Theorem difpr 3715

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 3583	. . 3 |- { B , C } = ( { B } u. { C } )
2	1	difeq2i 3237	. 2 |- ( A \ { B , C } ) = ( A \ ( { B } u. { C } ) )
3		difun1 3382	. 2 |- ( A \ ( { B } u. { C } ) ) = ( ( A \ { B } ) \ { C } )
4	2, 3	eqtri 2186	1 |- ( A \ { B , C } ) = ( ( A \ { B } ) \ { C } )

Syntax hints:   = wceq 1343   \ cdif 3113   u. cun 3114   {csn 3576   {cpr 3577

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 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rab 2453  df-v 2728  df-dif 3118  df-un 3120  df-in 3122  df-pr 3583

This theorem is referenced by:  hashdifpr  10733