Theorem psseq1d 1564

Description: An equality deduction for the proper subclass relationship.

Hypothesis

Ref	Expression
psseq1d.1	|- (ph -> A = B)

Assertion

Ref	Expression
psseq1d	|- (ph -> (A (. C <-> B (. C))

Proof of Theorem psseq1d

Step	Hyp	Ref	Expression
1		psseq1d.1	. 2 |- (ph -> A = B)
2		psseq1 1559	. 2 |- (A = B -> (A (. C <-> B (. C))
3	1, 2	syl 12	1 |- (ph -> (A (. C <-> B (. C))

Syntax hints:   -> wi 2   <-> wb 127   = wceq 1091   (. wpss 1488

This theorem is referenced by:  psseq12d 1566  chpsscon2t 5422

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ne 1192  df-in 1491  df-ss 1492  df-pss 1494

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