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Theorem difsnss 3666
 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 6403. (Contributed by Jim Kingdon, 10-Aug-2018.)
Assertion
Ref Expression
difsnss (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Proof of Theorem difsnss
StepHypRef Expression
1 uncom 3220 . 2 ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2 snssi 3664 . . 3 (𝐵𝐴 → {𝐵} ⊆ 𝐴)
3 undifss 3443 . . 3 ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
42, 3sylib 121 . 2 (𝐵𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
51, 4eqsstrid 3143 1 (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∈ wcel 1480   ∖ cdif 3068   ∪ cun 3069   ⊆ wss 3071  {csn 3527 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121 This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-dif 3073  df-un 3075  df-in 3077  df-ss 3084  df-sn 3533 This theorem is referenced by:  fnsnsplitss  5619  dcdifsnid  6400
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