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Metamath Proof Explorer

Theorem 3sstr4g

Description: Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 16-Aug-1994) (Proof shortened by Eric Schmidt, 26-Jan-2007)

	Ref	Expression
Hypotheses	3sstr4g.1	$⊢ φ \to A \subseteq B$
	3sstr4g.2	$⊢ C = A$
	3sstr4g.3	$⊢ D = B$
Assertion	3sstr4g	$⊢ φ \to C \subseteq D$

Proof

Step	Hyp	Ref	Expression
1		3sstr4g.1	$⊢ φ \to A \subseteq B$
2		3sstr4g.2	$⊢ C = A$
3		3sstr4g.3	$⊢ D = B$
4	2 1	eqsstrid	$⊢ φ \to C \subseteq B$
5	4 3	sseqtrrdi	$⊢ φ \to C \subseteq D$