Metamath Proof Explorer

Theorem fsumsub

Description: Split a finite sum over a subtraction. (Contributed by Scott Fenton, 12-Jun-2013) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses fsumneg.1 
|- ( ph -> A e. Fin )
fsumneg.2 
|- ( ( ph /\ k e. A ) -> B e. CC )
fsumsub.3 
|- ( ( ph /\ k e. A ) -> C e. CC )
Assertion fsumsub 
|- ( ph -> sum_ k e. A ( B - C ) = ( sum_ k e. A B - sum_ k e. A C ) )

Proof

Step Hyp Ref Expression
1 fsumneg.1 
 |-  ( ph -> A e. Fin )
2 fsumneg.2 
 |-  ( ( ph /\ k e. A ) -> B e. CC )
3 fsumsub.3 
 |-  ( ( ph /\ k e. A ) -> C e. CC )
4 3 negcld 
 |-  ( ( ph /\ k e. A ) -> -u C e. CC )
5 1 2 4 fsumadd 
 |-  ( ph -> sum_ k e. A ( B + -u C ) = ( sum_ k e. A B + sum_ k e. A -u C ) )
6 1 3 fsumneg 
 |-  ( ph -> sum_ k e. A -u C = -u sum_ k e. A C )
7 6 oveq2d 
 |-  ( ph -> ( sum_ k e. A B + sum_ k e. A -u C ) = ( sum_ k e. A B + -u sum_ k e. A C ) )
8 5 7 eqtrd 
 |-  ( ph -> sum_ k e. A ( B + -u C ) = ( sum_ k e. A B + -u sum_ k e. A C ) )
9 2 3 negsubd 
 |-  ( ( ph /\ k e. A ) -> ( B + -u C ) = ( B - C ) )
10 9 sumeq2dv 
 |-  ( ph -> sum_ k e. A ( B + -u C ) = sum_ k e. A ( B - C ) )
11 1 2 fsumcl 
 |-  ( ph -> sum_ k e. A B e. CC )
12 1 3 fsumcl 
 |-  ( ph -> sum_ k e. A C e. CC )
13 11 12 negsubd 
 |-  ( ph -> ( sum_ k e. A B + -u sum_ k e. A C ) = ( sum_ k e. A B - sum_ k e. A C ) )
14 8 10 13 3eqtr3d 
 |-  ( ph -> sum_ k e. A ( B - C ) = ( sum_ k e. A B - sum_ k e. A C ) )