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Metamath Proof Explorer

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 ) )