Metamath Proof Explorer


Theorem isumnn0nn

Description: Sum from 0 to infinity in terms of sum from 1 to infinity. (Contributed by NM, 2-Jan-2006) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses isumnn0nn.1 k = 0 A = B
isumnn0nn.2 φ k 0 F k = A
isumnn0nn.3 φ k 0 A
isumnn0nn.4 φ seq 0 + F dom
Assertion isumnn0nn φ k 0 A = B + k A

Proof

Step Hyp Ref Expression
1 isumnn0nn.1 k = 0 A = B
2 isumnn0nn.2 φ k 0 F k = A
3 isumnn0nn.3 φ k 0 A
4 isumnn0nn.4 φ seq 0 + F dom
5 nn0uz 0 = 0
6 0zd φ 0
7 5 6 2 3 4 isum1p φ k 0 A = F 0 + k 0 + 1 A
8 fveq2 k = 0 F k = F 0
9 8 1 eqeq12d k = 0 F k = A F 0 = B
10 2 ralrimiva φ k 0 F k = A
11 0nn0 0 0
12 11 a1i φ 0 0
13 9 10 12 rspcdva φ F 0 = B
14 0p1e1 0 + 1 = 1
15 14 fveq2i 0 + 1 = 1
16 nnuz = 1
17 15 16 eqtr4i 0 + 1 =
18 17 sumeq1i k 0 + 1 A = k A
19 18 a1i φ k 0 + 1 A = k A
20 13 19 oveq12d φ F 0 + k 0 + 1 A = B + k A
21 7 20 eqtrd φ k 0 A = B + k A