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Theorem difpr 3574
Description: Removing two elements as pair of elements corresponds to removing each of the two elements as singletons. (Contributed by Alexander van der Vekens, 13-Jul-2018.)
Assertion
Ref Expression
difpr (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})

Proof of Theorem difpr
StepHypRef Expression
1 df-pr 3448 . . 3 {𝐵, 𝐶} = ({𝐵} ∪ {𝐶})
21difeq2i 3113 . 2 (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶}))
3 difun1 3257 . 2 (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
42, 3eqtri 2108 1 (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
Colors of variables: wff set class
Syntax hints:   = wceq 1289  cdif 2994  cun 2995  {csn 3441  {cpr 3442
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 579  ax-in2 580  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ral 2364  df-rab 2368  df-v 2621  df-dif 2999  df-un 3001  df-in 3003  df-pr 3448
This theorem is referenced by:  hashdifpr  10193
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