Theorem baib 871

Description: Move conjunction outside of biconditional. (Contributed by NM, 13-May-1999.)

Hypothesis

Ref	Expression
baib.1	⊢ (φ ↔ (ψ ∧ χ))

Assertion

Ref	Expression
baib	⊢ (ψ → (φ ↔ χ))

Proof of Theorem baib

Step	Hyp	Ref	Expression
1		ibar 490	. 2 ⊢ (ψ → (χ ↔ (ψ ∧ χ)))
2		baib.1	. 2 ⊢ (φ ↔ (ψ ∧ χ))
3	1, 2	syl6rbbr 255	1 ⊢ (ψ → (φ ↔ χ))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-an 360

This theorem is referenced by:  baibr  872  rbaib  873  ceqsrexbv  2973  elrab3  2995  dfpss3  3355  rabsn  3790  elrint2  3968  fnres  5199  fvmpti  5699