Theorem: Substitution for Equivalent Subformula is Equivalent

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 $\mathbf B' = \map {\mathbf B} {\mathbf A \,//\, \mathbf A'}$ 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 $\map v {\mathbf A} = \map v {\mathbf A'}$.
It is to be shown that $\map v {\mathbf B} = \map v {\mathbf B'}$.
We proceed by induction.
Let $\map n {\mathbf B}$ 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 $\map n {\mathbf B} \ne 0$, for $\mathbf C = \mathbf A$ is a valid choice.
Suppose now that $\map n {\mathbf B} = 1$.
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:
:$\map v {\mathbf B} = \map v {\mathbf B'}$
Suppose now that the assertion is true for all $\mathbf B$ with $\map n {\mathbf B} \le n$.
Let $\map n {\mathbf B} = n + 1$.
Suppose $\mathbf B = \neg \mathbf B_1$.
Then obviously $\map n {\mathbf B_1} = n$, so by hypothesis:
:$\map v {\mathbf B_1} = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }$
Also, by definition of substitution:
:$\mathbf B' = \neg \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'}$
Now, by definition of boolean interpretation:
{{begin-eqn}}
{{eqn | l = \map v {\mathbf B} 
      | r = \map {f^\neg} {\map v {\mathbf B_1} }
}}
{{eqn | r = \map {f^\neg} {\map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} } }
}}
{{eqn | r = \map v {\neg \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }
}}
{{eqn | r = \map v {\mathbf B'}
}}
{{end-eqn}}
Suppose now that $\mathbf B = \mathbf B_1 \mathbin{\mathsf B} \mathbf B_2$ for a binary connective $\mathsf B$.
Then $\map n {\mathbf B_1}, \map n {\mathbf B_2} \le n$, so:
:$\map v {\mathbf B_1} = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }$
:$\map v {\mathbf B_2} = \map v {\map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} }$
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:
:$\mathbf B' = \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} \mathbin {\mathsf B} \map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'}$
Now, by definition of boolean interpretation:
{{begin-eqn}}
{{eqn | l = \map v {\mathbf B}
      | r = \map {f^{\mathsf B} } {\map v {\mathbf B_1}, \map v {\mathbf B_2} }
}}
{{eqn | r = \map {f^{\mathsf B} } {\map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }, \map v {\map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} } }
}}
{{eqn | r = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} \mathbin {\mathsf B} \map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} }
}}
{{eqn | r = \map v {\mathbf B'}
}}
{{end-eqn}}
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.
{{qed}}