Theorem baibr 915

Description: Move conjunction outside of biconditional. (Contributed by NM, 11-Jul-1994.)

Hypothesis

Ref	Expression
baib.1	⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))

Assertion

Ref	Expression
baibr	⊢ (𝜓 → (𝜒 ↔ 𝜑))

Proof of Theorem baibr

Step	Hyp	Ref	Expression
1		baib.1	. . 3 ⊢ (𝜑 ↔ (𝜓 ∧ 𝜒))
2	1	baib 914	. 2 ⊢ (𝜓 → (𝜑 ↔ 𝜒))
3	2	bicomd 140	1 ⊢ (𝜓 → (𝜒 ↔ 𝜑))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  rbaibr  917  pm5.44  920  exmoeu2  2067  r19.9rmv  3506  dfopg  3763  brinxp  4679  elioo5  9890  prmind2  12074  eulerthlemfi  12182  pcelnn  12274  bj-charfundcALT  13844