Theorem rbaibd

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

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

Proof

Step	Hyp	Ref	Expression
1		baibd.1	⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) )
2	1	biancomd	⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜃 ∧ 𝜒 ) ) )
3	2	baibd	⊢ ( ( 𝜑 ∧ 𝜃 ) → ( 𝜓 ↔ 𝜒 ) )