Show that $(exists x. P_1(x) to P_2(x)) to ((forall y.P_1(y)) to exists z.P_2(z))$ is a true statement.












1














We're covering Prepositional Logic in our class and I'm a little confused as to how the entire concept actually works. My understanding is that prepositional logic is used to study ways of combining or altering statements or propositions to form more complicated statements or propositions. Therefore when we have logical expression, we have to show that the statement is true or false.



Let's say I want to show that the following logical statement is true:



$$(exists x. P_1(x) to P_2(x)) to ((forall y.P_1(y)) to exists z.P_2(z))$$




  1. Assumption ($A1$): $(exists x. P_1(x) to P_2(x))$

  2. To Show ($S1$): $((forall y.P_1(y)) to exists z.P_2(z))$

  3. Assumption ($A2$): $((forall y.P_1(y))$

  4. To Show ($S2$): $exists z.P_2(z))$


Now I know that at this point I should choose $x$ as arbitrary but fixed in A1, but I don't know the exact reason for this, instead of lets says choosing $z equiv x$.




  1. Assumption ($A3$): $P_1(x) to P_2(x)$

  2. Choose $y equiv x$ in A2.

  3. Assumption ($A4$): $P_1(x)$


After this point, I'm uncertain with what to do.



PS - I know that "To Show" and "Assumption" may not be the correct English terms but they're the closest translation from German "Annahme" and "Zu Zeigen" respectively.






proof-verification logic predicate-logic







share|cite|improve this question
























  • This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
    – Git Gud
    Dec 1 at 18:25
















1














We're covering Prepositional Logic in our class and I'm a little confused as to how the entire concept actually works. My understanding is that prepositional logic is used to study ways of combining or altering statements or propositions to form more complicated statements or propositions. Therefore when we have logical expression, we have to show that the statement is true or false.



Let's say I want to show that the following logical statement is true:



$$(exists x. P_1(x) to P_2(x)) to ((forall y.P_1(y)) to exists z.P_2(z))$$




  1. Assumption ($A1$): $(exists x. P_1(x) to P_2(x))$

  2. To Show ($S1$): $((forall y.P_1(y)) to exists z.P_2(z))$

  3. Assumption ($A2$): $((forall y.P_1(y))$

  4. To Show ($S2$): $exists z.P_2(z))$


Now I know that at this point I should choose $x$ as arbitrary but fixed in A1, but I don't know the exact reason for this, instead of lets says choosing $z equiv x$.




  1. Assumption ($A3$): $P_1(x) to P_2(x)$

  2. Choose $y equiv x$ in A2.

  3. Assumption ($A4$): $P_1(x)$


After this point, I'm uncertain with what to do.



PS - I know that "To Show" and "Assumption" may not be the correct English terms but they're the closest translation from German "Annahme" and "Zu Zeigen" respectively.






proof-verification logic predicate-logic







share|cite|improve this question
























  • This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
    – Git Gud
    Dec 1 at 18:25














1












1








1







We're covering Prepositional Logic in our class and I'm a little confused as to how the entire concept actually works. My understanding is that prepositional logic is used to study ways of combining or altering statements or propositions to form more complicated statements or propositions. Therefore when we have logical expression, we have to show that the statement is true or false.



Let's say I want to show that the following logical statement is true:



$$(exists x. P_1(x) to P_2(x)) to ((forall y.P_1(y)) to exists z.P_2(z))$$




  1. Assumption ($A1$): $(exists x. P_1(x) to P_2(x))$

  2. To Show ($S1$): $((forall y.P_1(y)) to exists z.P_2(z))$

  3. Assumption ($A2$): $((forall y.P_1(y))$

  4. To Show ($S2$): $exists z.P_2(z))$


Now I know that at this point I should choose $x$ as arbitrary but fixed in A1, but I don't know the exact reason for this, instead of lets says choosing $z equiv x$.




  1. Assumption ($A3$): $P_1(x) to P_2(x)$

  2. Choose $y equiv x$ in A2.

  3. Assumption ($A4$): $P_1(x)$


After this point, I'm uncertain with what to do.



PS - I know that "To Show" and "Assumption" may not be the correct English terms but they're the closest translation from German "Annahme" and "Zu Zeigen" respectively.






proof-verification logic predicate-logic







share|cite|improve this question















We're covering Prepositional Logic in our class and I'm a little confused as to how the entire concept actually works. My understanding is that prepositional logic is used to study ways of combining or altering statements or propositions to form more complicated statements or propositions. Therefore when we have logical expression, we have to show that the statement is true or false.



Let's say I want to show that the following logical statement is true:



$$(exists x. P_1(x) to P_2(x)) to ((forall y.P_1(y)) to exists z.P_2(z))$$




  1. Assumption ($A1$): $(exists x. P_1(x) to P_2(x))$

  2. To Show ($S1$): $((forall y.P_1(y)) to exists z.P_2(z))$

  3. Assumption ($A2$): $((forall y.P_1(y))$

  4. To Show ($S2$): $exists z.P_2(z))$


Now I know that at this point I should choose $x$ as arbitrary but fixed in A1, but I don't know the exact reason for this, instead of lets says choosing $z equiv x$.




  1. Assumption ($A3$): $P_1(x) to P_2(x)$

  2. Choose $y equiv x$ in A2.

  3. Assumption ($A4$): $P_1(x)$


After this point, I'm uncertain with what to do.



PS - I know that "To Show" and "Assumption" may not be the correct English terms but they're the closest translation from German "Annahme" and "Zu Zeigen" respectively.






proof-verification logic predicate-logic




proof-verification logic predicate-logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 18:23









Git Gud

28.7k1049100




28.7k1049100










asked Dec 1 at 18:11









xxxtentacion

372112




372112












  • This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
    – Git Gud
    Dec 1 at 18:25


















  • This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
    – Git Gud
    Dec 1 at 18:25
















This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
– Git Gud
Dec 1 at 18:25




This isn't Propositional Logic anymore, but rather Predicate Logic. You're basically done, You have $P_1(x)$ and you have $P_1(x)to P_2(x)$. You can now, presumably, use modus ponens. Is this familiar to you? Otherwise it is fine, although I don't like the way you phrased 5. at all.
– Git Gud
Dec 1 at 18:25










1 Answer
1






active

oldest

votes


















1














Here is an outline of an argument I assume you can turn into whatever rigor you need:



1) If $exists x. P_1(x)rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)rightarrow P_2(c)$.



2) If $forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.



3) Since we are in the case that $P_1(c)rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.



4) Since $P_2(c)$ is true, it must be the case that $exists z. P_2(z)$ is also true.



5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.






share|cite|improve this answer























    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%2f3021642%2fshow-that-exists-x-p-1x-to-p-2x-to-forall-y-p-1y-to-exists-z%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









    1














    Here is an outline of an argument I assume you can turn into whatever rigor you need:



    1) If $exists x. P_1(x)rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)rightarrow P_2(c)$.



    2) If $forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.



    3) Since we are in the case that $P_1(c)rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.



    4) Since $P_2(c)$ is true, it must be the case that $exists z. P_2(z)$ is also true.



    5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.






    share|cite|improve this answer




























      1














      Here is an outline of an argument I assume you can turn into whatever rigor you need:



      1) If $exists x. P_1(x)rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)rightarrow P_2(c)$.



      2) If $forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.



      3) Since we are in the case that $P_1(c)rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.



      4) Since $P_2(c)$ is true, it must be the case that $exists z. P_2(z)$ is also true.



      5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.






      share|cite|improve this answer


























        1












        1








        1






        Here is an outline of an argument I assume you can turn into whatever rigor you need:



        1) If $exists x. P_1(x)rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)rightarrow P_2(c)$.



        2) If $forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.



        3) Since we are in the case that $P_1(c)rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.



        4) Since $P_2(c)$ is true, it must be the case that $exists z. P_2(z)$ is also true.



        5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.






        share|cite|improve this answer














        Here is an outline of an argument I assume you can turn into whatever rigor you need:



        1) If $exists x. P_1(x)rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)rightarrow P_2(c)$.



        2) If $forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.



        3) Since we are in the case that $P_1(c)rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.



        4) Since $P_2(c)$ is true, it must be the case that $exists z. P_2(z)$ is also true.



        5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 at 18:39

























        answered Dec 1 at 18:32









        AVeryTiredGuy

        112




        112






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3021642%2fshow-that-exists-x-p-1x-to-p-2x-to-forall-y-p-1y-to-exists-z%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more