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Theorem difsnid 3854
 Description: If we remove a single element from a class then put it back in, we end up with the original class. (Contributed by NM, 2-Oct-2006.)
Assertion
Ref Expression
difsnid (B A → ((A {B}) ∪ {B}) = A)

Proof of Theorem difsnid
StepHypRef Expression
1 uncom 3408 . 2 ((A {B}) ∪ {B}) = ({B} ∪ (A {B}))
2 snssi 3852 . . 3 (B A → {B} A)
3 undif 3630 . . 3 ({B} A ↔ ({B} ∪ (A {B})) = A)
42, 3sylib 188 . 2 (B A → ({B} ∪ (A {B})) = A)
51, 4syl5eq 2397 1 (B A → ((A {B}) ∪ {B}) = A)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1642   ∈ wcel 1710   ∖ cdif 3206   ∪ cun 3207   ⊆ wss 3257  {csn 3737 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-un 3214  df-dif 3215  df-ss 3259  df-nul 3551  df-sn 3741 This theorem is referenced by:  pwadjoin  4119  phiall  4618
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