Theorem rbaibr 912

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

Hypothesis

Ref	Expression
baib.1	⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))

Assertion

Ref	Expression
rbaibr	⊢ (𝜒 → (𝜓 ↔ 𝜑))

Proof of Theorem rbaibr

Step	Hyp	Ref	Expression
1		baib.1	. . 3 ⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))
2		ancom 264	. . 3 ⊢ ((𝜓 ∧ 𝜒) ↔ (𝜒 ∧ 𝜓))
3	1, 2	bitri 183	. 2 ⊢ (𝜑 ↔ (𝜒 ∧ 𝜓))
4	3	baibr 910	1 ⊢ (𝜒 → (𝜓 ↔ 𝜑))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by: (None)