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 C. A <-> C C. B )

Proof

Step Hyp Ref Expression
1 psseq1i.1
 |-  A = B
2 psseq2
 |-  ( A = B -> ( C C. A <-> C C. B ) )
3 1 2 ax-mp
 |-  ( C C. A <-> C C. B )