Metamath Proof Explorer


Theorem esumf1o

Description: Re-index an extended sum using a bijection. (Contributed by Thierry Arnoux, 6-Apr-2017)

Ref Expression
Hypotheses esumf1o.0 nφ
esumf1o.b _nB
esumf1o.d _kD
esumf1o.a _nA
esumf1o.c _nC
esumf1o.f _nF
esumf1o.1 k=GB=D
esumf1o.2 φAV
esumf1o.3 φF:C1-1 ontoA
esumf1o.4 φnCFn=G
esumf1o.5 φkAB0+∞
Assertion esumf1o φ*kAB=*nCD

Proof

Step Hyp Ref Expression
1 esumf1o.0 nφ
2 esumf1o.b _nB
3 esumf1o.d _kD
4 esumf1o.a _nA
5 esumf1o.c _nC
6 esumf1o.f _nF
7 esumf1o.1 k=GB=D
8 esumf1o.2 φAV
9 esumf1o.3 φF:C1-1 ontoA
10 esumf1o.4 φnCFn=G
11 esumf1o.5 φkAB0+∞
12 xrge0base 0+∞=Base𝑠*𝑠0+∞
13 xrge0cmn 𝑠*𝑠0+∞CMnd
14 13 a1i φ𝑠*𝑠0+∞CMnd
15 xrge0tps 𝑠*𝑠0+∞TopSp
16 15 a1i φ𝑠*𝑠0+∞TopSp
17 11 fmpttd φkAB:A0+∞
18 12 14 16 8 17 9 tsmsf1o φ𝑠*𝑠0+∞tsumskAB=𝑠*𝑠0+∞tsumskABF
19 f1of F:C1-1 ontoAF:CA
20 9 19 syl φF:CA
21 20 ffvelcdmda φnCFnA
22 10 21 eqeltrrd φnCGA
23 22 ex φnCGA
24 1 23 ralrimi φnCGA
25 5 6 20 feqmptdf φF=nCFn
26 1 10 mpteq2da φnCFn=nCG
27 25 26 eqtrd φF=nCG
28 eqidd φkAB=kAB
29 2 3 5 4 1 24 27 28 7 fmptcof2 φkABF=nCD
30 29 oveq2d φ𝑠*𝑠0+∞tsumskABF=𝑠*𝑠0+∞tsumsnCD
31 18 30 eqtrd φ𝑠*𝑠0+∞tsumskAB=𝑠*𝑠0+∞tsumsnCD
32 31 unieqd φ𝑠*𝑠0+∞tsumskAB=𝑠*𝑠0+∞tsumsnCD
33 df-esum *kAB=𝑠*𝑠0+∞tsumskAB
34 df-esum *nCD=𝑠*𝑠0+∞tsumsnCD
35 32 33 34 3eqtr4g φ*kAB=*nCD