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Theorem difsnss 3507
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 6043. (Contributed by Jim Kingdon, 10-Aug-2018.)
Assertion
Ref Expression
difsnss (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Proof of Theorem difsnss
StepHypRef Expression
1 uncom 3084 . 2 ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2 snssi 3505 . . 3 (𝐵𝐴 → {𝐵} ⊆ 𝐴)
3 undifss 3300 . . 3 ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
42, 3sylib 127 . 2 (𝐵𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
51, 4syl5eqss 2986 1 (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)
Colors of variables: wff set class
Syntax hints:  wi 4  wcel 1393  cdif 2911  cun 2912  wss 2914  {csn 3372
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 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2556  df-dif 2917  df-un 2919  df-in 2921  df-ss 2928  df-sn 3378
This theorem is referenced by:  nndifsnid  6043  fidifsnid  6295
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