Theorem baib 506

Description: Move conjunction outside of biconditional.

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 487	. 2 ⊢ (ψ → (χ ↔ (ψ ∧ χ)))
2		baib.1	. 2 ⊢ (φ ↔ (ψ ∧ χ))
3	1, 2	syl6rbbr 417	1 ⊢ (ψ → (φ ↔ χ))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196

This theorem is referenced by:  baibr 507  nlimon 2369  ltresr 4052  leltt 4278  nn0subt 4587  znnsubt 4595  seqcauchy 5106  sh2 5126  dmdbr2 5733

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198

metamath.org