Metamath Proof Explorer


Theorem bicomd

Description: Commute two sides of a biconditional in a deduction. (Contributed by NM, 14-May-1993)

Ref Expression
Hypothesis bicomd.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion bicomd ( 𝜑 → ( 𝜒 ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 bicomd.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2 bicom ⊢ ( ( 𝜓 ↔ 𝜒 ) ↔ ( 𝜒 ↔ 𝜓 ) )
3 1 2 sylib ⊢ ( 𝜑 → ( 𝜒 ↔ 𝜓 ) )