Metamath Proof Explorer


Theorem psseq2i

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

Ref Expression
Hypothesis psseq1i.1 ⊢ A = B
Assertion psseq2i ⊢ C ⊂ A ↔ C ⊂ B

Proof

Step Hyp Ref Expression
1 psseq1i.1 ⊢ A = B
2 psseq2 ⊢ A = B → C ⊂ A ↔ C ⊂ B
3 1 2 ax-mp ⊢ C ⊂ A ↔ C ⊂ B