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Metamath Proof Explorer

Theorem biadan

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 and elelb . (Contributed by BJ, 4-Mar-2023) (Proof shortened by Wolf Lammen, 8-Mar-2023)

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

Proof

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