# Truth values give unique answer

I wanted to ask the proof for uniqueness of answer given by truth tables. I am reading Kleene's "Introduction to Metamathematics" Chapter 6 Section 28 on evaluation and consistency. There he introduces truth tables for operations and, implies, or and not. Then he defines the value of the propositional letter formula $$A$$ that has $$P_1, ..., P_m$$ as distinct propositional letters with chosen values of $$t$$ or $$f$$ for each of them as the result of applying truth tables to each pair of $$t$$'s and $$f$$'s. But nowhere does it say that such evaluation yields only one, unique result.

My attempt:

Proof by induction. From the definition of formula by Kleene there are always $$2n$$ parentheses in a propositional letter formula (which can also be proved by induction on $$n$$). We want to prove that for each formula $$A$$ having $$2n$$ parentheses it is true that the evaluation procedure yields a unique answer, and that is true for all $$n$$, and therefore for all formulas.

Basis: $$n=0$$. Then, the formulas have the form $$P$$ where $$P$$ is some propositional letter. Then, if $$P$$ is $$t$$ then the formula has value $$t$$, and if $$P$$ is $$f$$ then the formula has value $$f$$. So, in all cases the result is unique.

Step: Assume that it is true for $$n=k$$. Then, consider formula $$A$$ which has $$2k+2$$ parentheses. It means that it was formed from $$M$$ and $$N$$ by an operator implies, and,or or not. Both $$M$$ and $$N$$ each have at most $$2k$$ parentheses, which means that by induction hypothesis their truth value is unique. Then, because truth tables give unique answer given two values, then the formula $$A$$ has its truth value determined uniquely.

Is this proof correct?

I would appreciate any advice and suggestions!

In your step: When the $$\neg$$ operator is the main operator of formula $$A$$, then it is not built up from two formulas $$M$$ and $$N$$ plus one operator.
Now, that issue seems easily repaired by saying that $$A$$ can also be built up from one smaller formula $$M$$ and any unary operator like the $$\neg$$.
However, you could still be in trouble: if your syntax is defined in such a way that the result of applying a binary operator will result in an expression that starts and ends with parentheses (that is, if your syntax is such that $$A \lor B$$ is not synatactically correct, for it demands it to be $$(A \lor B)$$), then note that the formula $$\neg (A \lor B)$$ is built up from formula $$(A \lor B)$$ and the $$\neg$$ operator, and note that $$(A \lor B)$$ does not have less parentheses than $$\neg (A \lor B)$$.
So, I would suggest that in your proof you deal with the $$\neg$$ case separately. Otherwise, it works.