Theorem eqeq1d

Description: Deduction from equality to equivalence of equalities. (Contributed by NM, 27-Dec-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 5-Dec-2019)

	Ref	Expression
Hypothesis	eqeq1d.1	|- ( ph -> A = B )
Assertion	eqeq1d	|- ( ph -> ( A = C <-> B = C ) )

Proof

Step	Hyp	Ref	Expression
1		eqeq1d.1	|- ( ph -> A = B )
2		dfcleq	|- ( A = B <-> A. x ( x e. A <-> x e. B ) )
3	2	biimpi	|- ( A = B -> A. x ( x e. A <-> x e. B ) )
4		bibi1	|- ( ( x e. A <-> x e. B ) -> ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) )
5	4	alimi	|- ( A. x ( x e. A <-> x e. B ) -> A. x ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) )
6		albi	|- ( A. x ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) -> ( A. x ( x e. A <-> x e. C ) <-> A. x ( x e. B <-> x e. C ) ) )
7	1 3 5 6	4syl	|- ( ph -> ( A. x ( x e. A <-> x e. C ) <-> A. x ( x e. B <-> x e. C ) ) )
8		dfcleq	|- ( A = C <-> A. x ( x e. A <-> x e. C ) )
9		dfcleq	|- ( B = C <-> A. x ( x e. B <-> x e. C ) )
10	7 8 9	3bitr4g	|- ( ph -> ( A = C <-> B = C ) )