Prove formula using skolemization and resolution












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$begingroup$


I have the following formula that I am trying to prove is valid:



$$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$



By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:



$$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
$$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
$$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
$$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
$$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
$$text{(no change)} qquad qquad text{Distribute matrix}$$
$$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$



I'm not sure how to reach the empty clause from here. I'd appreciate a hint.






first-order-logic quantifiers







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$endgroup$

















    1












    $begingroup$


    I have the following formula that I am trying to prove is valid:



    $$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$



    By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:



    $$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
    $$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
    $$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
    $$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
    $$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
    $$text{(no change)} qquad qquad text{Distribute matrix}$$
    $$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$



    I'm not sure how to reach the empty clause from here. I'd appreciate a hint.






    first-order-logic quantifiers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have the following formula that I am trying to prove is valid:



      $$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$



      By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:



      $$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
      $$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
      $$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
      $$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
      $$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
      $$text{(no change)} qquad qquad text{Distribute matrix}$$
      $$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$



      I'm not sure how to reach the empty clause from here. I'd appreciate a hint.






      first-order-logic quantifiers







      share|cite|improve this question











      $endgroup$




      I have the following formula that I am trying to prove is valid:



      $$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$



      By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:



      $$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
      $$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
      $$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
      $$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
      $$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
      $$text{(no change)} qquad qquad text{Distribute matrix}$$
      $$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$



      I'm not sure how to reach the empty clause from here. I'd appreciate a hint.






      first-order-logic quantifiers




      first-order-logic quantifiers






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      share|cite|improve this question













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      edited Dec 4 '18 at 2:32







      Steve

















      asked Dec 4 '18 at 2:26









      SteveSteve

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          $begingroup$

          When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$



          That is, you get:



          $exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$



          $exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$



          $exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$



          $exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$



          $exists x exists w forall y forall z (q(x,y) land neg q(z,w))$



          Skolemizing now gets you:



          $forall y forall z (q(a,y) land neg q(z,b))$



          Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$



          And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.






          share|cite|improve this answer











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            $begingroup$

            When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$



            That is, you get:



            $exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$



            $exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$



            $exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$



            $exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$



            $exists x exists w forall y forall z (q(x,y) land neg q(z,w))$



            Skolemizing now gets you:



            $forall y forall z (q(a,y) land neg q(z,b))$



            Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$



            And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$



              That is, you get:



              $exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$



              $exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$



              $exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$



              $exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$



              $exists x exists w forall y forall z (q(x,y) land neg q(z,w))$



              Skolemizing now gets you:



              $forall y forall z (q(a,y) land neg q(z,b))$



              Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$



              And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$



                That is, you get:



                $exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$



                $exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$



                $exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$



                $exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$



                $exists x exists w forall y forall z (q(x,y) land neg q(z,w))$



                Skolemizing now gets you:



                $forall y forall z (q(a,y) land neg q(z,b))$



                Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$



                And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.






                share|cite|improve this answer











                $endgroup$



                When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$



                That is, you get:



                $exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$



                $exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$



                $exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$



                $exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$



                $exists x exists w forall y forall z (q(x,y) land neg q(z,w))$



                Skolemizing now gets you:



                $forall y forall z (q(a,y) land neg q(z,b))$



                Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$



                And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 3:16

























                answered Dec 4 '18 at 3:05









                Bram28Bram28

                61.6k44793




                61.6k44793






























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