Why does skolemization *not* preserve validity?

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I'm wondering what exactly is meant when people say "Skolemization preserves satisfiability but not validity". I'm having trouble wrapping my brain around it because I think of Skolemization, when considered as an inference rule, to simply "be an okay thing to do" when working with a second-order formula.



Skolemization is said to preserve satisfiability but not validity. What exactly do satisfiability and validity mean in the context? Is there a way to think about what Skolemization is that makes it obvious that it doesn't preserve validity? Does "co-Skolemization" described below have the same property?




Skolemization, as far as I can tell, is based on the equivalence between the first-order formula (and therefore degenerate second-order formula)



$$ forall x_1 cdots x_n . exists y . P(x_1, cdots, x_n, y) $$



and the following second order formula, that moves the $exists$ in front of the $forall$.



$$ exists f. forall x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n))$$



It makes perfect sense when the expression is thought of as a game. In the first expression, the opponent first makes $n$ moves and then the player makes 1. In the second expression, the player moves first but commits to a response function of sorts rather than committing to a single move. That's not a proof, but it seems like good intuitive justification for why rewriting an expression like this is okay in the first place.



It seems to me that performing this translation at all only makes sense if we're talking about a closed expression. That is, $P(cdots)$ must have no free variables or, equivalently, $P$ must be an $n+1$-place predicate (primitive or defined).




In propositional logic, satisfiability and validity can be though of by describing what happens to free variables. All variables are free because there's no way to bind them.



$ P lor P $ is satisfiable because the assignment $ P = top $ satisfies is.



$ P lor lnot P $ is a tautology for obvious reasons. I don't know whether it's right to say $ P lor lnot P $ is valid.



An inference rule that maps $P$ to $P lor Q$ is valid (disjunction introduction).



A rule that maps $P$ to $P land Q$ preserves satisfiability. We can take any old assignment $mu$ and construct $mu cup Q = top $ which satisfies the new expression.



In the context of Skolemization, are the meanings of satisfiability and validity similar?




Let "co-Skolemization" (probably has a real name) be the translation from



$$ exists x_1 cdots x_n . forall y. P(x_1, cdots, x_n, y) $$



to



$$ forall f exists x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n)) $$



That seems just as legitimate as Skolemization, considered as an inference rule in a second-order setting. There's a caveat that in a first order theory definitions are provided for all fully saturated combinations of function symbols and arguments, so therefore
co-Skolemization" definitely takes us out of first-order land, but I don't know how relevant that is.










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    I'm wondering what exactly is meant when people say "Skolemization preserves satisfiability but not validity". I'm having trouble wrapping my brain around it because I think of Skolemization, when considered as an inference rule, to simply "be an okay thing to do" when working with a second-order formula.



    Skolemization is said to preserve satisfiability but not validity. What exactly do satisfiability and validity mean in the context? Is there a way to think about what Skolemization is that makes it obvious that it doesn't preserve validity? Does "co-Skolemization" described below have the same property?




    Skolemization, as far as I can tell, is based on the equivalence between the first-order formula (and therefore degenerate second-order formula)



    $$ forall x_1 cdots x_n . exists y . P(x_1, cdots, x_n, y) $$



    and the following second order formula, that moves the $exists$ in front of the $forall$.



    $$ exists f. forall x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n))$$



    It makes perfect sense when the expression is thought of as a game. In the first expression, the opponent first makes $n$ moves and then the player makes 1. In the second expression, the player moves first but commits to a response function of sorts rather than committing to a single move. That's not a proof, but it seems like good intuitive justification for why rewriting an expression like this is okay in the first place.



    It seems to me that performing this translation at all only makes sense if we're talking about a closed expression. That is, $P(cdots)$ must have no free variables or, equivalently, $P$ must be an $n+1$-place predicate (primitive or defined).




    In propositional logic, satisfiability and validity can be though of by describing what happens to free variables. All variables are free because there's no way to bind them.



    $ P lor P $ is satisfiable because the assignment $ P = top $ satisfies is.



    $ P lor lnot P $ is a tautology for obvious reasons. I don't know whether it's right to say $ P lor lnot P $ is valid.



    An inference rule that maps $P$ to $P lor Q$ is valid (disjunction introduction).



    A rule that maps $P$ to $P land Q$ preserves satisfiability. We can take any old assignment $mu$ and construct $mu cup Q = top $ which satisfies the new expression.



    In the context of Skolemization, are the meanings of satisfiability and validity similar?




    Let "co-Skolemization" (probably has a real name) be the translation from



    $$ exists x_1 cdots x_n . forall y. P(x_1, cdots, x_n, y) $$



    to



    $$ forall f exists x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n)) $$



    That seems just as legitimate as Skolemization, considered as an inference rule in a second-order setting. There's a caveat that in a first order theory definitions are provided for all fully saturated combinations of function symbols and arguments, so therefore
    co-Skolemization" definitely takes us out of first-order land, but I don't know how relevant that is.










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      up vote
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      up vote
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      I'm wondering what exactly is meant when people say "Skolemization preserves satisfiability but not validity". I'm having trouble wrapping my brain around it because I think of Skolemization, when considered as an inference rule, to simply "be an okay thing to do" when working with a second-order formula.



      Skolemization is said to preserve satisfiability but not validity. What exactly do satisfiability and validity mean in the context? Is there a way to think about what Skolemization is that makes it obvious that it doesn't preserve validity? Does "co-Skolemization" described below have the same property?




      Skolemization, as far as I can tell, is based on the equivalence between the first-order formula (and therefore degenerate second-order formula)



      $$ forall x_1 cdots x_n . exists y . P(x_1, cdots, x_n, y) $$



      and the following second order formula, that moves the $exists$ in front of the $forall$.



      $$ exists f. forall x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n))$$



      It makes perfect sense when the expression is thought of as a game. In the first expression, the opponent first makes $n$ moves and then the player makes 1. In the second expression, the player moves first but commits to a response function of sorts rather than committing to a single move. That's not a proof, but it seems like good intuitive justification for why rewriting an expression like this is okay in the first place.



      It seems to me that performing this translation at all only makes sense if we're talking about a closed expression. That is, $P(cdots)$ must have no free variables or, equivalently, $P$ must be an $n+1$-place predicate (primitive or defined).




      In propositional logic, satisfiability and validity can be though of by describing what happens to free variables. All variables are free because there's no way to bind them.



      $ P lor P $ is satisfiable because the assignment $ P = top $ satisfies is.



      $ P lor lnot P $ is a tautology for obvious reasons. I don't know whether it's right to say $ P lor lnot P $ is valid.



      An inference rule that maps $P$ to $P lor Q$ is valid (disjunction introduction).



      A rule that maps $P$ to $P land Q$ preserves satisfiability. We can take any old assignment $mu$ and construct $mu cup Q = top $ which satisfies the new expression.



      In the context of Skolemization, are the meanings of satisfiability and validity similar?




      Let "co-Skolemization" (probably has a real name) be the translation from



      $$ exists x_1 cdots x_n . forall y. P(x_1, cdots, x_n, y) $$



      to



      $$ forall f exists x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n)) $$



      That seems just as legitimate as Skolemization, considered as an inference rule in a second-order setting. There's a caveat that in a first order theory definitions are provided for all fully saturated combinations of function symbols and arguments, so therefore
      co-Skolemization" definitely takes us out of first-order land, but I don't know how relevant that is.










      share|cite|improve this question













      I'm wondering what exactly is meant when people say "Skolemization preserves satisfiability but not validity". I'm having trouble wrapping my brain around it because I think of Skolemization, when considered as an inference rule, to simply "be an okay thing to do" when working with a second-order formula.



      Skolemization is said to preserve satisfiability but not validity. What exactly do satisfiability and validity mean in the context? Is there a way to think about what Skolemization is that makes it obvious that it doesn't preserve validity? Does "co-Skolemization" described below have the same property?




      Skolemization, as far as I can tell, is based on the equivalence between the first-order formula (and therefore degenerate second-order formula)



      $$ forall x_1 cdots x_n . exists y . P(x_1, cdots, x_n, y) $$



      and the following second order formula, that moves the $exists$ in front of the $forall$.



      $$ exists f. forall x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n))$$



      It makes perfect sense when the expression is thought of as a game. In the first expression, the opponent first makes $n$ moves and then the player makes 1. In the second expression, the player moves first but commits to a response function of sorts rather than committing to a single move. That's not a proof, but it seems like good intuitive justification for why rewriting an expression like this is okay in the first place.



      It seems to me that performing this translation at all only makes sense if we're talking about a closed expression. That is, $P(cdots)$ must have no free variables or, equivalently, $P$ must be an $n+1$-place predicate (primitive or defined).




      In propositional logic, satisfiability and validity can be though of by describing what happens to free variables. All variables are free because there's no way to bind them.



      $ P lor P $ is satisfiable because the assignment $ P = top $ satisfies is.



      $ P lor lnot P $ is a tautology for obvious reasons. I don't know whether it's right to say $ P lor lnot P $ is valid.



      An inference rule that maps $P$ to $P lor Q$ is valid (disjunction introduction).



      A rule that maps $P$ to $P land Q$ preserves satisfiability. We can take any old assignment $mu$ and construct $mu cup Q = top $ which satisfies the new expression.



      In the context of Skolemization, are the meanings of satisfiability and validity similar?




      Let "co-Skolemization" (probably has a real name) be the translation from



      $$ exists x_1 cdots x_n . forall y. P(x_1, cdots, x_n, y) $$



      to



      $$ forall f exists x_1 cdots x_n . P(x_1, cdots, x_n, f(x_1, cdots, x_n)) $$



      That seems just as legitimate as Skolemization, considered as an inference rule in a second-order setting. There's a caveat that in a first order theory definitions are provided for all fully saturated combinations of function symbols and arguments, so therefore
      co-Skolemization" definitely takes us out of first-order land, but I don't know how relevant that is.







      logic






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      asked 1 hour ago









      Gregory Nisbet

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          Consider the formula



          $$exists x (P(a) lor neg P(x))$$



          This is a valid formula, since in any domain there is always an object that will satisfy the formula $P(a) lor neg P(x)$, namely whatever object $a$ refers to.



          However, if we Skolemize, we get:



          $$P(a) lor neg P(b)$$



          which is not a valid formula.






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            When you Skolemize, you drop the existential quantifier in front of $f$. Every structure that is a model of the original formula can be extended by providing an interpretation of $f$. Such an extension may or may not be a model of the Skolemized formula. Even though the original formula may be valid, the Skolemized formula doesn't have to be. What is guaranteed is that the two formulae are equisatisfiable.






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              Consider the formula



              $$exists x (P(a) lor neg P(x))$$



              This is a valid formula, since in any domain there is always an object that will satisfy the formula $P(a) lor neg P(x)$, namely whatever object $a$ refers to.



              However, if we Skolemize, we get:



              $$P(a) lor neg P(b)$$



              which is not a valid formula.






              share|cite|improve this answer
























                up vote
                3
                down vote













                Consider the formula



                $$exists x (P(a) lor neg P(x))$$



                This is a valid formula, since in any domain there is always an object that will satisfy the formula $P(a) lor neg P(x)$, namely whatever object $a$ refers to.



                However, if we Skolemize, we get:



                $$P(a) lor neg P(b)$$



                which is not a valid formula.






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                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Consider the formula



                  $$exists x (P(a) lor neg P(x))$$



                  This is a valid formula, since in any domain there is always an object that will satisfy the formula $P(a) lor neg P(x)$, namely whatever object $a$ refers to.



                  However, if we Skolemize, we get:



                  $$P(a) lor neg P(b)$$



                  which is not a valid formula.






                  share|cite|improve this answer












                  Consider the formula



                  $$exists x (P(a) lor neg P(x))$$



                  This is a valid formula, since in any domain there is always an object that will satisfy the formula $P(a) lor neg P(x)$, namely whatever object $a$ refers to.



                  However, if we Skolemize, we get:



                  $$P(a) lor neg P(b)$$



                  which is not a valid formula.







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                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 41 mins ago









                  Bram28

                  55.8k33982




                  55.8k33982




















                      up vote
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                      down vote













                      When you Skolemize, you drop the existential quantifier in front of $f$. Every structure that is a model of the original formula can be extended by providing an interpretation of $f$. Such an extension may or may not be a model of the Skolemized formula. Even though the original formula may be valid, the Skolemized formula doesn't have to be. What is guaranteed is that the two formulae are equisatisfiable.






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        When you Skolemize, you drop the existential quantifier in front of $f$. Every structure that is a model of the original formula can be extended by providing an interpretation of $f$. Such an extension may or may not be a model of the Skolemized formula. Even though the original formula may be valid, the Skolemized formula doesn't have to be. What is guaranteed is that the two formulae are equisatisfiable.






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          When you Skolemize, you drop the existential quantifier in front of $f$. Every structure that is a model of the original formula can be extended by providing an interpretation of $f$. Such an extension may or may not be a model of the Skolemized formula. Even though the original formula may be valid, the Skolemized formula doesn't have to be. What is guaranteed is that the two formulae are equisatisfiable.






                          share|cite|improve this answer












                          When you Skolemize, you drop the existential quantifier in front of $f$. Every structure that is a model of the original formula can be extended by providing an interpretation of $f$. Such an extension may or may not be a model of the Skolemized formula. Even though the original formula may be valid, the Skolemized formula doesn't have to be. What is guaranteed is that the two formulae are equisatisfiable.







                          share|cite|improve this answer












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                          answered 41 mins ago









                          Fabio Somenzi

                          6,06721221




                          6,06721221



























                               

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