Theorem psseq1d 2702

Description: An equality deduction for the proper subclass relationship.

Hypothesis

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

Assertion

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

Proof of Theorem psseq1d

Step	Hyp	Ref	Expression
1		psseq1d.1	. 2 |- (ph -> A = B)
2		psseq1 2697	. 2 |- (A = B -> (A C. C <-> B C. C))
3	1, 2	syl 12	1 |- (ph -> (A C. C <-> B C. C))

Syntax hints:   -> wi 3   <-> wb 163   = wceq 1298   C. wpss 2594

This theorem is referenced by:  psseq12d 2704  chpsscon2 11061

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-ne 2019  df-in 2603  df-ss 2605  df-pss 2607

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