A simple question about quantifiers and negation in conditionals












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$begingroup$


In classical propositional logic $neg(A to B) equiv A land neg B$. In predicate logic, therefore, do we have $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))$?



Or, instead, would $neg (forall x (phi(x) to psi(x)) equiv exists x (phi(x) land neg psi(x))$?



Or do we have, in fact, $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$. If this is the case, then why do we have $forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$ ?



More generally, I don't understand how we understand a formula like (1), as opposed to one like (2):



$$(1);;forall x (phi(x) land neg psi(x))$$



$$(2); ;forall x (phi(x) to neg psi(x))$$




Edit: It is still unclear to me what the precise difference between (1) and (2) is (and their variants $sans$ negation). (1) requires the truth of $phi(x)$ for all substitution instances, whereas (2) does not. Is that all?



But then, what might be a natural sentence of English which expresses (1)? Would it be something like (1')?



(1') Every man dances and he doesn't play football






logic predicate-logic







share|cite|improve this question











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    0












    $begingroup$


    In classical propositional logic $neg(A to B) equiv A land neg B$. In predicate logic, therefore, do we have $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))$?



    Or, instead, would $neg (forall x (phi(x) to psi(x)) equiv exists x (phi(x) land neg psi(x))$?



    Or do we have, in fact, $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$. If this is the case, then why do we have $forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$ ?



    More generally, I don't understand how we understand a formula like (1), as opposed to one like (2):



    $$(1);;forall x (phi(x) land neg psi(x))$$



    $$(2); ;forall x (phi(x) to neg psi(x))$$




    Edit: It is still unclear to me what the precise difference between (1) and (2) is (and their variants $sans$ negation). (1) requires the truth of $phi(x)$ for all substitution instances, whereas (2) does not. Is that all?



    But then, what might be a natural sentence of English which expresses (1)? Would it be something like (1')?



    (1') Every man dances and he doesn't play football






    logic predicate-logic







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In classical propositional logic $neg(A to B) equiv A land neg B$. In predicate logic, therefore, do we have $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))$?



      Or, instead, would $neg (forall x (phi(x) to psi(x)) equiv exists x (phi(x) land neg psi(x))$?



      Or do we have, in fact, $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$. If this is the case, then why do we have $forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$ ?



      More generally, I don't understand how we understand a formula like (1), as opposed to one like (2):



      $$(1);;forall x (phi(x) land neg psi(x))$$



      $$(2); ;forall x (phi(x) to neg psi(x))$$




      Edit: It is still unclear to me what the precise difference between (1) and (2) is (and their variants $sans$ negation). (1) requires the truth of $phi(x)$ for all substitution instances, whereas (2) does not. Is that all?



      But then, what might be a natural sentence of English which expresses (1)? Would it be something like (1')?



      (1') Every man dances and he doesn't play football






      logic predicate-logic







      share|cite|improve this question











      $endgroup$




      In classical propositional logic $neg(A to B) equiv A land neg B$. In predicate logic, therefore, do we have $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))$?



      Or, instead, would $neg (forall x (phi(x) to psi(x)) equiv exists x (phi(x) land neg psi(x))$?



      Or do we have, in fact, $neg (forall x (phi(x) to psi(x)) equiv forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$. If this is the case, then why do we have $forall x (phi(x) land neg psi(x))equiv exists x (phi(x) land neg psi(x))$ ?



      More generally, I don't understand how we understand a formula like (1), as opposed to one like (2):



      $$(1);;forall x (phi(x) land neg psi(x))$$



      $$(2); ;forall x (phi(x) to neg psi(x))$$




      Edit: It is still unclear to me what the precise difference between (1) and (2) is (and their variants $sans$ negation). (1) requires the truth of $phi(x)$ for all substitution instances, whereas (2) does not. Is that all?



      But then, what might be a natural sentence of English which expresses (1)? Would it be something like (1')?



      (1') Every man dances and he doesn't play football






      logic predicate-logic




      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 11 '18 at 10:47







      Edward.Lin

















      asked Dec 11 '18 at 10:12









      Edward.LinEdward.Lin

      83




      83






















          2 Answers
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          1












          $begingroup$

          Take it step by step, first apply quantifier duality, and next negate the predicate.



          $$begin{align}&lnotforall x~big(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~lnotbig(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~big(phi(x)landlnotpsi(x)big)end{align}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, yes. What about the difference between (1) and (2)?
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:40










          • $begingroup$
            I have revised my question
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:48



















          1












          $begingroup$

          In general $negforall x;P(x)$ and $exists x;neg P(x)$ are the equivalent.



          In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".



          So if $P(x)$ is $phi(x)topsi(x)$ then it appears that $negforall x;[phi(x)topsi(x)]$ and $exists x;neg [phi(x)topsi(x)]$ are equivalent.



          Since $neg [phi(x)topsi(x)]$ and $phi(x)wedgenegpsi(x)$ are equivalent, there is also equivalence with $exists x;[phi(x)wedgenegpsi(x)]$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            1












            $begingroup$

            Take it step by step, first apply quantifier duality, and next negate the predicate.



            $$begin{align}&lnotforall x~big(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~lnotbig(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~big(phi(x)landlnotpsi(x)big)end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ah, yes. What about the difference between (1) and (2)?
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:40










            • $begingroup$
              I have revised my question
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:48
















            1












            $begingroup$

            Take it step by step, first apply quantifier duality, and next negate the predicate.



            $$begin{align}&lnotforall x~big(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~lnotbig(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~big(phi(x)landlnotpsi(x)big)end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ah, yes. What about the difference between (1) and (2)?
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:40










            • $begingroup$
              I have revised my question
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:48














            1












            1








            1





            $begingroup$

            Take it step by step, first apply quantifier duality, and next negate the predicate.



            $$begin{align}&lnotforall x~big(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~lnotbig(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~big(phi(x)landlnotpsi(x)big)end{align}$$






            share|cite|improve this answer









            $endgroup$



            Take it step by step, first apply quantifier duality, and next negate the predicate.



            $$begin{align}&lnotforall x~big(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~lnotbig(phi(x)topsi(x)big)\[1ex]equiv ~&exists x~big(phi(x)landlnotpsi(x)big)end{align}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 10:25









            Graham KempGraham Kemp

            86k43378




            86k43378












            • $begingroup$
              Ah, yes. What about the difference between (1) and (2)?
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:40










            • $begingroup$
              I have revised my question
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:48


















            • $begingroup$
              Ah, yes. What about the difference between (1) and (2)?
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:40










            • $begingroup$
              I have revised my question
              $endgroup$
              – Edward.Lin
              Dec 11 '18 at 10:48
















            $begingroup$
            Ah, yes. What about the difference between (1) and (2)?
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:40




            $begingroup$
            Ah, yes. What about the difference between (1) and (2)?
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:40












            $begingroup$
            I have revised my question
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:48




            $begingroup$
            I have revised my question
            $endgroup$
            – Edward.Lin
            Dec 11 '18 at 10:48











            1












            $begingroup$

            In general $negforall x;P(x)$ and $exists x;neg P(x)$ are the equivalent.



            In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".



            So if $P(x)$ is $phi(x)topsi(x)$ then it appears that $negforall x;[phi(x)topsi(x)]$ and $exists x;neg [phi(x)topsi(x)]$ are equivalent.



            Since $neg [phi(x)topsi(x)]$ and $phi(x)wedgenegpsi(x)$ are equivalent, there is also equivalence with $exists x;[phi(x)wedgenegpsi(x)]$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              In general $negforall x;P(x)$ and $exists x;neg P(x)$ are the equivalent.



              In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".



              So if $P(x)$ is $phi(x)topsi(x)$ then it appears that $negforall x;[phi(x)topsi(x)]$ and $exists x;neg [phi(x)topsi(x)]$ are equivalent.



              Since $neg [phi(x)topsi(x)]$ and $phi(x)wedgenegpsi(x)$ are equivalent, there is also equivalence with $exists x;[phi(x)wedgenegpsi(x)]$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                In general $negforall x;P(x)$ and $exists x;neg P(x)$ are the equivalent.



                In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".



                So if $P(x)$ is $phi(x)topsi(x)$ then it appears that $negforall x;[phi(x)topsi(x)]$ and $exists x;neg [phi(x)topsi(x)]$ are equivalent.



                Since $neg [phi(x)topsi(x)]$ and $phi(x)wedgenegpsi(x)$ are equivalent, there is also equivalence with $exists x;[phi(x)wedgenegpsi(x)]$






                share|cite|improve this answer









                $endgroup$



                In general $negforall x;P(x)$ and $exists x;neg P(x)$ are the equivalent.



                In words: "not all $x$ have property $P$" is the same as "some $x$ exists that has not property $P$".



                So if $P(x)$ is $phi(x)topsi(x)$ then it appears that $negforall x;[phi(x)topsi(x)]$ and $exists x;neg [phi(x)topsi(x)]$ are equivalent.



                Since $neg [phi(x)topsi(x)]$ and $phi(x)wedgenegpsi(x)$ are equivalent, there is also equivalence with $exists x;[phi(x)wedgenegpsi(x)]$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 10:24









                drhabdrhab

                102k545136




                102k545136






























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