Theorem baibr 862

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 861	. 2 ⊢ (𝜓 → (𝜑 ↔ 𝜒))
3	2	bicomd 139	1 ⊢ (𝜓 → (𝜒 ↔ 𝜑))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  rbaibr  864  pm5.44  867  exmoeu2  1989  r19.9rmv  3333  dfopg  3568  brinxp  4426  elioo5  8956  prmind2  10502