Metamath Proof Explorer


Theorem aks6d1c2p1

Description: In the AKS-theorem the subset defined by E takes values in the positive integers. (Contributed by metakunt, 7-Jan-2025)

Ref Expression
Hypotheses aks6d1c2p1.1 ⊢ φ → N ∈ ℕ
aks6d1c2p1.2 ⊢ φ → P ∈ ℙ
aks6d1c2p1.3 ⊢ φ → P ∥ N
aks6d1c2p1.4 ⊢ E = k ∈ ℕ 0 , l ∈ ℕ 0 ⟼ P k ⁢ N P l
Assertion aks6d1c2p1 ⊢ φ → E : ℕ 0 × ℕ 0 ⟶ ℕ

Proof

Step Hyp Ref Expression
1 aks6d1c2p1.1 ⊢ φ → N ∈ ℕ
2 aks6d1c2p1.2 ⊢ φ → P ∈ ℙ
3 aks6d1c2p1.3 ⊢ φ → P ∥ N
4 aks6d1c2p1.4 ⊢ E = k ∈ ℕ 0 , l ∈ ℕ 0 ⟼ P k ⁢ N P l
5 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
6 2 5 syl ⊢ φ → P ∈ ℕ
7 6 adantr ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → P ∈ ℕ
8 simpr ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → a ∈ ℕ 0 × ℕ 0
9 xp1st ⊢ a ∈ ℕ 0 × ℕ 0 → 1 st ⁡ a ∈ ℕ 0
10 8 9 syl ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → 1 st ⁡ a ∈ ℕ 0
11 7 10 nnexpcld ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → P 1 st ⁡ a ∈ ℕ
12 1 6 jca ⊢ φ → N ∈ ℕ ∧ P ∈ ℕ
13 nndivdvds ⊢ N ∈ ℕ ∧ P ∈ ℕ → P ∥ N ↔ N P ∈ ℕ
14 12 13 syl ⊢ φ → P ∥ N ↔ N P ∈ ℕ
15 3 14 mpbid ⊢ φ → N P ∈ ℕ
16 15 adantr ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → N P ∈ ℕ
17 xp2nd ⊢ a ∈ ℕ 0 × ℕ 0 → 2 nd ⁡ a ∈ ℕ 0
18 8 17 syl ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → 2 nd ⁡ a ∈ ℕ 0
19 16 18 nnexpcld ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → N P 2 nd ⁡ a ∈ ℕ
20 11 19 nnmulcld ⊢ φ ∧ a ∈ ℕ 0 × ℕ 0 → P 1 st ⁡ a ⁢ N P 2 nd ⁡ a ∈ ℕ
21 vex ⊢ k ∈ V
22 vex ⊢ l ∈ V
23 21 22 op1std ⊢ a = k l → 1 st ⁡ a = k
24 23 oveq2d ⊢ a = k l → P 1 st ⁡ a = P k
25 21 22 op2ndd ⊢ a = k l → 2 nd ⁡ a = l
26 25 oveq2d ⊢ a = k l → N P 2 nd ⁡ a = N P l
27 24 26 oveq12d ⊢ a = k l → P 1 st ⁡ a ⁢ N P 2 nd ⁡ a = P k ⁢ N P l
28 27 mpompt ⊢ a ∈ ℕ 0 × ℕ 0 ⟼ P 1 st ⁡ a ⁢ N P 2 nd ⁡ a = k ∈ ℕ 0 , l ∈ ℕ 0 ⟼ P k ⁢ N P l
29 28 eqcomi ⊢ k ∈ ℕ 0 , l ∈ ℕ 0 ⟼ P k ⁢ N P l = a ∈ ℕ 0 × ℕ 0 ⟼ P 1 st ⁡ a ⁢ N P 2 nd ⁡ a
30 4 29 eqtri ⊢ E = a ∈ ℕ 0 × ℕ 0 ⟼ P 1 st ⁡ a ⁢ N P 2 nd ⁡ a
31 20 30 fmptd ⊢ φ → E : ℕ 0 × ℕ 0 ⟶ ℕ