Елена Леонтьева
страница заблокирована администрацией ВКонтакте
Мария Павлова

2 подписчика · 26 лет, Владивосток, Россия
Zzzz Zzzz
Zzzz Zzzz

18 лет, Tokyo, Япония
Ольга Демкина
Ольга Демкина

7 подписчиков
Lelka Bezymnaja
Lelka Bezymnaja

16 подписчиков · Симферополь, Россия
Луиза Стокер-Эльбэ
Луиза Стокер-Эльбэ

4 подписчика · Барнаул, Россия
Hornuu Kartiba

20 лет, Ереван, Армения
DELETED
DELETED
страница удалена пользователем
Юсуф Ашуров
Юсуф Ашуров

5 подписчиков · 17 лет, Санкт-Петербург, Россия
Alexa Baderina
Alexa Baderina

1 подписчик · 23 года
Елена Варюхина
Елена Варюхина

Вологда, Россия
Стас Бузынин

8 подписчиков · 17 лет, Самара, Россия
Антонха Ивченко

1 подписчик · Россия
Eeze Lahori

4 подписчика · Lahore, Пакистан
Арсений Анохин

закрытая страница
Равшан Ламирович
Равшан Ламирович

Евлах, Азербайджан
Амелия Гасанова

закрытая страница
Катарина Журомская

закрытая страница
New Cards
New Cards

1 подписчик · 29 лет
Дмитрий Назаров
Дмитрий Назаров

16 подписчиков · Россия
Маша Щеголева
Маша Щеголева
страница заблокирована администрацией ВКонтакте
Анечка Дроздова

закрытая страница
DELETED
DELETED
страница удалена пользователем
Алена Александровна

6 подписчиков · Тулун, Россия
Саша Кирович
Саша Кирович

1 подписчик · 19 лет, Rīga, Латвия
DELETED
DELETED
страница удалена пользователем
Наталья Нигматуллина
Наталья Нигматуллина

1 подписчик · Костанай, Казахстан
DELETED
DELETED
страница удалена пользователем
Аля Фиалка
Аля Фиалка

5 подписчиков · Киев, Украина
Ксения Михайлова
DELETED
DELETED
страница удалена пользователем
Дима Булкин
Дима Булкин

Москва, Россия
Алина Царева
Алина Царева

181 подписчик · 16 лет, Владимир, Россия
Настя Степа
Настя Степа

Курск, Россия
Егор Крид
Егор Крид

Москва, Россия
DELETED
DELETED
страница удалена пользователем
Лидия Ефремова
Лидия Ефремова
страница заблокирована администрацией ВКонтакте
Alex Pozitive
Alex Pozitive

21 подписчик · 32 года, Череповец, Россия
Алла Личман
Алла Личман

2 подписчика · Antwerpen, Бельгия
Арсен Царикаев
Арсен Царикаев
страница заблокирована администрацией ВКонтакте
Дамир Шубаров
Дамир Шубаров

14 подписчиков · 19 лет
Vlad Vladislav
Vlad Vladislav

1 подписчик
Сабина Султан
Сабина Султан

14 подписчиков · 26 лет, Шымкент, Казахстан
Валерия Волкова
Валерия Волкова

21 подписчик · Казахстан
Дионис Вольф

страница недавно удалена пользователем
Murat Kara
Murat Kara

İstanbul, Турция
Нина Владимировна

закрытая страница
Вика Люкшина
Вика Люкшина

1 подписчик · 19 лет
Ева Ева
Ева Ева

1 подписчик
Павел Бойцов
Павел Бойцов

109 подписчиков · Сортавала, Россия
Дацунова Елена

1 подписчик · Тихорецк, Россия
Маша Фомина
Маша Фомина
страница заблокирована администрацией ВКонтакте
Азамат Бозиев
DELETED
DELETED
страница удалена пользователем
Софья Романюк
Софья Романюк
страница заблокирована администрацией ВКонтакте
Тимур Казюлин
Тимур Казюлин
страница заблокирована администрацией ВКонтакте
Emil Sarqsian

Ереван, Армения

Someone just viewed: TO BILL CLINTON FRO AJAY : BUBBA, I DID NOT JEW DOWN BARACK OBAMA OR DAVD AXELROD OR RAHM EMANUE AND YES I CAN – ND THEY WOULD NOT WIN – OK BUABBA DN U DO MENS ODO KNOWTHAT OK BUBAB AND I NWO THAT U KNOW THAT IKNOW THAT U KNOW OK BUBBA, BUT ALSO – I CANT MENAS CANNOT JEW DOWN BIBI – THANK U BUBBA, FI U WANT EXPLAIN IN YORU SPOPHISTCTED LAGAUEGA DN SOUTHERN DARWL IF AND WHEN SOMEONE ASKS OR SOMEONE WANTS Inbox x QUINCY WA, PROBABLY HOTMAN x YONI NETANYAHU x Streak 5:31 PM (42 minutes ago) to me Someone just viewed: “TO BILL CLINTON FRO AJAY : BUBBA, I DID NOT JEW DOWN BARACK OBAMA OR DAVD AXELROD OR RAHM EMANUE AND YES I CAN – ND THEY WOULD NOT WIN – OK BUABBA DN U DO MENS ODO KNOWTHAT OK BUBAB AND I NWO THAT U KNOW THAT IKNOW THAT U KNOW OK BUBBA, BUT ALSO – I CANT MENAS CANNOT JEW DOWN BIBI – THANK U BUBBA, FI U WANT EXPLAIN IN YORU SPOPHISTCTED LAGAUEGA DN SOUTHERN DARWL IF AND WHEN SOMEONE ASKS OR SOMEONE WANTS ” People on thread: Yoni Netanyahu Olga Blog Post By Email Device: Unknown Device Location: Quincy, WA

HOWZ 6--- BUBBA

BUBBA: SHIVA HAS HOW MANY FORMS -? T LEATS 25- ONE OFTHEM IS BHOLE NATH – ONE IS NEEKANATH MEANS MY MOM IS WHORE – BUT U KNOW BARACK ON TIMES CVER SID NATARAJ OR DID HE EXMBARCE PEACE EANS NEELKATH DAVID AND RAHM -> DOES THIS QAULIFY AS SHORT BUT LONG ENOUGH TO CHEW ON ? – ME IS MEANS – WHEN CHOICE WAS PRESENTED TO ACCEPT HE NOBLE AWARD U RAHM SAID – ITS CALL FOR ACTION AND AT ACCEPTANCE DID OUR BOY SAID AND VEOKED MAHABHARAT CALED THE JUST WAR – I MENA THE JIHAD TRSNALTION IN HINDU IS CALELD GEETA HAS MAHABHARAT AND DARHMA MEANS JUST WAR – MEANS – YADA YADA HI DAHRMASAY GLAIR BHAVAITA BAHARTA – ABYUTHANAMA MA DAHRMA’S TADATMA NUM SRAJA MYHAUM – IS CALED SHANK IN HINDU – AND SHOFAR IN HEBEW – RAHM

– ALSO CALLLED – THERE IS SEASON U SAID DEADFISH AND SCHOED YOUR FINGERS IN A SHARPED TWIST – OK RAHM- BUT DO U KNOW LIKE SENATOR FROM PUNAJAB – IS PAOSIITNINING ITEM SONG – WENT HARD ON SOME PEOPLES -MOM – THERE IS SIMILALRLY ALSO CALED WORD CALELD PAUL SAMUELSON THE JEW WHO WAS FIRST AMERICAN WHO WON THE NOBLE IN ECON – ME MEANS ARAHM EMANUEL BHAI MERE A- WREVALED PREFENRECS

IN HINDU OAND ENGLSIH IS ACELD WHEN PSUH COMES TO SHOVE DID U CHOSOE NEEKHANT OR DID U SAY -YEAH – ON TIME FRONT PGE COVER – ME IS CAELD NATARAJ ? – > IF I SY MORE – THEN – MAYBE DAVID CANT UNDERATND

– BUT RAHM – DO U KNOW WOR D YONI – ? OK DO U KNOW WORD YONI NETANAYHU IS CALED RASPOUTIN IN RUSSHIA AND GANDHI IN LUCKNOW – RAHM ? I MEAN U MOVED FURNITURE AT IDF – AND GUESS RAHM – WHERE’S MY CAR ?

I CANT SAY MORE BUBBA – I SPECIALLY CANT BECAUSE WHEN GANDHI AROVED TO SUA LIKE EEEDIE MUROHY – DID ONCEOUPON – HE WAS MEEMERIZED BY BNOT JUST BUSINESS SPEPLES JARGON UT LASO HOW JIMMY :) AND U GUYS POLITICIAIANS PEROSN SPOKE –

FROM 643361467 AJAY MISHRA TO PRESSIENT CLINTON N FOR YONI NETANAYHU N FOR NEEKANTH OR N FOR NATARAJ OR N FOR NESHER OR N FOR NOSHER
I MEANS – REALLY BUBBA – TELL BARCK AND HIS CAMP – THAT SHIVA HAS 25 MORPHOLOGIES T LEAST – SO, NEEKATH AND NATRAJ ARE TWO N FWORDS – U CAN ADD IN THE VOCBULARY OF BARACK

AND TELL HIM THAT AGANDHI SAID – – WHAT? – YEAH – LIKE MLK INTERPREETE – GANDHI SAID – WHAT BUBBA ?WHEN MK WS ASKED – WHAT IS THE ONLY ALETRNATIVE?

LET ME EXPLAIN LIKE BRAHMINS DO BUBBA

[A] U KNOW BLOLLEAN AND NON BOOLEAN AND HOSEIAN AND NON HOBESIAN CHOICE DINENR MENU?

[B] N FOR NEEKANTH OR N FOR NARTARAJ

[C] N FORWHICH NUBER OF SHIVA IS U WANTS – MY MAN ? TEL ME

N FOR WHAT BUBBA

HOWZ 6--- BUBBA

HOWZ 6--- BUBBA

OLGA LEDNIHENKO ARSHAD VARSI SANJAY DUTT AJAY MISHRA BILL CLINTON HILALRY JI - NUMBER 12 ND WHO AND WHICH ONE

AJAY MISHRA 643361467 BILL CLINTON MOVIE POSTER (1)

ME IS MEANS N FOR NESHER

OR

N FOR N FOR NO SHER

IS A CHOICE

AND

A MENU IN

LUNCH

BREAK FAST AND  IN DINNER