Metamath Proof Explorer


Theorem sticksstones23

Description: Non-exhaustive sticks and stones. (Contributed by metakunt, 7-May-2025)

Ref Expression
Hypotheses sticksstones23.1 ⊢ φ → N ∈ ℕ 0
sticksstones23.2 ⊢ φ → S ∈ Fin
sticksstones23.3 ⊢ φ → S ≠ ∅
sticksstones23.4 ⊢ A = f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N
Assertion sticksstones23 ⊢ φ → A = ( N + S S)

Proof

Step Hyp Ref Expression
1 sticksstones23.1 ⊢ φ → N ∈ ℕ 0
2 sticksstones23.2 ⊢ φ → S ∈ Fin
3 sticksstones23.3 ⊢ φ → S ≠ ∅
4 sticksstones23.4 ⊢ A = f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N
5 4 a1i ⊢ φ → A = f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N
6 df-rab ⊢ f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N = f | f ∈ ℕ 0 S ∧ ∑ i ∈ S f ⁡ i ≤ N
7 6 a1i ⊢ φ → f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N = f | f ∈ ℕ 0 S ∧ ∑ i ∈ S f ⁡ i ≤ N
8 nn0ex ⊢ ℕ 0 ∈ V
9 8 a1i ⊢ φ → ℕ 0 ∈ V
10 elmapg ⊢ ℕ 0 ∈ V ∧ S ∈ Fin → f ∈ ℕ 0 S ↔ f : S ⟶ ℕ 0
11 9 2 10 syl2anc ⊢ φ → f ∈ ℕ 0 S ↔ f : S ⟶ ℕ 0
12 11 anbi1d ⊢ φ → f ∈ ℕ 0 S ∧ ∑ i ∈ S f ⁡ i ≤ N ↔ f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
13 12 abbidv ⊢ φ → f | f ∈ ℕ 0 S ∧ ∑ i ∈ S f ⁡ i ≤ N = f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
14 7 13 eqtrd ⊢ φ → f ∈ ℕ 0 S | ∑ i ∈ S f ⁡ i ≤ N = f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
15 5 14 eqtrd ⊢ φ → A = f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
16 15 fveq2d ⊢ φ → A = f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
17 eqid ⊢ f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N = f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N
18 1 2 3 17 sticksstones22 ⊢ φ → f | f : S ⟶ ℕ 0 ∧ ∑ i ∈ S f ⁡ i ≤ N = ( N + S S)
19 16 18 eqtrd ⊢ φ → A = ( N + S S)