First Order Logic: how to prove the formula?












1












$begingroup$


How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?



$textbf{My work}:$



1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality



2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom



3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)



4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$



5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.



I dont know what to do next...






logic first-order-logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
    $endgroup$
    – Aleksandra
    Dec 9 '18 at 10:53


















1












$begingroup$


How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?



$textbf{My work}:$



1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality



2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom



3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)



4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$



5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.



I dont know what to do next...






logic first-order-logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
    $endgroup$
    – Aleksandra
    Dec 9 '18 at 10:53
















1












1








1





$begingroup$


How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?



$textbf{My work}:$



1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality



2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom



3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)



4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$



5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.



I dont know what to do next...






logic first-order-logic







share|cite|improve this question









$endgroup$




How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?



$textbf{My work}:$



1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality



2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom



3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)



4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$



5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.



I dont know what to do next...






logic first-order-logic




logic first-order-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 9:12









Aleksandra Aleksandra

545




545












  • $begingroup$
    I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
    $endgroup$
    – Aleksandra
    Dec 9 '18 at 10:53




















  • $begingroup$
    I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
    $endgroup$
    – Aleksandra
    Dec 9 '18 at 10:53


















$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53






$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53












1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint



The formula :




$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$




is a simple "variant" of the substitution axiom for equality :




$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$




where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.



Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.



Then apply the above axiom schema to get :




$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.




Finally, we have to use the Deduction Theorem followed by Generalization thrice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032185%2ffirst-order-logic-how-to-prove-the-formula%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint



The formula :




$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$




is a simple "variant" of the substitution axiom for equality :




$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$




where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.



Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.



Then apply the above axiom schema to get :




$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.




Finally, we have to use the Deduction Theorem followed by Generalization thrice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12
















0












$begingroup$

Hint



The formula :




$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$




is a simple "variant" of the substitution axiom for equality :




$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$




where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.



Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.



Then apply the above axiom schema to get :




$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.




Finally, we have to use the Deduction Theorem followed by Generalization thrice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12














0












0








0





$begingroup$

Hint



The formula :




$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$




is a simple "variant" of the substitution axiom for equality :




$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$




where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.



Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.



Then apply the above axiom schema to get :




$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.




Finally, we have to use the Deduction Theorem followed by Generalization thrice.






share|cite|improve this answer









$endgroup$



Hint



The formula :




$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$




is a simple "variant" of the substitution axiom for equality :




$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$




where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.



Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.



Then apply the above axiom schema to get :




$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.




Finally, we have to use the Deduction Theorem followed by Generalization thrice.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 11:27









Mauro ALLEGRANZAMauro ALLEGRANZA

67.2k449115




67.2k449115












  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12


















  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12
















$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12




$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032185%2ffirst-order-logic-how-to-prove-the-formula%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more