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Metamath Proof Explorer

Theorem eqimss2i

Description: Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007)

		Ref	Expression
	Hypothesis	eqimssi.1	$⊢ A = B$
	Assertion	eqimss2i	$⊢ B \subseteq A$

Step	Hyp	Ref	Expression
1		eqimssi.1	$⊢ A = B$
2		ssid	$⊢ B \subseteq B$
3	2 1	sseqtrri	$⊢ B \subseteq A$