Are these logic statements equivalent?












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I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.




  1. $forall{x}.(P(x) iff Q(x))$

  2. $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$

  3. $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$










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    $begingroup$


    I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.




    1. $forall{x}.(P(x) iff Q(x))$

    2. $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$

    3. $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$










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      0





      $begingroup$


      I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.




      1. $forall{x}.(P(x) iff Q(x))$

      2. $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$

      3. $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$










      share|cite|improve this question









      $endgroup$




      I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.




      1. $forall{x}.(P(x) iff Q(x))$

      2. $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$

      3. $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$







      first-order-logic






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      asked Dec 2 '18 at 2:57









      logically_confusedlogically_confused

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          I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.



          Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):



          First, universal quantifiers distribute over conjunctions, so:
          $$
          forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
          $$



          Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
          $$
          (forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
          $$

          Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.



          Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.






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            $begingroup$

            I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.



            Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):



            First, universal quantifiers distribute over conjunctions, so:
            $$
            forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
            $$



            Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
            $$
            (forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
            $$

            Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.



            Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.



              Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):



              First, universal quantifiers distribute over conjunctions, so:
              $$
              forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
              $$



              Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
              $$
              (forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
              $$

              Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.



              Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.



                Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):



                First, universal quantifiers distribute over conjunctions, so:
                $$
                forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
                $$



                Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
                $$
                (forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
                $$

                Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.



                Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.






                share|cite|improve this answer









                $endgroup$



                I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.



                Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):



                First, universal quantifiers distribute over conjunctions, so:
                $$
                forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
                $$



                Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
                $$
                (forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
                $$

                Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.



                Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 3:48









                FranklinBashFranklinBash

                1012




                1012






























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