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: {{begin-eqn}} {{eqn|l = v \left({\mathbf B}\right)       |r = f^\neg \left({v \left({\mathbf B_1}\right)}\right) }} {{eqn|r = f^\neg \left({v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right)}\right) }} {{eqn|r = v \left({\neg \mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right)}\right) }} {{eqn|r = v \left({\mathbf B'}\right) }} {{end-eqn}} 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: {{begin-eqn}} {{eqn|l = v \left({\mathbf B}\right)       |r = f^{\mathsf B} \left({v \left({\mathbf B_1}\right), v \left({\mathbf B_2}\right)}\right) }} {{eqn|r = 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) }} {{eqn|r = v \left({\mathbf B_1 \left({\mathbf A \,//\, \mathbf A'}\right) \mathbin{\mathsf B} \mathbf B_2 \left({\mathbf A \,//\, \mathbf A'}\right)}\right) }} {{eqn|r = v \left({\mathbf B'}\right) }} {{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}}