Deduction Theorem

Theorem
Let $\mathscr H$ be instance 1 of a Hilbert proof system.

Then the deduction rule:


 * $\dfrac{U,\mathbf A \vdash \mathbf B}{U \vdash \mathbf A \implies \mathbf B}$

is a derived rule for $\mathscr H$.

Proof
For any proof of $U, \mathbf A \vdash \mathbf B$, we indicate how to transform it into a proof of $U \vdash \mathbf A \implies \mathbf B$ without using the deduction rule.

This is done by applying the Second Principle of Mathematical Induction to the length $n$ of the proof of $U,\mathbf A \vdash \mathbf B$.

If $n = 1$, then one of the following must occur:


 * $\mathbf B \in U$
 * $\mathbf B = \mathbf A$
 * $\mathbf B$ is an axiom or a theorem of $\mathscr H$

In the first case, obviously $U \vdash \mathbf B$.

By Axiom 1, $U \vdash \mathbf B \implies \paren {\mathbf A \implies \mathbf B}$.

By Modus Ponens, $U \vdash \mathbf A \implies \mathbf B$.

In the second case, $U \vdash \mathbf A \implies \mathbf A$ by the Law of Identity.

Finally, in the third case, we have $U \vdash \mathbf B$.

As in the first case, we conclude $U \vdash \mathbf A \implies \mathbf B$.

If $n > 1$, the only other option for arriving at $U, \mathbf A \vdash \mathbf B$ is through Modus Ponens.

That is to say, two earlier lines of the proof contain:


 * $U, \mathbf A \vdash \mathbf C$
 * $U, \mathbf A \vdash \mathbf C \implies \mathbf B$

for some WFF $\mathbf C$.

But then these sequents have shorter proofs.

Hence, they satisfy the induction hypothesis.

Thus, we may infer:


 * $U \vdash \mathbf A \implies \mathbf C$
 * $U \vdash \mathbf A \implies \paren {\mathbf C \implies \mathbf B}$

This allows us to give the following proof of $U \vdash \mathbf A \implies \mathbf B$:

The result follows by the Second Principle of Mathematical Induction.