Metamath Proof Explorer


Theorem bicom

Description: Commutative law for the biconditional. Theorem *4.21 of WhiteheadRussell p. 117. (Contributed by NM, 11-May-1993)

Ref Expression
Assertion bicom ( ( 𝜑 ↔ 𝜓 ) ↔ ( 𝜓 ↔ 𝜑 ) )

Proof

Step Hyp Ref Expression
1 bicom1 ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜓 ↔ 𝜑 ) )
2 bicom1 ⊢ ( ( 𝜓 ↔ 𝜑 ) → ( 𝜑 ↔ 𝜓 ) )
3 1 2 impbii ⊢ ( ( 𝜑 ↔ 𝜓 ) ↔ ( 𝜓 ↔ 𝜑 ) )