Metamath Proof Explorer


Theorem eleq2i

Description: Inference from equality to equivalence of membership. (Contributed by NM, 26-May-1993)

Ref Expression
Hypothesis eleq1i.1 𝐴 = 𝐵
Assertion eleq2i ( 𝐶𝐴𝐶𝐵 )

Proof

Step Hyp Ref Expression
1 eleq1i.1 𝐴 = 𝐵
2 eleq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴𝐶𝐵 ) )
3 1 2 ax-mp ( 𝐶𝐴𝐶𝐵 )