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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