Metamath Proof Explorer


Theorem ineqri

Description: Inference from membership to intersection. (Contributed by NM, 21-Jun-1993)

Ref Expression
Hypothesis ineqri.1 x A x B x C
Assertion ineqri A B = C

Proof

Step Hyp Ref Expression
1 ineqri.1 x A x B x C
2 elin x A B x A x B
3 2 1 bitri x A B x C
4 3 eqriv A B = C