Theorem baibr 928

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 927	. 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  930  pm5.44  933  exmoeu2  2128  r19.9rmv  3588  dfopg  3865  brinxp  4800  infidc  7176  elioo5  10212  prmind2  12755  eulerthlemfi  12863  phisum  12876  pcelnn  12957  bj-charfundcALT  16508