Theorem baibr 829

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 828	. 2 ⊢ (𝜓 → (𝜑 ↔ 𝜒))
3	2	bicomd 129	1 ⊢ (𝜓 → (𝜒 ↔ 𝜑))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  rbaibr  831  pm5.44  834  exmoeu2  1948  ssnelpss  3289  r19.9rmv  3313  dfopg  3547  brinxp  4408  elioo5  8802