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Metamath Proof Explorer

Theorem disjeqi

Description: Equality theorem for disjoints, inference version. (Contributed by Peter Mazsa, 22-Sep-2021)

		Ref	Expression
	Hypothesis	disjeqi.1	$⊢ A = B$
	Assertion	disjeqi	$⊢ Disj A \leftrightarrow Disj B$

Step	Hyp	Ref	Expression
1		disjeqi.1	$⊢ A = B$
2		disjeq	$⊢ A = B \to (Disj A \leftrightarrow Disj B)$
3	1 2	ax-mp	$⊢ Disj A \leftrightarrow Disj B$