Theorem difpr 3852

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 3712	. . 3 |- { B , C } = ( { B } u. { C } )
2	1	difeq2i 3344	. 2 |- ( A \ { B , C } ) = ( A \ ( { B } u. { C } ) )
3		difun1 3491	. 2 |- ( A \ ( { B } u. { C } ) ) = ( ( A \ { B } ) \ { C } )
4	2, 3	eqtri 2259	1 |- ( A \ { B , C } ) = ( ( A \ { B } ) \ { C } )

Syntax hints:   = wceq 1402   \ cdif 3217   u. cun 3218   {csn 3705   {cpr 3706

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 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rab 2537  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-pr 3712

This theorem is referenced by:  hashdifpr  11239