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Metamath Proof Explorer

Theorem fsumle

Description: If all of the terms of finite sums compare, so do the sums. (Contributed by NM, 11-Dec-2005) (Proof shortened by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses fsumle.1 
|- ( ph -> A e. Fin )
fsumle.2 
|- ( ( ph /\ k e. A ) -> B e. RR )
fsumle.3 
|- ( ( ph /\ k e. A ) -> C e. RR )
fsumle.4 
|- ( ( ph /\ k e. A ) -> B <_ C )
Assertion fsumle 
|- ( ph -> sum_ k e. A B <_ sum_ k e. A C )

Proof

Step Hyp Ref Expression
1 fsumle.1 
 |-  ( ph -> A e. Fin )
2 fsumle.2 
 |-  ( ( ph /\ k e. A ) -> B e. RR )
3 fsumle.3 
 |-  ( ( ph /\ k e. A ) -> C e. RR )
4 fsumle.4 
 |-  ( ( ph /\ k e. A ) -> B <_ C )
5 3 2 resubcld 
 |-  ( ( ph /\ k e. A ) -> ( C - B ) e. RR )
6 3 2 subge0d 
 |-  ( ( ph /\ k e. A ) -> ( 0 <_ ( C - B ) <-> B <_ C ) )
7 4 6 mpbird 
 |-  ( ( ph /\ k e. A ) -> 0 <_ ( C - B ) )
8 1 5 7 fsumge0 
 |-  ( ph -> 0 <_ sum_ k e. A ( C - B ) )
9 3 recnd 
 |-  ( ( ph /\ k e. A ) -> C e. CC )
10 2 recnd 
 |-  ( ( ph /\ k e. A ) -> B e. CC )
11 1 9 10 fsumsub 
 |-  ( ph -> sum_ k e. A ( C - B ) = ( sum_ k e. A C - sum_ k e. A B ) )
12 8 11 breqtrd 
 |-  ( ph -> 0 <_ ( sum_ k e. A C - sum_ k e. A B ) )
13 1 3 fsumrecl 
 |-  ( ph -> sum_ k e. A C e. RR )
14 1 2 fsumrecl 
 |-  ( ph -> sum_ k e. A B e. RR )
15 13 14 subge0d 
 |-  ( ph -> ( 0 <_ ( sum_ k e. A C - sum_ k e. A B ) <-> sum_ k e. A B <_ sum_ k e. A C ) )
16 12 15 mpbid 
 |-  ( ph -> sum_ k e. A B <_ sum_ k e. A C )