Theorem bicom1 130

Description: Commutative law for equivalence. (Contributed by Wolf Lammen, 10-Nov-2012.)

Assertion

Ref	Expression
bicom1	⊢ ((𝜑 ↔ 𝜓) → (𝜓 ↔ 𝜑))

Proof of Theorem bicom1

Step	Hyp	Ref	Expression
1		bi2 129	. 2 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 → 𝜑))
2		bi1 117	. 2 ⊢ ((𝜑 ↔ 𝜓) → (𝜑 → 𝜓))
3	1, 2	impbid 128	1 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  bicomi  131  bicom  139  pm5.21ndd  694  cbvexdh  1896  elabgf2  12976