Metamath Proof Explorer


Theorem ineqcomi

Description: Two ways of expressing that two classes have a given intersection. Inference form of ineqcom . Disjointness inference when C = (/) . (Contributed by Peter Mazsa, 26-Mar-2017) (Proof shortened by SN, 20-Sep-2024)

Ref Expression
Hypothesis ineqcomi.1
|- ( A i^i B ) = C
Assertion ineqcomi
|- ( B i^i A ) = C

Proof

Step Hyp Ref Expression
1 ineqcomi.1
 |-  ( A i^i B ) = C
2 incom
 |-  ( B i^i A ) = ( A i^i B )
3 2 1 eqtri
 |-  ( B i^i A ) = C