Theorem rbaibd 550

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	. . 3 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))
2	1	biancomd 469	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜃 ∧ 𝜒)))
3	2	baibd 549	1 ⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401

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

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by:  qsqueeze  13331  o1lo12  15705  incexc2  16007  gexdvds  19798  ssdifidlprm  21642  fsumvma  27540  subsdrg  33860  qusker  33910  0funclem  50193