Proof using natural deduction (Tautology)












2












$begingroup$


I've been asked to prove the following tautology via natural deduction:



$forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$



I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.



First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.



Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.



Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



Is there a way to make this third deduction or did I just start my whole proof wrong?






logic first-order-logic predicate-logic natural-deduction formal-proofs







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I've been asked to prove the following tautology via natural deduction:



    $forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$



    I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.



    First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.



    Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.



    Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



    If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



    Is there a way to make this third deduction or did I just start my whole proof wrong?






    logic first-order-logic predicate-logic natural-deduction formal-proofs







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I've been asked to prove the following tautology via natural deduction:



      $forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$



      I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.



      First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.



      Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.



      Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



      If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



      Is there a way to make this third deduction or did I just start my whole proof wrong?






      logic first-order-logic predicate-logic natural-deduction formal-proofs







      share|cite|improve this question











      $endgroup$




      I've been asked to prove the following tautology via natural deduction:



      $forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$



      I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.



      First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.



      Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.



      Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



      If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.



      Is there a way to make this third deduction or did I just start my whole proof wrong?






      logic first-order-logic predicate-logic natural-deduction formal-proofs




      logic first-order-logic predicate-logic natural-deduction formal-proofs






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 15:51









      Taroccoesbrocco

      5,14761839




      5,14761839










      asked Jan 5 at 13:43









      NaborDHNaborDH

      154




      154






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).



          Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.



          begin{equation}
          dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
          end{equation}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the detailed explanation.
            $endgroup$
            – NaborDH
            Jan 5 at 17:18










          • $begingroup$
            @NaborDH - You are welcome!
            $endgroup$
            – Taroccoesbrocco
            Jan 5 at 19:03











          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%2f3062730%2fproof-using-natural-deduction-tautology%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









          2












          $begingroup$

          To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).



          Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.



          begin{equation}
          dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
          end{equation}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the detailed explanation.
            $endgroup$
            – NaborDH
            Jan 5 at 17:18










          • $begingroup$
            @NaborDH - You are welcome!
            $endgroup$
            – Taroccoesbrocco
            Jan 5 at 19:03
















          2












          $begingroup$

          To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).



          Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.



          begin{equation}
          dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
          end{equation}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the detailed explanation.
            $endgroup$
            – NaborDH
            Jan 5 at 17:18










          • $begingroup$
            @NaborDH - You are welcome!
            $endgroup$
            – Taroccoesbrocco
            Jan 5 at 19:03














          2












          2








          2





          $begingroup$

          To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).



          Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.



          begin{equation}
          dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
          end{equation}






          share|cite|improve this answer









          $endgroup$



          To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).



          Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.



          begin{equation}
          dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
          end{equation}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 15:45









          TaroccoesbroccoTaroccoesbrocco

          5,14761839




          5,14761839












          • $begingroup$
            Thanks for the detailed explanation.
            $endgroup$
            – NaborDH
            Jan 5 at 17:18










          • $begingroup$
            @NaborDH - You are welcome!
            $endgroup$
            – Taroccoesbrocco
            Jan 5 at 19:03


















          • $begingroup$
            Thanks for the detailed explanation.
            $endgroup$
            – NaborDH
            Jan 5 at 17:18










          • $begingroup$
            @NaborDH - You are welcome!
            $endgroup$
            – Taroccoesbrocco
            Jan 5 at 19:03
















          $begingroup$
          Thanks for the detailed explanation.
          $endgroup$
          – NaborDH
          Jan 5 at 17:18




          $begingroup$
          Thanks for the detailed explanation.
          $endgroup$
          – NaborDH
          Jan 5 at 17:18












          $begingroup$
          @NaborDH - You are welcome!
          $endgroup$
          – Taroccoesbrocco
          Jan 5 at 19:03




          $begingroup$
          @NaborDH - You are welcome!
          $endgroup$
          – Taroccoesbrocco
          Jan 5 at 19:03


















          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%2f3062730%2fproof-using-natural-deduction-tautology%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

          1300-talet

          Image
          Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
          Read more

          1300-talet

          Image
          Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
          Read more

          Display a custom attribute below product name in the front-end Magento 1.9.3.8

          Image
          0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
          Read more