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Theorem difpr 3722
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 3590 . . 3 {𝐵, 𝐶} = ({𝐵} ∪ {𝐶})
21difeq2i 3242 . 2 (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶}))
3 difun1 3387 . 2 (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
42, 3eqtri 2191 1 (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
Colors of variables: wff set class
Syntax hints:   = wceq 1348  cdif 3118  cun 3119  {csn 3583  {cpr 3584
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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rab 2457  df-v 2732  df-dif 3123  df-un 3125  df-in 3127  df-pr 3590
This theorem is referenced by:  hashdifpr  10755
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