Is it possible to prove (P→Q)↔[(P ∨ Q)↔Q] without using truth tables?

[Ogisto54]

Homework Statement

Need to demonstrate this proposition: (P→Q)↔[(P ∨ Q)↔Q] . My textbook use truth tables, but I'd like to do without it. It asks me if it's always truth

The Attempt at a Solution

Im unable to demonstrate the Tautology and obtain (¬Q) as solution.
I start by facing the right side in this way: [(PvQ) → Q ∧ Q → (PvQ)] and apply the same concept
with the other " ↔ " . Is this correct?

<Moderator's note: Type setting edited. Boldface is considered shouting.>

 

[fresh_42]

Why don't you proceed step by step. You want to start from right to left, so we have $(P \vee Q) \leftrightarrow Q$ as a given fact. Now what can be concluded from $P\,$? Can we deduce $Q\,$?

The same in the other direction. Here $P \rightarrow Q$ is given as a fact. Now we need to show that both directions of the right hand side are valid, first $\rightarrow$ and then $\leftarrow$ just with the help of the theorem $P \rightarrow Q$.