Verification of proof of propositional logic I made a proof for the following theorem. But I'm not completely certain that it's fully correct.

Suppose $\phi$ is a propositional formula and that the two evaluations $v$ and $w$ are equal for all atoms that appear in $\phi$. Then is $\overline v(\phi)$ = $\overline w(\phi)$.

Proof: We deliver a proof by induction on the length of $\phi$.
Induction basis: Suppose $l(\phi) = 1$. Then $\phi$ is an atom. Thus $\overline v(\phi) = \overline w(\phi)$.
Induction step: Suppose $l(\phi) > 1$. Then we must distinguish various cases:
1) Suppose $\phi = \phi_1 \wedge \phi_2$. Then $l(\phi_1) < l(\phi)$ and $l(\phi_2) < l(\phi)$. By induction we find that $\overline v(\phi_1) = \overline w(\phi_1)$ and $\overline v(\phi_2) = \overline w(\phi_2)$. Successively we find that $\overline v(\phi) = \overline v(\phi_1 \wedge \phi_2) \equiv \overline v(\phi_1) \wedge \overline v(\phi_2) \equiv \overline w(\phi_1) \wedge \overline w(\phi_2) \equiv \overline w(\phi_1 \wedge \phi_2) \equiv \overline w(\phi)$.
We find the other cases completely analog to 1).
Now I'm not sure whether it's true that $\overline v(\phi) = \overline w(\phi)$ in my induction basis? And if it's legal to write $\overline v(\phi) = \overline v(\phi_1 \wedge \phi_2)$?
 A: It's true, and it's legal — respectively. 
There's no need to be uneasy about the base case:
$$\overline{v}(\phi) = v(\phi) = w(\phi) = \overline{w}(\phi).$$
However, in the chain of identities for the induction step, there is one concern, though actually it's not the one you ask about. If $\phi = \phi_1\land\phi_2$, then of course $\overline{v}(\phi) = \overline{v}(\phi_1\land\phi_2)$ — no worry there. Here's the worry: it's at best confusing and at worst just incorrect to write "$\overline{v}(\phi_1\land\phi_2) = \overline{v}(\phi_1)\land\overline{v}(\phi_2)$".
The $\land$ on the righthand side should be another symbol or, for example, $\mathsf{and}$, because there, you're speaking in the metalanguage whereas $\land$ is a symbol of the language. On the righthand side, you're combining values of $\overline{v}$, which are truth values, so you must combine them with an operator on truth values. On the lefthand side, $\land$, like $\lor$ and $\neg$, is a connective, a mere symbol. It's important to distinguish between these domains and levels of discourse. Thus, the identity should be 
$$
\overline{v}(\phi_1\land\phi_2) = \overline{v}(\phi_1) \operatorname{\mathsf{and}} \overline{v}(\phi_2)
$$
or something like it. You might use $\&$, for example, instead of $\operatorname{\mathsf{and}}$; but it's best not to overload $\land$.
