Translation into predicate calculus












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$begingroup$


I want to translate the following sentence into predicate calculus:



"Anything taller than something Alice is taller than is taller than Alice."



Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$



Which is the correct one, and why?










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    0












    $begingroup$


    I want to translate the following sentence into predicate calculus:



    "Anything taller than something Alice is taller than is taller than Alice."



    Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
    But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$



    Which is the correct one, and why?










    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      I want to translate the following sentence into predicate calculus:



      "Anything taller than something Alice is taller than is taller than Alice."



      Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
      But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$



      Which is the correct one, and why?










      share|cite|improve this question











      $endgroup$




      I want to translate the following sentence into predicate calculus:



      "Anything taller than something Alice is taller than is taller than Alice."



      Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
      But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$



      Which is the correct one, and why?







      logic predicate-logic logic-translation






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      edited Dec 18 '18 at 1:45









      Bram28

      63.2k44793




      63.2k44793










      asked Oct 15 '16 at 22:46









      satokunsatokun

      403412




      403412






















          2 Answers
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          $begingroup$

          In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.






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          $endgroup$





















            1












            $begingroup$

            The direct translation would be (with $>$ instead of $T$):
            $$ forall x( exists y(x > y land a > y) to x > a ) $$
            where the $exists y$ is inside the premise of the implication.



            This is equivalent to
            $$ forall x forall y( (x>y land a>y) to x > a ) $$
            where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".



            Note, though, that
            $$ forall x exists y( (x>y land a>y) to x > a ) $$
            is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              2












              $begingroup$

              In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.






                  share|cite|improve this answer









                  $endgroup$



                  In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 15 '16 at 22:59









                  Rob ArthanRob Arthan

                  29.3k42966




                  29.3k42966























                      1












                      $begingroup$

                      The direct translation would be (with $>$ instead of $T$):
                      $$ forall x( exists y(x > y land a > y) to x > a ) $$
                      where the $exists y$ is inside the premise of the implication.



                      This is equivalent to
                      $$ forall x forall y( (x>y land a>y) to x > a ) $$
                      where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".



                      Note, though, that
                      $$ forall x exists y( (x>y land a>y) to x > a ) $$
                      is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The direct translation would be (with $>$ instead of $T$):
                        $$ forall x( exists y(x > y land a > y) to x > a ) $$
                        where the $exists y$ is inside the premise of the implication.



                        This is equivalent to
                        $$ forall x forall y( (x>y land a>y) to x > a ) $$
                        where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".



                        Note, though, that
                        $$ forall x exists y( (x>y land a>y) to x > a ) $$
                        is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The direct translation would be (with $>$ instead of $T$):
                          $$ forall x( exists y(x > y land a > y) to x > a ) $$
                          where the $exists y$ is inside the premise of the implication.



                          This is equivalent to
                          $$ forall x forall y( (x>y land a>y) to x > a ) $$
                          where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".



                          Note, though, that
                          $$ forall x exists y( (x>y land a>y) to x > a ) $$
                          is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.






                          share|cite|improve this answer









                          $endgroup$



                          The direct translation would be (with $>$ instead of $T$):
                          $$ forall x( exists y(x > y land a > y) to x > a ) $$
                          where the $exists y$ is inside the premise of the implication.



                          This is equivalent to
                          $$ forall x forall y( (x>y land a>y) to x > a ) $$
                          where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".



                          Note, though, that
                          $$ forall x exists y( (x>y land a>y) to x > a ) $$
                          is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Oct 15 '16 at 22:58









                          Henning MakholmHenning Makholm

                          241k17307546




                          241k17307546






























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