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Metamath Proof Explorer

Theorem eqimssd

Description: Equality implies inclusion, deduction version. (Contributed by SN, 6-Nov-2024)

		Ref	Expression
	Hypothesis	eqimssd.1	$⊢ φ \to A = B$
	Assertion	eqimssd	$⊢ φ \to A \subseteq B$

Step	Hyp	Ref	Expression
1		eqimssd.1	$⊢ φ \to A = B$
2		ssid	$⊢ B \subseteq B$
3	1 2	eqsstrdi	$⊢ φ \to A \subseteq B$