Theorem difpr 3722

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 3590	. . 3 |- { B , C } = ( { B } u. { C } )
2	1	difeq2i 3242	. 2 |- ( A \ { B , C } ) = ( A \ ( { B } u. { C } ) )
3		difun1 3387	. 2 |- ( A \ ( { B } u. { C } ) ) = ( ( A \ { B } ) \ { C } )
4	2, 3	eqtri 2191	1 |- ( A \ { B , C } ) = ( ( A \ { B } ) \ { C } )

Syntax hints:   = wceq 1348   \ cdif 3118   u. cun 3119   {csn 3583   {cpr 3584

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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rab 2457  df-v 2732  df-dif 3123  df-un 3125  df-in 3127  df-pr 3590

This theorem is referenced by:  hashdifpr  10755