Metamath Proof Explorer


Theorem esumss

Description: Change the index set to a subset by adding zeroes. (Contributed by Thierry Arnoux, 19-Jun-2017)

Ref Expression
Hypotheses esumss.p k φ
esumss.a _ k A
esumss.b _ k B
esumss.1 φ A B
esumss.2 φ B V
esumss.3 φ k B C 0 +∞
esumss.4 φ k B A C = 0
Assertion esumss φ * k A C = * k B C

Proof

Step Hyp Ref Expression
1 esumss.p k φ
2 esumss.a _ k A
3 esumss.b _ k B
4 esumss.1 φ A B
5 esumss.2 φ B V
6 esumss.3 φ k B C 0 +∞
7 esumss.4 φ k B A C = 0
8 3 2 resmptf A B k B C A = k A C
9 4 8 syl φ k B C A = k A C
10 9 oveq2d φ 𝑠 * 𝑠 0 +∞ tsums k B C A = 𝑠 * 𝑠 0 +∞ tsums k A C
11 xrge0base 0 +∞ = Base 𝑠 * 𝑠 0 +∞
12 xrge00 0 = 0 𝑠 * 𝑠 0 +∞
13 xrge0cmn 𝑠 * 𝑠 0 +∞ CMnd
14 13 a1i φ 𝑠 * 𝑠 0 +∞ CMnd
15 xrge0tps 𝑠 * 𝑠 0 +∞ TopSp
16 15 a1i φ 𝑠 * 𝑠 0 +∞ TopSp
17 nfcv _ k 0 +∞
18 eqid k B C = k B C
19 1 3 17 6 18 fmptdF φ k B C : B 0 +∞
20 1 3 2 7 5 suppss2f φ k B C supp 0 A
21 11 12 14 16 5 19 20 tsmsres φ 𝑠 * 𝑠 0 +∞ tsums k B C A = 𝑠 * 𝑠 0 +∞ tsums k B C
22 10 21 eqtr3d φ 𝑠 * 𝑠 0 +∞ tsums k A C = 𝑠 * 𝑠 0 +∞ tsums k B C
23 22 unieqd φ 𝑠 * 𝑠 0 +∞ tsums k A C = 𝑠 * 𝑠 0 +∞ tsums k B C
24 df-esum * k A C = 𝑠 * 𝑠 0 +∞ tsums k A C
25 df-esum * k B C = 𝑠 * 𝑠 0 +∞ tsums k B C
26 23 24 25 3eqtr4g φ * k A C = * k B C