Theorem bicom1 129

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 128	. 2 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 → 𝜑))
2		bi1 116	. 2 ⊢ ((𝜑 ↔ 𝜓) → (𝜑 → 𝜓))
3	1, 2	impbid 127	1 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 ↔ 𝜑))

Syntax hints:   → wi 4   ↔ wb 103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  bicomi  130  bicom  138  pm5.21ndd  653  cbvexdh  1842  elabgf2  10590