Metamath Proof Explorer


Theorem ackval0012

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

Ref Expression
Assertion ackval0012 ⊢ Ack ⁡ 0 ⁡ 0 Ack ⁡ 0 ⁡ 1 Ack ⁡ 0 ⁡ 2 = 1 2 3

Proof

Step Hyp Ref Expression
1 ackval0 ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1
2 oveq1 ⊢ n = 0 → n + 1 = 0 + 1
3 0p1e1 ⊢ 0 + 1 = 1
4 2 3 eqtrdi ⊢ n = 0 → n + 1 = 1
5 0nn0 ⊢ 0 ∈ ℕ 0
6 5 a1i ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → 0 ∈ ℕ 0
7 1nn0 ⊢ 1 ∈ ℕ 0
8 7 a1i ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → 1 ∈ ℕ 0
9 1 4 6 8 fvmptd3 ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → Ack ⁡ 0 ⁡ 0 = 1
10 oveq1 ⊢ n = 1 → n + 1 = 1 + 1
11 1p1e2 ⊢ 1 + 1 = 2
12 10 11 eqtrdi ⊢ n = 1 → n + 1 = 2
13 2nn0 ⊢ 2 ∈ ℕ 0
14 13 a1i ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → 2 ∈ ℕ 0
15 1 12 8 14 fvmptd3 ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → Ack ⁡ 0 ⁡ 1 = 2
16 oveq1 ⊢ n = 2 → n + 1 = 2 + 1
17 2p1e3 ⊢ 2 + 1 = 3
18 16 17 eqtrdi ⊢ n = 2 → n + 1 = 3
19 3nn0 ⊢ 3 ∈ ℕ 0
20 19 a1i ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → 3 ∈ ℕ 0
21 1 18 14 20 fvmptd3 ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → Ack ⁡ 0 ⁡ 2 = 3
22 9 15 21 oteq123d ⊢ Ack ⁡ 0 = n ∈ ℕ 0 ⟼ n + 1 → Ack ⁡ 0 ⁡ 0 Ack ⁡ 0 ⁡ 1 Ack ⁡ 0 ⁡ 2 = 1 2 3
23 1 22 ax-mp ⊢ Ack ⁡ 0 ⁡ 0 Ack ⁡ 0 ⁡ 1 Ack ⁡ 0 ⁡ 2 = 1 2 3