Metamath Proof Explorer


Theorem ackval41a

Description: The Ackermann function at (4,1). (Contributed by AV, 9-May-2024)

Ref Expression
Assertion ackval41a ⊢ Ack ⁡ 4 ⁡ 1 = 2 16 − 3

Proof

Step Hyp Ref Expression
1 df-4 ⊢ 4 = 3 + 1
2 1 fveq2i ⊢ Ack ⁡ 4 = Ack ⁡ 3 + 1
3 1e0p1 ⊢ 1 = 0 + 1
4 2 3 fveq12i ⊢ Ack ⁡ 4 ⁡ 1 = Ack ⁡ 3 + 1 ⁡ 0 + 1
5 3nn0 ⊢ 3 ∈ ℕ 0
6 0nn0 ⊢ 0 ∈ ℕ 0
7 ackvalsucsucval ⊢ 3 ∈ ℕ 0 ∧ 0 ∈ ℕ 0 → Ack ⁡ 3 + 1 ⁡ 0 + 1 = Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 0
8 5 6 7 mp2an ⊢ Ack ⁡ 3 + 1 ⁡ 0 + 1 = Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 0
9 3p1e4 ⊢ 3 + 1 = 4
10 9 fveq2i ⊢ Ack ⁡ 3 + 1 = Ack ⁡ 4
11 10 fveq1i ⊢ Ack ⁡ 3 + 1 ⁡ 0 = Ack ⁡ 4 ⁡ 0
12 ackval40 ⊢ Ack ⁡ 4 ⁡ 0 = 13
13 11 12 eqtri ⊢ Ack ⁡ 3 + 1 ⁡ 0 = 13
14 13 fveq2i ⊢ Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 0 = Ack ⁡ 3 ⁡ 13
15 1nn0 ⊢ 1 ∈ ℕ 0
16 15 5 deccl ⊢ 13 ∈ ℕ 0
17 oveq1 ⊢ n = 13 → n + 3 = 13 + 3
18 17 oveq2d ⊢ n = 13 → 2 n + 3 = 2 13 + 3
19 18 oveq1d ⊢ n = 13 → 2 n + 3 − 3 = 2 13 + 3 − 3
20 eqid ⊢ 13 = 13
21 3p3e6 ⊢ 3 + 3 = 6
22 15 5 5 20 21 decaddi ⊢ 13 + 3 = 16
23 22 oveq2i ⊢ 2 13 + 3 = 2 16
24 23 oveq1i ⊢ 2 13 + 3 − 3 = 2 16 − 3
25 19 24 eqtrdi ⊢ n = 13 → 2 n + 3 − 3 = 2 16 − 3
26 ackval3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3
27 ovex ⊢ 2 16 − 3 ∈ V
28 25 26 27 fvmpt ⊢ 13 ∈ ℕ 0 → Ack ⁡ 3 ⁡ 13 = 2 16 − 3
29 16 28 ax-mp ⊢ Ack ⁡ 3 ⁡ 13 = 2 16 − 3
30 14 29 eqtri ⊢ Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 0 = 2 16 − 3
31 8 30 eqtri ⊢ Ack ⁡ 3 + 1 ⁡ 0 + 1 = 2 16 − 3
32 4 31 eqtri ⊢ Ack ⁡ 4 ⁡ 1 = 2 16 − 3