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. C <-> B C. C )

Proof

Step Hyp Ref Expression
1 psseq1i.1
 |-  A = B
2 psseq1
 |-  ( A = B -> ( A C. C <-> B C. C ) )
3 1 2 ax-mp
 |-  ( A C. C <-> B C. C )