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Theorem 0dif 3500
Description: The difference between the empty set and a class. Part of Exercise 4.4 of [Stoll] p. 16. (Contributed by NM, 17-Aug-2004.)
Assertion
Ref Expression
0dif  |-  ( (/)  \  A )  =  (/)

Proof of Theorem 0dif
StepHypRef Expression
1 difss 3278 . 2  |-  ( (/)  \  A )  C_  (/)
2 ss0 3460 . 2  |-  ( (
(/)  \  A )  C_  (/)  ->  ( (/)  \  A
)  =  (/) )
31, 2ax-mp 10 1  |-  ( (/)  \  A )  =  (/)
Colors of variables: wff set class
Syntax hints:    = wceq 1619    \ cdif 3124    C_ wss 3127   (/)c0 3430
This theorem is referenced by:  fresaun  5350  dffv2  5526  ablfac1eulem  15269  itgioo  19132  ballotlemfval0  23015  ballotlemgun  23044  symdif0  23744  sssu  24508  bwt2  24959
This theorem was proved from axioms:  ax-1 7  ax-2 8  ax-3 9  ax-mp 10  ax-5 1533  ax-6 1534  ax-7 1535  ax-gen 1536  ax-8 1623  ax-11 1624  ax-17 1628  ax-12o 1664  ax-10 1678  ax-9 1684  ax-4 1692  ax-16 1927  ax-ext 2239
This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-tru 1315  df-ex 1538  df-nf 1540  df-sb 1884  df-clab 2245  df-cleq 2251  df-clel 2254  df-nfc 2383  df-v 2765  df-dif 3130  df-in 3134  df-ss 3141  df-nul 3431
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