Theorem difpr 3764

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

Step	Hyp	Ref	Expression
1		df-pr 3629	. . 3 ⊢ {𝐵, 𝐶} = ({𝐵} ∪ {𝐶})
2	1	difeq2i 3278	. 2 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶}))
3		difun1 3423	. 2 ⊢ (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
4	2, 3	eqtri 2217	1 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})

Syntax hints:   = wceq 1364   ∖ cdif 3154   ∪ cun 3155  {csn 3622  {cpr 3623

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-rab 2484  df-v 2765  df-dif 3159  df-un 3161  df-in 3163  df-pr 3629

This theorem is referenced by:  hashdifpr  10912