Theorem difpr 3820

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 3680	. . 3 ⊢ {𝐵, 𝐶} = ({𝐵} ∪ {𝐶})
2	1	difeq2i 3324	. 2 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶}))
3		difun1 3469	. 2 ⊢ (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})
4	2, 3	eqtri 2252	1 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶})

Syntax hints:   = wceq 1398   ∖ cdif 3198   ∪ cun 3199  {csn 3673  {cpr 3674

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-rab 2520  df-v 2805  df-dif 3203  df-un 3205  df-in 3207  df-pr 3680

This theorem is referenced by:  hashdifpr  11130