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
|- ( ph -> A = B )
Assertion eqcomd
|- ( ph -> B = A )

Proof

Step Hyp Ref Expression
1 eqcomd.1
 |-  ( ph -> A = B )
2 eqid
 |-  A = A
3 1 eqeq1d
 |-  ( ph -> ( A = A <-> B = A ) )
4 2 3 mpbii
 |-  ( ph -> B = A )