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Theorem difpr 4166
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
StepHypRef Expression
1 df-pr 4030 . . 3  |-  { B ,  C }  =  ( { B }  u.  { C } )
21difeq2i 3619 . 2  |-  ( A 
\  { B ,  C } )  =  ( A  \  ( { B }  u.  { C } ) )
3 difun1 3758 . 2  |-  ( A 
\  ( { B }  u.  { C } ) )  =  ( ( A  \  { B } )  \  { C } )
42, 3eqtri 2496 1  |-  ( A 
\  { B ,  C } )  =  ( ( A  \  { B } )  \  { C } )
Colors of variables: wff setvar class
Syntax hints:    = wceq 1379    \ cdif 3473    u. cun 3474   {csn 4027   {cpr 4029
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1601  ax-4 1612  ax-5 1680  ax-6 1719  ax-7 1739  ax-10 1786  ax-11 1791  ax-12 1803  ax-13 1968  ax-ext 2445
This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-tru 1382  df-ex 1597  df-nf 1600  df-sb 1712  df-clab 2453  df-cleq 2459  df-clel 2462  df-nfc 2617  df-ral 2819  df-rab 2823  df-v 3115  df-dif 3479  df-un 3481  df-in 3483  df-pr 4030
This theorem is referenced by:  nbgrassvwo2  24114
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