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Metamath Proof Explorer

Theorem eqimssi

Description: Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007)

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

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