Metamath Proof Explorer


Theorem bicom1

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

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

Proof

Step Hyp Ref Expression
1 biimpr ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜓 → 𝜑 ) )
2 biimp ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜑 → 𝜓 ) )
3 1 2 impbid ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜓 ↔ 𝜑 ) )