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Theorem difpr 3700
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 3568 . . 3 {𝐵, 𝐶} = ({𝐵} ∪ {𝐶})
21difeq2i 3223 . 2 (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶}))
3 difun1 3368 . 2 (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
42, 3eqtri 2178 1 (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
Colors of variables: wff set class
Syntax hints:   = wceq 1335  cdif 3099  cun 3100  {csn 3561  {cpr 3562
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 604  ax-in2 605  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-ral 2440  df-rab 2444  df-v 2714  df-dif 3104  df-un 3106  df-in 3108  df-pr 3568
This theorem is referenced by:  hashdifpr  10705
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