Metamath Proof Explorer


Theorem ackval1012

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

Ref Expression
Assertion ackval1012 ⊢ Ack ⁡ 1 ⁡ 0 Ack ⁡ 1 ⁡ 1 Ack ⁡ 1 ⁡ 2 = 2 3 4

Proof

Step Hyp Ref Expression
1 ackval1 ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2
2 oveq1 ⊢ n = 0 → n + 2 = 0 + 2
3 2cn ⊢ 2 ∈ ℂ
4 3 addlidi ⊢ 0 + 2 = 2
5 2 4 eqtrdi ⊢ n = 0 → n + 2 = 2
6 0nn0 ⊢ 0 ∈ ℕ 0
7 6 a1i ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → 0 ∈ ℕ 0
8 2nn0 ⊢ 2 ∈ ℕ 0
9 8 a1i ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → 2 ∈ ℕ 0
10 1 5 7 9 fvmptd3 ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → Ack ⁡ 1 ⁡ 0 = 2
11 oveq1 ⊢ n = 1 → n + 2 = 1 + 2
12 1p2e3 ⊢ 1 + 2 = 3
13 11 12 eqtrdi ⊢ n = 1 → n + 2 = 3
14 1nn0 ⊢ 1 ∈ ℕ 0
15 14 a1i ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → 1 ∈ ℕ 0
16 3nn0 ⊢ 3 ∈ ℕ 0
17 16 a1i ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → 3 ∈ ℕ 0
18 1 13 15 17 fvmptd3 ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → Ack ⁡ 1 ⁡ 1 = 3
19 oveq1 ⊢ n = 2 → n + 2 = 2 + 2
20 2p2e4 ⊢ 2 + 2 = 4
21 19 20 eqtrdi ⊢ n = 2 → n + 2 = 4
22 4nn0 ⊢ 4 ∈ ℕ 0
23 22 a1i ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → 4 ∈ ℕ 0
24 1 21 9 23 fvmptd3 ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → Ack ⁡ 1 ⁡ 2 = 4
25 10 18 24 oteq123d ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2 → Ack ⁡ 1 ⁡ 0 Ack ⁡ 1 ⁡ 1 Ack ⁡ 1 ⁡ 2 = 2 3 4
26 1 25 ax-mp ⊢ Ack ⁡ 1 ⁡ 0 Ack ⁡ 1 ⁡ 1 Ack ⁡ 1 ⁡ 2 = 2 3 4