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Theorem difsnss 3740
Description: If we remove a single element from a class then put it back in, we end up with a subset of the original class. If equality is decidable, we can replace subset with equality as seen in nndifsnid 6510. (Contributed by Jim Kingdon, 10-Aug-2018.)
Assertion
Ref Expression
difsnss  |-  ( B  e.  A  ->  (
( A  \  { B } )  u.  { B } )  C_  A
)

Proof of Theorem difsnss
StepHypRef Expression
1 uncom 3281 . 2  |-  ( ( A  \  { B } )  u.  { B } )  =  ( { B }  u.  ( A  \  { B } ) )
2 snssi 3738 . . 3  |-  ( B  e.  A  ->  { B }  C_  A )
3 undifss 3505 . . 3  |-  ( { B }  C_  A  <->  ( { B }  u.  ( A  \  { B } ) )  C_  A )
42, 3sylib 122 . 2  |-  ( B  e.  A  ->  ( { B }  u.  ( A  \  { B }
) )  C_  A
)
51, 4eqsstrid 3203 1  |-  ( B  e.  A  ->  (
( A  \  { B } )  u.  { B } )  C_  A
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    e. wcel 2148    \ cdif 3128    u. cun 3129    C_ wss 3131   {csn 3594
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-dif 3133  df-un 3135  df-in 3137  df-ss 3144  df-sn 3600
This theorem is referenced by:  fnsnsplitss  5717  dcdifsnid  6507
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