Theorem baib

Description: Move conjunction outside of biconditional. (Contributed by NM, 13-May-1999)

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

Proof

Step	Hyp	Ref	Expression
1		baib.1	⊢ ( 𝜑 ↔ ( 𝜓 ∧ 𝜒 ) )
2		ibar	⊢ ( 𝜓 → ( 𝜒 ↔ ( 𝜓 ∧ 𝜒 ) ) )
3	2 1	syl6rbbr	⊢ ( 𝜓 → ( 𝜑 ↔ 𝜒 ) )