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Theorem difsnss 3538
 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 6111. (Contributed by Jim Kingdon, 10-Aug-2018.)
Assertion
Ref Expression
difsnss (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Proof of Theorem difsnss
StepHypRef Expression
1 uncom 3115 . 2 ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2 snssi 3536 . . 3 (𝐵𝐴 → {𝐵} ⊆ 𝐴)
3 undifss 3331 . . 3 ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
42, 3sylib 131 . 2 (𝐵𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
51, 4syl5eqss 3017 1 (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∈ wcel 1409   ∖ cdif 2942   ∪ cun 2943   ⊆ wss 2945  {csn 3403 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038 This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-v 2576  df-dif 2948  df-un 2950  df-in 2952  df-ss 2959  df-sn 3409 This theorem is referenced by:  nndifsnid  6111  fidifsnid  6363
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