Theorem difpr 3810

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 3673	. . 3 |- { B , C } = ( { B } u. { C } )
2	1	difeq2i 3319	. 2 |- ( A \ { B , C } ) = ( A \ ( { B } u. { C } ) )
3		difun1 3464	. 2 |- ( A \ ( { B } u. { C } ) ) = ( ( A \ { B } ) \ { C } )
4	2, 3	eqtri 2250	1 |- ( A \ { B , C } ) = ( ( A \ { B } ) \ { C } )

Syntax hints:   = wceq 1395   \ cdif 3194   u. cun 3195   {csn 3666   {cpr 3667

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 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rab 2517  df-v 2801  df-dif 3199  df-un 3201  df-in 3203  df-pr 3673

This theorem is referenced by:  hashdifpr  11042