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 )