(𝑝 → 𝑞) ↔ (𝑞 ∨ ~𝑝)

Let us construct the trush table for the formula $(𝑝 → 𝑞) ↔ (𝑞 ∨ \sim 𝑝):$

$\begin{array}{||c|c||c|c|c|c||} \hline\hline p & q & 𝑝 → 𝑞 & \sim 𝑝 & 𝑞 ∨ \sim 𝑝 & (𝑝 → 𝑞) ↔ (𝑞 ∨ \sim 𝑝)\\ \hline\hline 0 & 0 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 0 & 1 &1\\ \hline\hline \end{array}$

We conclude that this formula is tautology.