Theorem baibr 507

Description: Move conjunction outside of biconditional.

Hypothesis

Ref	Expression
baibr.1	⊢ (φ ↔ (ψ ∧ χ))

Assertion

Ref	Expression
baibr	⊢ (ψ → (χ ↔ φ))

Proof of Theorem baibr

Step	Hyp	Ref	Expression
1		baibr.1	. . 3 ⊢ (φ ↔ (ψ ∧ χ))
2	1	baib 506	. 2 ⊢ (ψ → (φ ↔ χ))
3	2	bicomd 399	1 ⊢ (ψ → (χ ↔ φ))

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

This theorem is referenced by:  exmoeu2 1040  ssnelpss 1751  canth 2945  kmlem14 3593  iscard 3659  cvexchlem 5759

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