Metamath Proof Explorer


Theorem gsumsplit1r

Description: Splitting off the rightmost summand of a group sum. This corresponds to the (inductive) definition of a (finite) product in Lang p. 4, first formula. (Contributed by AV, 26-Dec-2023)

Ref Expression
Hypotheses gsumsplit1r.b B=BaseG
gsumsplit1r.p +˙=+G
gsumsplit1r.g φGV
gsumsplit1r.m φM
gsumsplit1r.n φNM
gsumsplit1r.f φF:MN+1B
Assertion gsumsplit1r φGF=GFMN+˙FN+1

Proof

Step Hyp Ref Expression
1 gsumsplit1r.b B=BaseG
2 gsumsplit1r.p +˙=+G
3 gsumsplit1r.g φGV
4 gsumsplit1r.m φM
5 gsumsplit1r.n φNM
6 gsumsplit1r.f φF:MN+1B
7 peano2uz NMN+1M
8 5 7 syl φN+1M
9 1 2 3 8 6 gsumval2 φGF=seqM+˙FN+1
10 seqp1 NMseqM+˙FN+1=seqM+˙FN+˙FN+1
11 5 10 syl φseqM+˙FN+1=seqM+˙FN+˙FN+1
12 fzssp1 MNMN+1
13 12 a1i φMNMN+1
14 6 13 fssresd φFMN:MNB
15 1 2 3 5 14 gsumval2 φGFMN=seqM+˙FMNN
16 4 uzidd φMM
17 seq1 MseqM+˙FMNM=FMNM
18 4 17 syl φseqM+˙FMNM=FMNM
19 eluzfz1 NMMMN
20 5 19 syl φMMN
21 20 fvresd φFMNM=FM
22 18 21 eqtrd φseqM+˙FMNM=FM
23 fzp1ss MM+1NMN
24 4 23 syl φM+1NMN
25 24 sselda φxM+1NxMN
26 25 fvresd φxM+1NFMNx=Fx
27 16 22 5 26 seqfveq2 φseqM+˙FMNN=seqM+˙FN
28 15 27 eqtr2d φseqM+˙FN=GFMN
29 28 oveq1d φseqM+˙FN+˙FN+1=GFMN+˙FN+1
30 9 11 29 3eqtrd φGF=GFMN+˙FN+1