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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 )