Metamath Proof Explorer


Theorem ackval1

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

Ref Expression
Assertion ackval1 ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2

Proof

Step Hyp Ref Expression
1 1e0p1 ⊢ 1 = 0 + 1
2 1 fveq2i ⊢ Ack ⁡ 1 = Ack ⁡ 0 + 1
3 0nn0 ⊢ 0 ∈ ℕ 0
4 ackvalsuc1mpt ⊢ 0 ∈ ℕ 0 → Ack ⁡ 0 + 1 = n ∈ ℕ 0 ⟼ IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 ⁡ 1
5 3 4 ax-mp ⊢ Ack ⁡ 0 + 1 = n ∈ ℕ 0 ⟼ IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 ⁡ 1
6 peano2nn0 ⊢ n ∈ ℕ 0 → n + 1 ∈ ℕ 0
7 1nn0 ⊢ 1 ∈ ℕ 0
8 ackval0 ⊢ Ack ⁡ 0 = i ∈ ℕ 0 ⟼ i + 1
9 8 itcovalpc ⊢ n + 1 ∈ ℕ 0 ∧ 1 ∈ ℕ 0 → IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 = i ∈ ℕ 0 ⟼ i + 1 ⁢ n + 1
10 6 7 9 sylancl ⊢ n ∈ ℕ 0 → IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 = i ∈ ℕ 0 ⟼ i + 1 ⁢ n + 1
11 nn0cn ⊢ n + 1 ∈ ℕ 0 → n + 1 ∈ ℂ
12 6 11 syl ⊢ n ∈ ℕ 0 → n + 1 ∈ ℂ
13 12 mullidd ⊢ n ∈ ℕ 0 → 1 ⁢ n + 1 = n + 1
14 13 oveq2d ⊢ n ∈ ℕ 0 → i + 1 ⁢ n + 1 = i + n + 1
15 14 mpteq2dv ⊢ n ∈ ℕ 0 → i ∈ ℕ 0 ⟼ i + 1 ⁢ n + 1 = i ∈ ℕ 0 ⟼ i + n + 1
16 10 15 eqtrd ⊢ n ∈ ℕ 0 → IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 = i ∈ ℕ 0 ⟼ i + n + 1
17 16 fveq1d ⊢ n ∈ ℕ 0 → IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 ⁡ 1 = i ∈ ℕ 0 ⟼ i + n + 1 ⁡ 1
18 eqidd ⊢ n ∈ ℕ 0 → i ∈ ℕ 0 ⟼ i + n + 1 = i ∈ ℕ 0 ⟼ i + n + 1
19 oveq1 ⊢ i = 1 → i + n + 1 = 1 + n + 1
20 19 adantl ⊢ n ∈ ℕ 0 ∧ i = 1 → i + n + 1 = 1 + n + 1
21 7 a1i ⊢ n ∈ ℕ 0 → 1 ∈ ℕ 0
22 ovexd ⊢ n ∈ ℕ 0 → 1 + n + 1 ∈ V
23 18 20 21 22 fvmptd ⊢ n ∈ ℕ 0 → i ∈ ℕ 0 ⟼ i + n + 1 ⁡ 1 = 1 + n + 1
24 1cnd ⊢ n ∈ ℕ 0 → 1 ∈ ℂ
25 nn0cn ⊢ n ∈ ℕ 0 → n ∈ ℂ
26 peano2cn ⊢ n ∈ ℂ → n + 1 ∈ ℂ
27 25 26 syl ⊢ n ∈ ℕ 0 → n + 1 ∈ ℂ
28 24 27 addcomd ⊢ n ∈ ℕ 0 → 1 + n + 1 = n + 1 + 1
29 25 24 24 addassd ⊢ n ∈ ℕ 0 → n + 1 + 1 = n + 1 + 1
30 1p1e2 ⊢ 1 + 1 = 2
31 30 oveq2i ⊢ n + 1 + 1 = n + 2
32 31 a1i ⊢ n ∈ ℕ 0 → n + 1 + 1 = n + 2
33 28 29 32 3eqtrd ⊢ n ∈ ℕ 0 → 1 + n + 1 = n + 2
34 17 23 33 3eqtrd ⊢ n ∈ ℕ 0 → IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 ⁡ 1 = n + 2
35 34 mpteq2ia ⊢ n ∈ ℕ 0 ⟼ IterComp ⁡ Ack ⁡ 0 ⁡ n + 1 ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2
36 2 5 35 3eqtri ⊢ Ack ⁡ 1 = n ∈ ℕ 0 ⟼ n + 2