Proof using natural deduction (Tautology)












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$begingroup$


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







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    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






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      share|cite|improve this question













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      edited Jan 5 at 15:51









      Taroccoesbrocco

      5,14761839




      5,14761839










      asked Jan 5 at 13:43









      NaborDHNaborDH

      154




      154






















          1 Answer
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          $begingroup$

          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









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










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











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          $begingroup$

          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


















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