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Theorem difsnss 3484
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. In classical logic, we could replace subset with equality. (Contributed by Jim Kingdon, 10-Aug-2018.)
Assertion
Ref Expression
difsnss (B A → ((A ∖ {B}) ∪ {B}) ⊆ A)

Proof of Theorem difsnss
StepHypRef Expression
1 uncom 3064 . 2 ((A ∖ {B}) ∪ {B}) = ({B} ∪ (A ∖ {B}))
2 snssi 3482 . . 3 (B A → {B} ⊆ A)
3 undifss 3280 . . 3 ({B} ⊆ A ↔ ({B} ∪ (A ∖ {B})) ⊆ A)
42, 3sylib 127 . 2 (B A → ({B} ∪ (A ∖ {B})) ⊆ A)
51, 4syl5eqss 2966 1 (B A → ((A ∖ {B}) ∪ {B}) ⊆ A)
Colors of variables: wff set class
Syntax hints:  wi 4   wcel 1374  cdif 2891  cun 2892  wss 2894  {csn 3350
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537  df-dif 2897  df-un 2899  df-in 2901  df-ss 2908  df-sn 3356
This theorem is referenced by: (None)
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