Theorem baibd 925

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 462	1 ⊢ ((𝜑 ∧ 𝜒) → (𝜓 ↔ 𝜃))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  pw2f1odclem  6946  eluz  9681  elicc4  10082  s111  11108  divalgmodcl  12314  eqglact  13636  eqgid  13637  iscrng2  13852  issubrg3  14084  iscld2  14651  cncnp2m  14778  cnnei  14779  reopnap  15093  cnlimc  15219  2omap  16071  pw1map  16073