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Theorem difsnss 3501
 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 3081 . 2 ((A ∖ {B}) ∪ {B}) = ({B} ∪ (A ∖ {B}))
2 snssi 3499 . . 3 (B A → {B} ⊆ A)
3 undifss 3297 . . 3 ({B} ⊆ A ↔ ({B} ∪ (A ∖ {B})) ⊆ A)
42, 3sylib 127 . 2 (B A → ({B} ∪ (A ∖ {B})) ⊆ A)
51, 4syl5eqss 2983 1 (B A → ((A ∖ {B}) ∪ {B}) ⊆ A)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∈ wcel 1390   ∖ cdif 2908   ∪ cun 2909   ⊆ wss 2911  {csn 3367 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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-sn 3373 This theorem is referenced by: (None)
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