Metamath Proof Explorer


Theorem gsumws1

Description: A singleton composite recovers the initial symbol. (Contributed by Stefan O'Rear, 16-Aug-2015)

Ref Expression
Hypothesis gsumwcl.b ⊢ B = Base G
Assertion gsumws1 ⊢ S ∈ B → ∑ G ⟨“ S ”⟩ = S

Proof

Step Hyp Ref Expression
1 gsumwcl.b ⊢ B = Base G
2 s1val ⊢ S ∈ B → ⟨“ S ”⟩ = 0 S
3 2 oveq2d ⊢ S ∈ B → ∑ G ⟨“ S ”⟩ = ∑ G 0 S
4 eqid ⊢ + G = + G
5 elfvdm ⊢ S ∈ Base G → G ∈ dom ⁡ Base
6 5 1 eleq2s ⊢ S ∈ B → G ∈ dom ⁡ Base
7 0nn0 ⊢ 0 ∈ ℕ 0
8 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
9 7 8 eleqtri ⊢ 0 ∈ ℤ ≥ 0
10 9 a1i ⊢ S ∈ B → 0 ∈ ℤ ≥ 0
11 0z ⊢ 0 ∈ ℤ
12 f1osng ⊢ 0 ∈ ℤ ∧ S ∈ B → 0 S : 0 ⟶ 1-1 onto S
13 11 12 mpan ⊢ S ∈ B → 0 S : 0 ⟶ 1-1 onto S
14 f1of ⊢ 0 S : 0 ⟶ 1-1 onto S → 0 S : 0 ⟶ S
15 13 14 syl ⊢ S ∈ B → 0 S : 0 ⟶ S
16 snssi ⊢ S ∈ B → S ⊆ B
17 15 16 fssd ⊢ S ∈ B → 0 S : 0 ⟶ B
18 fz0sn ⊢ 0 … 0 = 0
19 18 feq2i ⊢ 0 S : 0 … 0 ⟶ B ↔ 0 S : 0 ⟶ B
20 17 19 sylibr ⊢ S ∈ B → 0 S : 0 … 0 ⟶ B
21 1 4 6 10 20 gsumval2 ⊢ S ∈ B → ∑ G 0 S = seq 0 + G 0 S ⁡ 0
22 fvsng ⊢ 0 ∈ ℤ ∧ S ∈ B → 0 S ⁡ 0 = S
23 11 22 mpan ⊢ S ∈ B → 0 S ⁡ 0 = S
24 11 23 seq1i ⊢ S ∈ B → seq 0 + G 0 S ⁡ 0 = S
25 3 21 24 3eqtrd ⊢ S ∈ B → ∑ G ⟨“ S ”⟩ = S