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Metamath Proof Explorer

Theorem psseq1i

Description: An equality inference for the proper subclass relationship. (Contributed by NM, 9-Jun-2004)

Ref Expression
Hypothesis psseq1i.1 ⊢ A = B
Assertion psseq1i ⊢ A ⊂ C ↔ B ⊂ C

Proof

Step Hyp Ref Expression
1 psseq1i.1 ⊢ A = B
2 psseq1 ⊢ A = B → A ⊂ C ↔ B ⊂ C
3 1 2 ax-mp ⊢ A ⊂ C ↔ B ⊂ C