Theorem biadan 815

Description: An implication is equivalent to the equivalence of some implied equivalence and some other equivalence involving a conjunction. A utility lemma as illustrated in biadanii 818 and elelb 35009. (Contributed by BJ, 4-Mar-2023.) (Proof shortened by Wolf Lammen, 8-Mar-2023.)

Assertion

Ref	Expression
biadan	⊢ ((𝜑 → 𝜓) ↔ ((𝜓 → (𝜑 ↔ 𝜒)) ↔ (𝜑 ↔ (𝜓 ∧ 𝜒))))

Proof of Theorem biadan

Step	Hyp	Ref	Expression
1		pm4.71r 558	. 2 ⊢ ((𝜑 → 𝜓) ↔ (𝜑 ↔ (𝜓 ∧ 𝜑)))
2		bicom 221	. 2 ⊢ ((𝜑 ↔ (𝜓 ∧ 𝜑)) ↔ ((𝜓 ∧ 𝜑) ↔ 𝜑))
3		bicom 221	. . . 4 ⊢ ((𝜑 ↔ (𝜓 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜒) ↔ 𝜑))
4		pm5.32 573	. . . 4 ⊢ ((𝜓 → (𝜑 ↔ 𝜒)) ↔ ((𝜓 ∧ 𝜑) ↔ (𝜓 ∧ 𝜒)))
5	3, 4	bibi12i 339	. . 3 ⊢ (((𝜑 ↔ (𝜓 ∧ 𝜒)) ↔ (𝜓 → (𝜑 ↔ 𝜒))) ↔ (((𝜓 ∧ 𝜒) ↔ 𝜑) ↔ ((𝜓 ∧ 𝜑) ↔ (𝜓 ∧ 𝜒))))
6		bicom 221	. . 3 ⊢ (((𝜓 → (𝜑 ↔ 𝜒)) ↔ (𝜑 ↔ (𝜓 ∧ 𝜒))) ↔ ((𝜑 ↔ (𝜓 ∧ 𝜒)) ↔ (𝜓 → (𝜑 ↔ 𝜒))))
7		biluk 386	. . 3 ⊢ (((𝜓 ∧ 𝜑) ↔ 𝜑) ↔ (((𝜓 ∧ 𝜒) ↔ 𝜑) ↔ ((𝜓 ∧ 𝜑) ↔ (𝜓 ∧ 𝜒))))
8	5, 6, 7	3bitr4ri 303	. 2 ⊢ (((𝜓 ∧ 𝜑) ↔ 𝜑) ↔ ((𝜓 → (𝜑 ↔ 𝜒)) ↔ (𝜑 ↔ (𝜓 ∧ 𝜒))))
9	1, 2, 8	3bitri 296	1 ⊢ ((𝜑 → 𝜓) ↔ ((𝜓 → (𝜑 ↔ 𝜒)) ↔ (𝜑 ↔ (𝜓 ∧ 𝜒))))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 206  df-an 396

This theorem is referenced by:  biadani  816