Metamath Proof Explorer


Theorem biadaniALT

Description: Alternate proof of biadani not using biadan . (Contributed by BJ, 4-Mar-2023) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis biadani.1 ⊢ ( 𝜑 → 𝜓 )
Assertion biadaniALT ( ( 𝜓 → ( 𝜑 ↔ 𝜒 ) ) ↔ ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 biadani.1 ⊢ ( 𝜑 → 𝜓 )
2 pm5.32 ⊢ ( ( 𝜓 → ( 𝜑 ↔ 𝜒 ) ) ↔ ( ( 𝜓 ∧ 𝜑 ) ↔ ( 𝜓 ∧ 𝜒 ) ) )
3 1 pm4.71ri ⊢ ( 𝜑 ↔ ( 𝜓 ∧ 𝜑 ) )
4 3 bibi1i ⊢ ( ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) ) ↔ ( ( 𝜓 ∧ 𝜑 ) ↔ ( 𝜓 ∧ 𝜒 ) ) )
5 2 4 bitr4i ⊢ ( ( 𝜓 → ( 𝜑 ↔ 𝜒 ) ) ↔ ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) ) )