Theorem eleq2d

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

	Ref	Expression
Hypothesis	eleq1d.1	|- ( ph -> A = B )
Assertion	eleq2d	|- ( ph -> ( C e. A <-> C e. B ) )

Proof

Step	Hyp	Ref	Expression
1		eleq1d.1	|- ( ph -> A = B )
2		dfcleq	|- ( A = B <-> A. x ( x e. A <-> x e. B ) )
3	1 2	sylib	|- ( ph -> A. x ( x e. A <-> x e. B ) )
4		anbi2	|- ( ( x e. A <-> x e. B ) -> ( ( x = C /\ x e. A ) <-> ( x = C /\ x e. B ) ) )
5	4	alexbii	|- ( A. x ( x e. A <-> x e. B ) -> ( E. x ( x = C /\ x e. A ) <-> E. x ( x = C /\ x e. B ) ) )
6	3 5	syl	|- ( ph -> ( E. x ( x = C /\ x e. A ) <-> E. x ( x = C /\ x e. B ) ) )
7		dfclel	|- ( C e. A <-> E. x ( x = C /\ x e. A ) )
8		dfclel	|- ( C e. B <-> E. x ( x = C /\ x e. B ) )
9	6 7 8	3bitr4g	|- ( ph -> ( C e. A <-> C e. B ) )