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