Prove logical equivalence of formulas with following properties Let $P$ be the set of all propositional variables and  $v : P \rightarrow \{ \top, \bot \}$ be a valuation such that $v(p) = \top$, for every $p \in P$. Prove that if $A \in \mathcal{Form}$ and $B \in \mathcal{Form(\{ \wedge, \Rightarrow\})}$ then $A$ is logically equivalent to $B$ if and only if $\tilde{v}(A) = \top$.
So, what I could prove is that if $A$ and $B$ are logically equivalent, then $B \vDash A$. It is not hard to prove that $\tilde{v}(B) = \top$ (using induction by complexity of formula $B$), so $\tilde{v}(A) = \top$, also.
However, I could not find a connection between the given hypotheses for the other direction. Every help is appreciated. 
 A: The converse is false. Let $B$ be $x_1\to x_2$ and $A$ be $x_1\lor x_2.$ Then $\tilde v(A) = \top,$ but $A$ is not logically equivalent to $B.$ What is probably intended is to show that a formula $A$ is equivalent to some $B$ that only involves the connectives $\to $ and $\land$ if and only if $\tilde v(A) = \top.$
Your proof still goes through for the one direction. For the converse, we can use the easy-to-see fact that if $\tilde v(A) = \top,$ then the conjunctive normal form of $A$ must have at least one un-negated disjunct in each clause. Then, consider a generic clause satisfying this requirement, $$ x\lor x_1\lor \ldots \lor x_n \lor \lnot y_1\lor \ldots \lor \lnot y_n$$ where the $x$'s and $y$'s are propositional variables. 
Using Boolean algebra, we can find the equivalence $x\lor  x_1 = (x\to x_1)\to x_1$ and if we iterate this for all the $x_i$ we can express the clause as $$ \phi\lor \lnot y_1\lor \ldots \lor \lnot y_n$$ where $\phi$ only uses $\to.$ Again using Boolean algebra $\phi\lor \lnot y_1 = y_1\to\phi,$ so repeating this, we express the entire clause only using the $\to$ connective. So our CNF is equivalent to a conjunction of clauses only using $\to,$ so only uses $\land$ and $\to.$
