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For clarity, let me take two examples: Shoenfield's Mathematical Logic, and Enderton's A Mathematical Introduction To Logic. Well, in those texts we have two different Deduction Theorems (we are in the context of predicate logic):

In Shoenfield's presentation:

  1. If $\phi$ is a sentence (closed formula), then $\Gamma, \phi \vdash \psi $ implies $\Gamma\vdash\phi\rightarrow\psi$.

While in Enderton's presentation, the hypotesis on $\phi$ is not necessary:

  1. For any formula $\phi$, we have that $\Gamma, \phi \vdash\psi$ implies $\Gamma\vdash\phi\rightarrow\psi$.

Now, what I can get is that this depends on the choice of the provability relation " $\vdash$ " choosen, i.e. on the calculus that we decide to use. Again, for clarity, let briefly explain the two different calculi:

Shoenfield's calculus:

Logical axioms:

  • every formula of the form A $\lor$ $\lnot$ A ;
  • every formula of the form A(x) $\rightarrow \exists$xA ;
  • the usual identity axioms.

Rules of inference:

  • [$\lor$] from A $\lor$ B infer A; from A $\lor$ A infer A; from A $\lor$ (B $\lor$ C) infer (A $\lor$ B) $\lor$ C ;
  • [Cut] from A $\lor$ B and $\lnot$ A $\lor$ C infer B $\lor$ C;
  • [$\exists$] if x is not free in B, from A $\rightarrow$ B infer $\exists$xA $\rightarrow$ B.

Enderton's calculus:

Here we have just a rule of inference, Modus Ponens, and we have the following logical axioms:

  • All propositional tautologies and usual axioms for the identity;
  • every formula of the form $\forall$x A $\rightarrow$ A[x\t] , for t substitutable for x in A;
  • every formula of the form $\forall$x (A $\rightarrow$ B) $\rightarrow$ ($\forall$x A $\rightarrow$ $\forall$x B) ;
  • every formula of the form A $\rightarrow$ $\forall$x A, for x not free in A.

Well, as Shoenfield observes, for his calculus we can obtain the following metatheorem: given A* instance of A (i.e. for some terms $a_1,...,a_n$ , A* is A[$x_1$ \ $a_1$,...,$x_n$ \ $a_n$]), $\Gamma \vdash $ A iff $\Gamma\vdash$ A*. This means, for example, that

PA $\cup$ {x=0} $\vdash$ x=0 implies PA $\cup$ {x=0} $\vdash$ y=0,

and then we can see why the Deduction Theorem can't work with open formula: otherwise, since it is truth that PA $\cup$ {x=0} $\vdash$ x=0, we would have PA $\vdash$ x=0 $\rightarrow$ y=0, wich could not be because x=0 $\rightarrow$ y=0 is not truth in $\mathbb{N}$ for all assignements to the variables.

So, I guess, the point (at least, one of the point) is that the above metatheorem is not available in Enderton's system, and so there we can have the 'liberalized' Deduction Theorem. But wich is the deep reason, in the choice of rules and axioms, for this difference?This is my first question.

My second question, then, is: why this doesen't create some kind of asymmetry between the two systems? For example, why this doesen't imply that there could be some formula $\phi$ wich is derivable from $\Gamma$ using Enderton's system, but wich is not derivable from $\Gamma$ using Shoenfield's system (or viceversa)?

Matteo __
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I don't have a copy of Shoenfield in front of me, but I think you will find that the semantics in the two books is different. Shoenfield treats a free variable in a premise or the conclusion of an argument as being implicitly universally quantified, whereas Enderton treats it as standing for a fixed but unspecified element of the universe of discourse. So for Shoenfield, to say that $PA \cup \{x=0\}$ logically implies $y=0$ means that if you have an interpretation in which the $PA$ axioms hold and everything is equal to 0, then in that interpretation, everything is equal to 0, which of course is correct. Whereas for Enderton, it means that if you have an interpretation, including assignments of particular values to the variables $x$ and $y$, in which the $PA$ axioms hold and $x$ is assigned the value 0, then $y$ is also assigned the value 0. And of course that is incorrect.

Dan Velleman
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