Theorem baibd

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

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

Proof

Step	Hyp	Ref	Expression
1		baibd.1	⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) )
2		ibar	⊢ ( 𝜒 → ( 𝜃 ↔ ( 𝜒 ∧ 𝜃 ) ) )
3	2	bicomd	⊢ ( 𝜒 → ( ( 𝜒 ∧ 𝜃 ) ↔ 𝜃 ) )
4	1 3	sylan9bb	⊢ ( ( 𝜑 ∧ 𝜒 ) → ( 𝜓 ↔ 𝜃 ) )