Metamath Proof Explorer


Theorem ineqri

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

Ref Expression
Hypothesis ineqri.1 xAxBxC
Assertion ineqri AB=C

Proof

Step Hyp Ref Expression
1 ineqri.1 xAxBxC
2 elin xABxAxB
3 2 1 bitri xABxC
4 3 eqriv AB=C