Theorem baibr 828

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 827	. 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  830  pm5.44  833  exmoeu2  1945  ssnelpss  3283  r19.9rmv  3307  dfopg  3538  brinxp  4351  elioo5  8572