Theorem difpr 3657

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 3529	. . 3 |- { B , C } = ( { B } u. { C } )
2	1	difeq2i 3186	. 2 |- ( A \ { B , C } ) = ( A \ ( { B } u. { C } ) )
3		difun1 3331	. 2 |- ( A \ ( { B } u. { C } ) ) = ( ( A \ { B } ) \ { C } )
4	2, 3	eqtri 2158	1 |- ( A \ { B , C } ) = ( ( A \ { B } ) \ { C } )

Syntax hints:   = wceq 1331   \ cdif 3063   u. cun 3064   {csn 3522   {cpr 3523

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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-rab 2423  df-v 2683  df-dif 3068  df-un 3070  df-in 3072  df-pr 3529

This theorem is referenced by:  hashdifpr  10559