Theorem ssnelpss 2328

Description: A subclass missing a member is a proper subclass.

Assertion

Ref	Expression
ssnelpss	|- (A (_ B -> ((C e. B /\ -. C e. A) -> A (. B))

Proof of Theorem ssnelpss

Step	Hyp	Ref	Expression
1		dfpss2 2131	. . 3 |- (A (. B <-> (A (_ B /\ -. A = B))
2	1	baibr 685	. 2 |- (A (_ B -> (-. A = B <-> A (. B))
3		nelneq2 1561	. . 3 |- ((C e. B /\ -. C e. A) -> -. B = A)
4		eqcom 1476	. . . 4 |- (B = A <-> A = B)
5	4	negbii 187	. . 3 |- (-. B = A <-> -. A = B)
6	3, 5	sylib 198	. 2 |- ((C e. B /\ -. C e. A) -> -. A = B)
7	2, 6	syl5bi 208	1 |- (A (_ B -> ((C e. B /\ -. C e. A) -> A (. B))

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   = wceq 955   e. wcel 957   (_ wss 2045   (. wpss 2046

This theorem is referenced by:  nthruc 6697  nthruz 6698

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 962  ax-17 970  ax-4 972  ax-5o 974  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-cleq 1469  df-clel 1472  df-ne 1586  df-pss 2053

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