Substitution for Equivalent Subformula is Equivalent

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Theorem

Let $\mathbf B$ a WFF of propositional logic.

Let $\mathbf A, \mathbf A'$ be equivalent WFFs.

Let $\mathbf A$ be a subformula of $\mathbf B$.

Let $\mathbf B' = \mathbf B \left({\mathbf A \,//\, \mathbf A'}\right)$ be the substitution of $\mathbf A'$ for $\mathbf A$ in $\mathbf B$.


Then $\mathbf B$ and $\mathbf B'$ are equivalent.


Proof

Let $v$ be an arbitrary boolean interpretation.

Then $v \left({\mathbf A}\right) = v \left({\mathbf A'}\right)$.


It is to be shown that $v \left({\mathbf B}\right) = v \left({\mathbf B'}\right)$.

We proceed by induction.

Let $n \left({\mathbf B}\right)$ be the number of WFFs $\mathbf C$ such that:

$\mathbf A$ is a subformula of $\mathbf C$, and $\mathbf C$ is a subformula of $\mathbf B$.

Note that $n \left({\mathbf B}\right) \ne 0$, for $\mathbf C = \mathbf A$ is a valid choice.


Suppose now that $n \left({\mathbf B}\right) = 1$.

Because we have the valid choices $\mathbf C = \mathbf A$ and $\mathbf C = \mathbf B$, it follows that these choices must be identical, i.e. $\mathbf A = \mathbf B$.

Hence $\mathbf B' = \mathbf A'$, and so:

$v \left({\mathbf B}\right) = v \left({\mathbf B'}\right)$


Suppose now that the assertion is true for all $\mathbf B$ with $n \left({\mathbf B}\right) \le n$.

Let $n \left({\mathbf B}\right) = n + 1$.


Suppose $\mathbf B = \neg \mathbf B_1$.

Then obviously $n \left({\mathbf B_1}\right) = n$, so by hypothesis:

$v \left({\mathbf B_1}\right) = v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)$

Also, by definition of substitution:

$\mathbf B' = \neg \mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)$

Now, by definition of boolean interpretation:

\(\displaystyle v \left({\mathbf B}\right)\) \(=\) \(\displaystyle f^\neg \left({v \left({\mathbf B_1}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f^\neg \left({v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle v \left({\neg \mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle v \left({\mathbf B'}\right)\)


Suppose now that $\mathbf B = \mathbf B_1 \mathbin{\mathsf B} \mathbf B_2$ for a binary connective $\mathsf B$.

Then $n \left({\mathbf B_1}\right), n \left({\mathbf B_2}\right) \le n$, so:

$v \left({\mathbf B_1}\right) = v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)$
$v \left({\mathbf B_2}\right) = v \left({\mathbf B_2 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)$

This follows by either the induction hypothesis, or when $\mathbf A$ is not a subformula of $\mathbf B_1$ or $\mathbf B_2$, is entirely trivial, considering the substitution does not change anything.

Also, by definition of substitution:

$\mathbf B' = \mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right) \mathbin{\mathsf B} \mathbf B_2 \left({\mathbf A \,//\, \mathbf A'}\right)$


Now, by definition of boolean interpretation:

\(\displaystyle v \left({\mathbf B}\right)\) \(=\) \(\displaystyle f^{\mathsf B} \left({v \left({\mathbf B_1}\right), v \left({\mathbf B_2}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f^{\mathsf B} \left({v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right), v \left({\mathbf B_2 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right) \mathbin{\mathsf B} \mathbf B_2 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle v \left({\mathbf B'}\right)\)


By definition of the language of propositional logic, $\mathbf B$ must have either of the above forms.

Hence the result, from the Second Principle of Mathematical Induction.

$\blacksquare$


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