Theorem baibd 875

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

Hypothesis

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

Assertion

Ref	Expression
baibd	⊢ ((φ ∧ χ) → (ψ ↔ θ))

Proof of Theorem baibd

Step	Hyp	Ref	Expression
1		baibd.1	. 2 ⊢ (φ → (ψ ↔ (χ ∧ θ)))
2		ibar 490	. . 3 ⊢ (χ → (θ ↔ (χ ∧ θ)))
3	2	bicomd 192	. 2 ⊢ (χ → ((χ ∧ θ) ↔ θ))
4	1, 3	sylan9bb 680	1 ⊢ ((φ ∧ χ) → (ψ ↔ θ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by: (None)