Theorem rbaibr 831

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 253	. . 3 ⊢ ((𝜓 ∧ 𝜒) ↔ (𝜒 ∧ 𝜓))
3	1, 2	bitri 173	. 2 ⊢ (𝜑 ↔ (𝜒 ∧ 𝜓))
4	3	baibr 829	1 ⊢ (𝜒 → (𝜓 ↔ 𝜑))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110

This theorem is referenced by: (None)