Metamath Proof Explorer


Theorem ackval3012

Description: The Ackermann function at (3,0), (3,1), (3,2). (Contributed by AV, 7-May-2024)

Ref Expression
Assertion ackval3012 ⊢ Ack ⁡ 3 ⁡ 0 Ack ⁡ 3 ⁡ 1 Ack ⁡ 3 ⁡ 2 = 5 13 29

Proof

Step Hyp Ref Expression
1 ackval3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3
2 oveq1 ⊢ n = 0 → n + 3 = 0 + 3
3 3cn ⊢ 3 ∈ ℂ
4 3 addlidi ⊢ 0 + 3 = 3
5 2 4 eqtrdi ⊢ n = 0 → n + 3 = 3
6 5 oveq2d ⊢ n = 0 → 2 n + 3 = 2 3
7 6 oveq1d ⊢ n = 0 → 2 n + 3 − 3 = 2 3 − 3
8 cu2 ⊢ 2 3 = 8
9 8 oveq1i ⊢ 2 3 − 3 = 8 − 3
10 5cn ⊢ 5 ∈ ℂ
11 5p3e8 ⊢ 5 + 3 = 8
12 11 eqcomi ⊢ 8 = 5 + 3
13 10 3 12 mvrraddi ⊢ 8 − 3 = 5
14 9 13 eqtri ⊢ 2 3 − 3 = 5
15 7 14 eqtrdi ⊢ n = 0 → 2 n + 3 − 3 = 5
16 0nn0 ⊢ 0 ∈ ℕ 0
17 16 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 0 ∈ ℕ 0
18 5nn0 ⊢ 5 ∈ ℕ 0
19 18 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 5 ∈ ℕ 0
20 1 15 17 19 fvmptd3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → Ack ⁡ 3 ⁡ 0 = 5
21 oveq1 ⊢ n = 1 → n + 3 = 1 + 3
22 ax-1cn ⊢ 1 ∈ ℂ
23 3p1e4 ⊢ 3 + 1 = 4
24 3 22 23 addcomli ⊢ 1 + 3 = 4
25 21 24 eqtrdi ⊢ n = 1 → n + 3 = 4
26 25 oveq2d ⊢ n = 1 → 2 n + 3 = 2 4
27 26 oveq1d ⊢ n = 1 → 2 n + 3 − 3 = 2 4 − 3
28 2exp4 ⊢ 2 4 = 16
29 28 oveq1i ⊢ 2 4 − 3 = 16 − 3
30 1nn0 ⊢ 1 ∈ ℕ 0
31 3nn0 ⊢ 3 ∈ ℕ 0
32 30 31 deccl ⊢ 13 ∈ ℕ 0
33 32 nn0cni ⊢ 13 ∈ ℂ
34 eqid ⊢ 13 = 13
35 3p3e6 ⊢ 3 + 3 = 6
36 30 31 31 34 35 decaddi ⊢ 13 + 3 = 16
37 36 eqcomi ⊢ 16 = 13 + 3
38 33 3 37 mvrraddi ⊢ 16 − 3 = 13
39 29 38 eqtri ⊢ 2 4 − 3 = 13
40 27 39 eqtrdi ⊢ n = 1 → 2 n + 3 − 3 = 13
41 30 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 1 ∈ ℕ 0
42 32 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 13 ∈ ℕ 0
43 1 40 41 42 fvmptd3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → Ack ⁡ 3 ⁡ 1 = 13
44 oveq1 ⊢ n = 2 → n + 3 = 2 + 3
45 2cn ⊢ 2 ∈ ℂ
46 3p2e5 ⊢ 3 + 2 = 5
47 3 45 46 addcomli ⊢ 2 + 3 = 5
48 44 47 eqtrdi ⊢ n = 2 → n + 3 = 5
49 48 oveq2d ⊢ n = 2 → 2 n + 3 = 2 5
50 49 oveq1d ⊢ n = 2 → 2 n + 3 − 3 = 2 5 − 3
51 2exp5 ⊢ 2 5 = 32
52 51 oveq1i ⊢ 2 5 − 3 = 32 − 3
53 2nn0 ⊢ 2 ∈ ℕ 0
54 9nn0 ⊢ 9 ∈ ℕ 0
55 53 54 deccl ⊢ 29 ∈ ℕ 0
56 55 nn0cni ⊢ 29 ∈ ℂ
57 eqid ⊢ 29 = 29
58 2p1e3 ⊢ 2 + 1 = 3
59 9p3e12 ⊢ 9 + 3 = 12
60 53 54 31 57 58 53 59 decaddci ⊢ 29 + 3 = 32
61 60 eqcomi ⊢ 32 = 29 + 3
62 56 3 61 mvrraddi ⊢ 32 − 3 = 29
63 52 62 eqtri ⊢ 2 5 − 3 = 29
64 50 63 eqtrdi ⊢ n = 2 → 2 n + 3 − 3 = 29
65 53 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 2 ∈ ℕ 0
66 55 a1i ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → 29 ∈ ℕ 0
67 1 64 65 66 fvmptd3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → Ack ⁡ 3 ⁡ 2 = 29
68 20 43 67 oteq123d ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3 → Ack ⁡ 3 ⁡ 0 Ack ⁡ 3 ⁡ 1 Ack ⁡ 3 ⁡ 2 = 5 13 29
69 1 68 ax-mp ⊢ Ack ⁡ 3 ⁡ 0 Ack ⁡ 3 ⁡ 1 Ack ⁡ 3 ⁡ 2 = 5 13 29