Theorem rbaibd 925

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015.)

Hypothesis

Ref	Expression
baibd.1	⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))

Assertion

Ref	Expression
rbaibd	⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

Proof of Theorem rbaibd

Step	Hyp	Ref	Expression
1		baibd.1	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))
2		iba 300	. . 3 ⊢ (𝜃 → (𝜒 ↔ (𝜒 ∧ 𝜃)))
3	2	bicomd 141	. 2 ⊢ (𝜃 → ((𝜒 ∧ 𝜃) ↔ 𝜒))
4	1, 3	sylan9bb 462	1 ⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by: (None)