Boolean Interpretation is Well-Defined/Proof 2

Proof
This is to be done by strong induction on the length of WFFs.

By definition of $v$ being a boolean interpretation, $v \left({p}\right)$ is well-defined for all $p \in \mathcal P_0$, the vocabulary of $\mathcal L_0$.

A WFF of length $1$ has (trivially) a unique parsing sequence.

Consequently, only a single defining rule for $v$ as a boolean interpretation applies.

So the result holds for all WFFs of length $1$.

Now, suppose the result is true for all WFFs of length $k$ or less.

Let $\mathbf A$ be a WFF of length $k+1$.

There are two possibilities:

Suppose $\mathbf A = \neg \mathbf B$ for some WFF $\mathbf B$.

Then $\mathbf B$ is of length $k$, so by the induction hypothesis has a unique value $v (\mathbf B)$ no matter what parsing sequence is used.

So as $v (\mathbf A) = f^\neg \left({ v (\mathbf B) }\right)$, it follows that $\mathbf A$ likewise has a unique value.

Hence the result holds for $k+1$ in this situation.

Suppose $\mathbf A = \left({\mathbf B * \mathbf C}\right)$ for some WFFs $\mathbf B$ and $\mathbf C$ and some connective $*$.

By the Unique Readability Theorem of Propositional Logic, $*$ must be the main connective.

So $\mathbf B$ and $\mathbf C$ are both WFFs shorter than $k+1$ and therefore by the induction hypothesis have unique values $v (\mathbf B)$ and $v (\mathbf C)$.

Since $\mathbf A = \left({\mathbf B * \mathbf C}\right)$ for a unique main connective $*$, it follows that:


 * $v (\mathbf A) = f^* \left({ v (\mathbf B), v (\mathbf C) }\right)$

is well-defined.

Hence the result holds for $k+1$ in this situation.

So the result follows by the Second Principle of Mathematical Induction.