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Metamath Proof Explorer

Theorem eqimss2

Description: Equality implies inclusion. (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 )