Theorem rbaibr

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015) (Proof shortened by Wolf Lammen, 19-Jan-2020)

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

Proof

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