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))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by:  rbaibr  934  pm5.44  937  exmoeu2  2135  r19.9rmv  3619  dfopg  3902  brinxp  4843  infidc  7248  elioo5  10346  prmind2  12917  eulerthlemfi  13029  phisum  13042  pcelnn  13123  prmorcht  16243  bj-charfundcALT  17001