Metamath Proof Explorer


Theorem sticksstones16

Description: Sticks and stones with collapsed definitions for positive integers. (Contributed by metakunt, 20-Oct-2024)

Ref Expression
Hypotheses sticksstones16.1 ⊢ φ → N ∈ ℕ 0
sticksstones16.2 ⊢ φ → K ∈ ℕ
sticksstones16.3 ⊢ A = g | g : 1 … K ⟶ ℕ 0 ∧ ∑ i = 1 K g ⁡ i = N
Assertion sticksstones16 ⊢ φ → A = ( N + K - 1 K − 1 )

Proof

Step Hyp Ref Expression
1 sticksstones16.1 ⊢ φ → N ∈ ℕ 0
2 sticksstones16.2 ⊢ φ → K ∈ ℕ
3 sticksstones16.3 ⊢ A = g | g : 1 … K ⟶ ℕ 0 ∧ ∑ i = 1 K g ⁡ i = N
4 fveq2 ⊢ i = j → g ⁡ i = g ⁡ j
5 4 cbvsumv ⊢ ∑ i = 1 K g ⁡ i = ∑ j = 1 K g ⁡ j
6 5 eqeq1i ⊢ ∑ i = 1 K g ⁡ i = N ↔ ∑ j = 1 K g ⁡ j = N
7 6 anbi2i ⊢ g : 1 … K ⟶ ℕ 0 ∧ ∑ i = 1 K g ⁡ i = N ↔ g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N
8 7 abbii ⊢ g | g : 1 … K ⟶ ℕ 0 ∧ ∑ i = 1 K g ⁡ i = N = g | g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N
9 3 8 eqtri ⊢ A = g | g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N
10 9 a1i ⊢ φ → A = g | g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N
11 nfv ⊢ Ⅎ g φ
12 2 nncnd ⊢ φ → K ∈ ℂ
13 1cnd ⊢ φ → 1 ∈ ℂ
14 12 13 npcand ⊢ φ → K - 1 + 1 = K
15 14 eqcomd ⊢ φ → K = K - 1 + 1
16 15 oveq2d ⊢ φ → 1 … K = 1 … K - 1 + 1
17 16 feq2d ⊢ φ → g : 1 … K ⟶ ℕ 0 ↔ g : 1 … K - 1 + 1 ⟶ ℕ 0
18 16 sumeq1d ⊢ φ → ∑ j = 1 K g ⁡ j = ∑ j = 1 K - 1 + 1 g ⁡ j
19 18 eqeq1d ⊢ φ → ∑ j = 1 K g ⁡ j = N ↔ ∑ j = 1 K - 1 + 1 g ⁡ j = N
20 17 19 anbi12d ⊢ φ → g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N ↔ g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N
21 11 20 abbid ⊢ φ → g | g : 1 … K ⟶ ℕ 0 ∧ ∑ j = 1 K g ⁡ j = N = g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N
22 10 21 eqtrd ⊢ φ → A = g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N
23 22 fveq2d ⊢ φ → A = g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N
24 nnm1nn0 ⊢ K ∈ ℕ → K − 1 ∈ ℕ 0
25 2 24 syl ⊢ φ → K − 1 ∈ ℕ 0
26 fveq2 ⊢ j = i → g ⁡ j = g ⁡ i
27 26 cbvsumv ⊢ ∑ j = 1 K - 1 + 1 g ⁡ j = ∑ i = 1 K - 1 + 1 g ⁡ i
28 27 eqeq1i ⊢ ∑ j = 1 K - 1 + 1 g ⁡ j = N ↔ ∑ i = 1 K - 1 + 1 g ⁡ i = N
29 28 anbi2i ⊢ g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N ↔ g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ i = 1 K - 1 + 1 g ⁡ i = N
30 29 abbii ⊢ g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N = g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ i = 1 K - 1 + 1 g ⁡ i = N
31 1 25 30 sticksstones15 ⊢ φ → g | g : 1 … K - 1 + 1 ⟶ ℕ 0 ∧ ∑ j = 1 K - 1 + 1 g ⁡ j = N = ( N + K - 1 K − 1 )
32 23 31 eqtrd ⊢ φ → A = ( N + K - 1 K − 1 )