Theorem sseq0 2308

Description: A subclass of an empty class is empty.

Assertion

Ref	Expression
sseq0	|- ((A (_ B /\ B = (/)) -> A = (/))

Proof of Theorem sseq0

Step	Hyp	Ref	Expression
1		sseq2 2086	. . 3 |- (B = (/) -> (A (_ B <-> A (_ (/)))
2	1	biimpac 420	. 2 |- ((A (_ B /\ B = (/)) -> A (_ (/))
3		ss0b 2306	. 2 |- (A (_ (/) <-> A = (/))
4	2, 3	sylib 198	1 |- ((A (_ B /\ B = (/)) -> A = (/))

Syntax hints:   -> wi 3   /\ wa 223   = wceq 958   (_ wss 2050  (/)c0 2283

This theorem is referenced by:  ssne0 2309  sncld 7784  lpbl 7877

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-v 1815  df-dif 2052  df-in 2054  df-ss 2056  df-nul 2284

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