Theorem baibd 931

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  7087  2omap  7269  eluz  9867  elicc4  10273  s111  11319  divalgmodcl  12614  eqglact  13942  eqgid  13943  iscrng2  14159  issubrg3  14392  iscld2  14969  cncnp2m  15096  cnnei  15097  reopnap  15411  cnlimc  15537  pw1map  16769