Metamath Proof Explorer


Theorem eqeq1dALT

Description: Alternate proof of eqeq1d , shorter but requiring ax-12 . (Contributed by NM, 27-Dec-1993) (Revised by Wolf Lammen, 19-Nov-2019) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis eqeq1d.1 φ A = B
Assertion eqeq1dALT φ A = C B = C

Proof

Step Hyp Ref Expression
1 eqeq1d.1 φ A = B
2 dfcleq A = B x x A x B
3 1 2 sylib φ x x A x B
4 3 19.21bi φ x A x B
5 4 bibi1d φ x A x C x B x C
6 5 albidv φ x x A x C x x B x C
7 dfcleq A = C x x A x C
8 dfcleq B = C x x B x C
9 6 7 8 3bitr4g φ A = C B = C