Metamath Proof Explorer


Theorem ackval2012

Description: The Ackermann function at (2,0), (2,1), (2,2). (Contributed by AV, 4-May-2024)

Ref Expression
Assertion ackval2012 ⊢ Ack ⁡ 2 ⁡ 0 Ack ⁡ 2 ⁡ 1 Ack ⁡ 2 ⁡ 2 = 3 5 7

Proof

Step Hyp Ref Expression
1 ackval2 ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3
2 oveq2 ⊢ n = 0 → 2 ⁢ n = 2 ⋅ 0
3 2 oveq1d ⊢ n = 0 → 2 ⁢ n + 3 = 2 ⋅ 0 + 3
4 2t0e0 ⊢ 2 ⋅ 0 = 0
5 4 oveq1i ⊢ 2 ⋅ 0 + 3 = 0 + 3
6 3cn ⊢ 3 ∈ ℂ
7 6 addlidi ⊢ 0 + 3 = 3
8 5 7 eqtri ⊢ 2 ⋅ 0 + 3 = 3
9 3 8 eqtrdi ⊢ n = 0 → 2 ⁢ n + 3 = 3
10 0nn0 ⊢ 0 ∈ ℕ 0
11 10 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 0 ∈ ℕ 0
12 3nn0 ⊢ 3 ∈ ℕ 0
13 12 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 3 ∈ ℕ 0
14 1 9 11 13 fvmptd3 ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → Ack ⁡ 2 ⁡ 0 = 3
15 oveq2 ⊢ n = 1 → 2 ⁢ n = 2 ⋅ 1
16 15 oveq1d ⊢ n = 1 → 2 ⁢ n + 3 = 2 ⋅ 1 + 3
17 2t1e2 ⊢ 2 ⋅ 1 = 2
18 17 oveq1i ⊢ 2 ⋅ 1 + 3 = 2 + 3
19 2cn ⊢ 2 ∈ ℂ
20 3p2e5 ⊢ 3 + 2 = 5
21 6 19 20 addcomli ⊢ 2 + 3 = 5
22 18 21 eqtri ⊢ 2 ⋅ 1 + 3 = 5
23 16 22 eqtrdi ⊢ n = 1 → 2 ⁢ n + 3 = 5
24 1nn0 ⊢ 1 ∈ ℕ 0
25 24 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 1 ∈ ℕ 0
26 5nn0 ⊢ 5 ∈ ℕ 0
27 26 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 5 ∈ ℕ 0
28 1 23 25 27 fvmptd3 ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → Ack ⁡ 2 ⁡ 1 = 5
29 oveq2 ⊢ n = 2 → 2 ⁢ n = 2 ⋅ 2
30 29 oveq1d ⊢ n = 2 → 2 ⁢ n + 3 = 2 ⋅ 2 + 3
31 2t2e4 ⊢ 2 ⋅ 2 = 4
32 31 oveq1i ⊢ 2 ⋅ 2 + 3 = 4 + 3
33 4p3e7 ⊢ 4 + 3 = 7
34 32 33 eqtri ⊢ 2 ⋅ 2 + 3 = 7
35 30 34 eqtrdi ⊢ n = 2 → 2 ⁢ n + 3 = 7
36 2nn0 ⊢ 2 ∈ ℕ 0
37 36 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 2 ∈ ℕ 0
38 7nn0 ⊢ 7 ∈ ℕ 0
39 38 a1i ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → 7 ∈ ℕ 0
40 1 35 37 39 fvmptd3 ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → Ack ⁡ 2 ⁡ 2 = 7
41 14 28 40 oteq123d ⊢ Ack ⁡ 2 = n ∈ ℕ 0 ⟼ 2 ⁢ n + 3 → Ack ⁡ 2 ⁡ 0 Ack ⁡ 2 ⁡ 1 Ack ⁡ 2 ⁡ 2 = 3 5 7
42 1 41 ax-mp ⊢ Ack ⁡ 2 ⁡ 0 Ack ⁡ 2 ⁡ 1 Ack ⁡ 2 ⁡ 2 = 3 5 7