Metamath Proof Explorer


Theorem eqcomd

Description: Deduction from commutative law for class equality. (Contributed by NM, 15-Aug-1994) Allow shortening of eqcom . (Revised by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis eqcomd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion eqcomd ( 𝜑 → 𝐵 = 𝐴 )

Proof

Step Hyp Ref Expression
1 eqcomd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 eqid ⊢ 𝐴 = 𝐴
3 1 eqeq1d ⊢ ( 𝜑 → ( 𝐴 = 𝐴 ↔ 𝐵 = 𝐴 ) )
4 2 3 mpbii ⊢ ( 𝜑 → 𝐵 = 𝐴 )