Theorem eqimss2

Description: Equality implies the subclass relation. (Contributed by NM, 23-Nov-2003)

	Ref	Expression
Assertion	eqimss2	⊢ ( 𝐵 = 𝐴 → 𝐴 ⊆ 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		eqimss	⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
2	1	eqcoms	⊢ ( 𝐵 = 𝐴 → 𝐴 ⊆ 𝐵 )