Theorem baibd 935

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 301	. . 3 ⊢ (𝜒 → (𝜃 ↔ (𝜒 ∧ 𝜃)))
3	2	bicomd 141	. 2 ⊢ (𝜒 → ((𝜒 ∧ 𝜃) ↔ 𝜃))
4	1, 3	sylan9bb 466	1 ⊢ ((𝜑 ∧ 𝜒) → (𝜓 ↔ 𝜃))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by:  pw2f1odclem  7134  2omap  7318  eluz  9944  elicc4  10352  s111  11413  divalgmodcl  12711  eqglact  14077  eqgid  14078  iscrng2  14368  issubrg3  14604  iscld2  15254  cncnp2m  15381  cnnei  15382  reopnap  15696  cnlimc  15822  pw1map  17123