Theorem baibd 928

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  6991  eluz  9731  elicc4  10132  s111  11159  divalgmodcl  12434  eqglact  13757  eqgid  13758  iscrng2  13973  issubrg3  14205  iscld2  14772  cncnp2m  14899  cnnei  14900  reopnap  15214  cnlimc  15340  2omap  16318  pw1map  16320