Theorem rbaibd 949

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
rbaibd	⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

Proof of Theorem rbaibd

Step	Hyp	Ref	Expression
1		baibd.1	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))
2		iba 524	. . 3 ⊢ (𝜃 → (𝜒 ↔ (𝜒 ∧ 𝜃)))
3	2	bicomd 213	. 2 ⊢ (𝜃 → ((𝜒 ∧ 𝜃) ↔ 𝜒))
4	1, 3	sylan9bb 736	1 ⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-an 386

This theorem is referenced by:  qsqueeze  12032  o1lo12  14269  incexc2  14570  gexdvds  17999  fsumvma  24938