User:Lord Farin/Sandbox/Completeness/Replacement Tautology

Theorem
Let $\mathcal L$ be the language of propositional logic.

Let $\mathbf A, \mathbf A'$ be WFFs of $\mathcal L$.

Let $\mathbf B$ be another WFF.

Suppose that $\mathbf A \iff \mathbf A'$ is a tautology for boolean interpretations.

Then:


 * $\mathbf B \iff \mathbf B \paren{ \mathbf A' \mathbin{//} \mathbf A }$

is also a tautology, where $\mathbf B \paren{ \mathbf A' \mathbin{//} \mathbf A }$ denotes the substitution of $\mathbf A'$ for $\mathbf A$ in $\mathbf B$.

Proof
Since $\mathbf A \iff \mathbf A'$ is a tautology, we have for all boolean interpretations $v: \mathcal L \to \set{ T, F }$:


 * $v \paren{ \mathbf A \iff \mathbf A' } = T$

By definition of $\iff$ under boolean interpretations, we have that:


 * $v \paren{ \mathbf A } = v \paren{ \mathbf A' }$

Now if $\mathbf B = \mathbf A$, then $\mathbf B \paren{ \mathbf A' \mathbin{//} \mathbf A } = \mathbf A'$, and the result is immediate.

Next, proceed by the Principle of Structural Induction on $\mathbf B$.

If $\mathbf B = p$ for a propositional symbol $p$, then $\mathbf A$ does not occur in $\mathbf B$ and hence:


 * $\mathbf B \iff \mathbf B \paren{ \mathbf A' \mathbin{//} \mathbf A }$

reduces to:


 * $p \iff p$

which is a tautology by Biconditional is Reflexive.

If $\mathbf B = \mathbf B_1 \circ \mathbf B_2$ for $\circ \in \set{ \land, \lor, \implies, \impliedby, \iff }$, then for any boolean interpretation $v$:

hence $\mathbf B \iff \mathbf B \paren{ \mathbf A' \mathbin{//} \mathbf A }$.

Lastly if $\mathbf B = \neg \mathbf B_1$, then for any boolean interpretation $v$:

Hence the result by the Principle of Structural Induction.