Deduction Theorem and Closed Formulas

Question

Why in the Deduction Theorem do we require a closed formula?

Deduction Theorem. Let $A$ be a closed formula in $T$. For every formula $B$ of $T$, $\vdash_T A \implies B$ iff $B$ is a theorem of $T[A]$.

I could not find any counterexample.

Can you explain me where is the problem?

Edit:

I found a counterexample.

if $A=C$ and $B=\forall(x)C$

when A is not a closed formula.

There are many (essentially equivalent)ways to define a deduction in first-order logic. Some ways use only "closed" formulas (sentences). Some ways allow open formulas (formulas with free occurrences of variables. — André Nicolas, Dec 31 '14 at 21:09
I have seen versions of the Deduction Theorem in which "$A$" need not be closed. They are a little more complicated. — André Nicolas, Dec 31 '14 at 21:49
Your question is not clear. Could you edit it to make it clear just what you are asking? — Rory Daulton, Jan 01 '15 at 00:18
It is right, for $C$ with $x$ free. Consider $C(x) := (x=0)$; from it, by generalization we can derive $\forall x (x=0)$. But we cannot conclude with : $(x=0) \rightarrow \forall x (x=0)$ because it is not true [For $0$ as value for $x$ the antecedent is true, while the consequent is clearly false; thus the conditional is false]. — Mauro ALLEGRANZA, Jan 01 '15 at 10:45
Also see this post: http://math.stackexchange.com/a/580692/630 — Carl Mummert, Jan 01 '15 at 12:47

score 3 · Answer 1 · answered Jan 01 '15 at 11:28

In the system of Joseph Shoenfield, Mathematical Logic (1967) the restriction on $A$ being closed is needed in the proof of the Deduction Theorem [see page 33] in order to apply the $\exists$-introduction rule [se page 21] :

if $x$ is not free in $B$, infer $\exists x A \rightarrow B$ from $A \rightarrow B$.

Without the proviso on $B$ in the rule, we can derive the invalid $\exists x (x=0) \rightarrow (x=0)$ from the tautology : $(x=0) \rightarrow (x=0)$.

Deduction Theorem and Closed Formulas

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