Metamath Proof Explorer


Theorem esumsup

Description: Express an extended sum as a supremum of extended sums. (Contributed by Thierry Arnoux, 24-May-2020)

Ref Expression
Hypotheses esumsup.1 ⊢ φ → B ∈ 0 +∞
esumsup.2 ⊢ φ ∧ k ∈ ℕ → A ∈ 0 +∞
Assertion esumsup ⊢ φ → ∑ * k ∈ ℕ A = sup ran ⁡ n ∈ ℕ ⟼ ∑ * k = 1 n A ℝ * <

Proof

Step Hyp Ref Expression
1 esumsup.1 ⊢ φ → B ∈ 0 +∞
2 esumsup.2 ⊢ φ ∧ k ∈ ℕ → A ∈ 0 +∞
3 2 fmpttd ⊢ φ → k ∈ ℕ ⟼ A : ℕ ⟶ 0 +∞
4 nfmpt1 ⊢ Ⅎ _ k k ∈ ℕ ⟼ A
5 4 esumfsup ⊢ k ∈ ℕ ⟼ A : ℕ ⟶ 0 +∞ → ∑ * k ∈ ℕ k ∈ ℕ ⟼ A ⁡ k = sup ran ⁡ seq 1 + 𝑒 k ∈ ℕ ⟼ A ℝ * <
6 3 5 syl ⊢ φ → ∑ * k ∈ ℕ k ∈ ℕ ⟼ A ⁡ k = sup ran ⁡ seq 1 + 𝑒 k ∈ ℕ ⟼ A ℝ * <
7 simpr ⊢ φ ∧ k ∈ ℕ → k ∈ ℕ
8 eqid ⊢ k ∈ ℕ ⟼ A = k ∈ ℕ ⟼ A
9 8 fvmpt2 ⊢ k ∈ ℕ ∧ A ∈ 0 +∞ → k ∈ ℕ ⟼ A ⁡ k = A
10 7 2 9 syl2anc ⊢ φ ∧ k ∈ ℕ → k ∈ ℕ ⟼ A ⁡ k = A
11 10 esumeq2dv ⊢ φ → ∑ * k ∈ ℕ k ∈ ℕ ⟼ A ⁡ k = ∑ * k ∈ ℕ A
12 1z ⊢ 1 ∈ ℤ
13 seqfn ⊢ 1 ∈ ℤ → seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℤ ≥ 1
14 12 13 ax-mp ⊢ seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℤ ≥ 1
15 nnuz ⊢ ℕ = ℤ ≥ 1
16 15 fneq2i ⊢ seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℕ ↔ seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℤ ≥ 1
17 14 16 mpbir ⊢ seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℕ
18 nfcv ⊢ Ⅎ _ n seq 1 + 𝑒 k ∈ ℕ ⟼ A
19 18 dffn5f ⊢ seq 1 + 𝑒 k ∈ ℕ ⟼ A Fn ℕ ↔ seq 1 + 𝑒 k ∈ ℕ ⟼ A = n ∈ ℕ ⟼ seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
20 17 19 mpbi ⊢ seq 1 + 𝑒 k ∈ ℕ ⟼ A = n ∈ ℕ ⟼ seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
21 20 a1i ⊢ φ → seq 1 + 𝑒 k ∈ ℕ ⟼ A = n ∈ ℕ ⟼ seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
22 fz1ssnn ⊢ 1 … n ⊆ ℕ
23 22 a1i ⊢ φ ∧ n ∈ ℕ → 1 … n ⊆ ℕ
24 23 sselda ⊢ φ ∧ n ∈ ℕ ∧ k ∈ 1 … n → k ∈ ℕ
25 simpll ⊢ φ ∧ n ∈ ℕ ∧ k ∈ 1 … n → φ
26 25 24 2 syl2anc ⊢ φ ∧ n ∈ ℕ ∧ k ∈ 1 … n → A ∈ 0 +∞
27 24 26 9 syl2anc ⊢ φ ∧ n ∈ ℕ ∧ k ∈ 1 … n → k ∈ ℕ ⟼ A ⁡ k = A
28 27 esumeq2dv ⊢ φ ∧ n ∈ ℕ → ∑ * k = 1 n k ∈ ℕ ⟼ A ⁡ k = ∑ * k = 1 n A
29 4 esumfzf ⊢ k ∈ ℕ ⟼ A : ℕ ⟶ 0 +∞ ∧ n ∈ ℕ → ∑ * k = 1 n k ∈ ℕ ⟼ A ⁡ k = seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
30 3 29 sylan ⊢ φ ∧ n ∈ ℕ → ∑ * k = 1 n k ∈ ℕ ⟼ A ⁡ k = seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
31 28 30 eqtr3d ⊢ φ ∧ n ∈ ℕ → ∑ * k = 1 n A = seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
32 31 mpteq2dva ⊢ φ → n ∈ ℕ ⟼ ∑ * k = 1 n A = n ∈ ℕ ⟼ seq 1 + 𝑒 k ∈ ℕ ⟼ A ⁡ n
33 21 32 eqtr4d ⊢ φ → seq 1 + 𝑒 k ∈ ℕ ⟼ A = n ∈ ℕ ⟼ ∑ * k = 1 n A
34 33 rneqd ⊢ φ → ran ⁡ seq 1 + 𝑒 k ∈ ℕ ⟼ A = ran ⁡ n ∈ ℕ ⟼ ∑ * k = 1 n A
35 34 supeq1d ⊢ φ → sup ran ⁡ seq 1 + 𝑒 k ∈ ℕ ⟼ A ℝ * < = sup ran ⁡ n ∈ ℕ ⟼ ∑ * k = 1 n A ℝ * <
36 6 11 35 3eqtr3d ⊢ φ → ∑ * k ∈ ℕ A = sup ran ⁡ n ∈ ℕ ⟼ ∑ * k = 1 n A ℝ * <