Negating a Symbolic Expression/Logic

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Negate the following and simplify as much as you can...



$$exists x ~forall y~(p(y) to forall z~q(z))$$



How would I negate this expression. Not sure how to start it.










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    up vote
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    Negate the following and simplify as much as you can...



    $$exists x ~forall y~(p(y) to forall z~q(z))$$



    How would I negate this expression. Not sure how to start it.










    share|cite|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Negate the following and simplify as much as you can...



      $$exists x ~forall y~(p(y) to forall z~q(z))$$



      How would I negate this expression. Not sure how to start it.










      share|cite|improve this question















      Negate the following and simplify as much as you can...



      $$exists x ~forall y~(p(y) to forall z~q(z))$$



      How would I negate this expression. Not sure how to start it.







      discrete-mathematics logic






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      edited Sep 6 at 3:16









      Graham Kemp

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      asked Sep 6 at 3:12









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          Begin with placing a negation sign.



          $qquadneg exists x ~forall y~(p(y) to forall z~q(z))$



          Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.



          $qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$






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            up vote
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            down vote













            Begin with placing a negation sign.



            $qquadneg exists x ~forall y~(p(y) to forall z~q(z))$



            Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.



            $qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$






            share|cite|improve this answer
























              up vote
              4
              down vote













              Begin with placing a negation sign.



              $qquadneg exists x ~forall y~(p(y) to forall z~q(z))$



              Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.



              $qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$






              share|cite|improve this answer






















                up vote
                4
                down vote










                up vote
                4
                down vote









                Begin with placing a negation sign.



                $qquadneg exists x ~forall y~(p(y) to forall z~q(z))$



                Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.



                $qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$






                share|cite|improve this answer












                Begin with placing a negation sign.



                $qquadneg exists x ~forall y~(p(y) to forall z~q(z))$



                Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.



                $qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 6 at 3:24









                Graham Kemp

                81.4k43275




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