Metamath Proof Explorer


Theorem esumdivc

Description: An extended sum divided by a constant. (Contributed by Thierry Arnoux, 6-Jul-2017)

Ref Expression
Hypotheses esumdivc.a φ A V
esumdivc.b φ k A B 0 +∞
esumdivc.c φ C +
Assertion esumdivc φ * k A B ÷ 𝑒 C = * k A B ÷ 𝑒 C

Proof

Step Hyp Ref Expression
1 esumdivc.a φ A V
2 esumdivc.b φ k A B 0 +∞
3 esumdivc.c φ C +
4 1red φ 1
5 3 rpred φ C
6 3 rpne0d φ C 0
7 rexdiv 1 C C 0 1 ÷ 𝑒 C = 1 C
8 4 5 6 7 syl3anc φ 1 ÷ 𝑒 C = 1 C
9 ioorp 0 +∞ = +
10 ioossico 0 +∞ 0 +∞
11 9 10 eqsstrri + 0 +∞
12 3 rpreccld φ 1 C +
13 11 12 sselid φ 1 C 0 +∞
14 8 13 eqeltrd φ 1 ÷ 𝑒 C 0 +∞
15 1 2 14 esummulc1 φ * k A B 𝑒 1 ÷ 𝑒 C = * k A B 𝑒 1 ÷ 𝑒 C
16 iccssxr 0 +∞ *
17 2 ralrimiva φ k A B 0 +∞
18 nfcv _ k A
19 18 esumcl A V k A B 0 +∞ * k A B 0 +∞
20 1 17 19 syl2anc φ * k A B 0 +∞
21 16 20 sselid φ * k A B *
22 xdivrec * k A B * C C 0 * k A B ÷ 𝑒 C = * k A B 𝑒 1 ÷ 𝑒 C
23 21 5 6 22 syl3anc φ * k A B ÷ 𝑒 C = * k A B 𝑒 1 ÷ 𝑒 C
24 16 2 sselid φ k A B *
25 5 adantr φ k A C
26 6 adantr φ k A C 0
27 xdivrec B * C C 0 B ÷ 𝑒 C = B 𝑒 1 ÷ 𝑒 C
28 24 25 26 27 syl3anc φ k A B ÷ 𝑒 C = B 𝑒 1 ÷ 𝑒 C
29 28 esumeq2dv φ * k A B ÷ 𝑒 C = * k A B 𝑒 1 ÷ 𝑒 C
30 15 23 29 3eqtr4d φ * k A B ÷ 𝑒 C = * k A B ÷ 𝑒 C