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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 → 𝐴 ⊆ 𝐵 )
	3sstr4g.2	⊢ 𝐶 = 𝐴
	3sstr4g.3	⊢ 𝐷 = 𝐵
Assertion	3sstr4g	⊢ ( 𝜑 → 𝐶 ⊆ 𝐷 )

Proof

Step	Hyp	Ref	Expression
1		3sstr4g.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2		3sstr4g.2	⊢ 𝐶 = 𝐴
3		3sstr4g.3	⊢ 𝐷 = 𝐵
4	2 1	eqsstrid	⊢ ( 𝜑 → 𝐶 ⊆ 𝐵 )
5	4 3	sseqtrrdi	⊢ ( 𝜑 → 𝐶 ⊆ 𝐷 )