Theorem baibr

Description: Move conjunction outside of biconditional. (Contributed by NM, 11-Jul-1994)

	Ref	Expression
Hypothesis	baib.1	⊢ ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) )
Assertion	baibr	⊢ ( 𝜓 → ( 𝜒 ↔ 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		baib.1	⊢ ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) )
2	1	baib	⊢ ( 𝜓 → ( 𝜑 ↔ 𝜒 ) )
3	2	bicomd	⊢ ( 𝜓 → ( 𝜒 ↔ 𝜑 ) )