Metamath Proof Explorer


Theorem eqeq2d

Description: Deduction from equality to equivalence of equalities. (Contributed by NM, 27-Dec-1993) Allow shortening of eqeq2 . (Revised by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis eqeq2d.1
|- ( ph -> A = B )
Assertion eqeq2d
|- ( ph -> ( C = A <-> C = B ) )

Proof

Step Hyp Ref Expression
1 eqeq2d.1
 |-  ( ph -> A = B )
2 1 eqeq1d
 |-  ( ph -> ( A = C <-> B = C ) )
3 eqcom
 |-  ( C = A <-> A = C )
4 eqcom
 |-  ( C = B <-> B = C )
5 2 3 4 3bitr4g
 |-  ( ph -> ( C = A <-> C = B ) )