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Metamath Proof Explorer

Theorem psseq2d

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

Ref Expression
Hypothesis psseq1d.1 ⊢ φ → A = B
Assertion psseq2d ⊢ φ → C ⊂ A ↔ C ⊂ B

Proof

Step Hyp Ref Expression
1 psseq1d.1 ⊢ φ → A = B
2 psseq2 ⊢ A = B → C ⊂ A ↔ C ⊂ B
3 1 2 syl ⊢ φ → C ⊂ A ↔ C ⊂ B