Let a, b, c, and d be propositions. Assume that the equivalences a ↔ (b ∨ ¬b)…

The expression b ∨ ¬b is a tautology, so a ↔ (b ∨ ¬b) forces a to be true.

The condition b ↔ c means b and c always have the same truth value.

Now simplify the formula: ((a ∧ b) → ((a ∧ c) ∨ d)). Since a is true, this becomes b → (c ∨ d). Since c has the same truth value as b, it becomes b → (b ∨ d).

If b is true, then b ∨ d is true. If b is false, the implication is true vacuously. Therefore the formula is always true.