Theorem baibr 932

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 931	. 2 ⊢ (𝜓 → (𝜑 ↔ 𝜒))
3	2	bicomd 141	1 ⊢ (𝜓 → (𝜒 ↔ 𝜑))

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:  rbaibr  934  pm5.44  937  exmoeu2  2135  r19.9rmv  3616  dfopg  3897  brinxp  4838  infidc  7238  elioo5  10314  prmind2  12876  eulerthlemfi  12984  phisum  12997  pcelnn  13078  bj-charfundcALT  16749