Metamath Proof Explorer


Theorem fmtno5

Description: The 5 th Fermat number. (Contributed by AV, 30-Jul-2021)

Ref Expression
Assertion fmtno5 ⊢ FermatNo ⁡ 5 = 4294967297

Proof

Step Hyp Ref Expression
1 df-5 ⊢ 5 = 4 + 1
2 1 fveq2i ⊢ FermatNo ⁡ 5 = FermatNo ⁡ 4 + 1
3 4nn0 ⊢ 4 ∈ ℕ 0
4 fmtnorec1 ⊢ 4 ∈ ℕ 0 → FermatNo ⁡ 4 + 1 = FermatNo ⁡ 4 − 1 2 + 1
5 3 4 ax-mp ⊢ FermatNo ⁡ 4 + 1 = FermatNo ⁡ 4 − 1 2 + 1
6 2 5 eqtri ⊢ FermatNo ⁡ 5 = FermatNo ⁡ 4 − 1 2 + 1
7 2nn0 ⊢ 2 ∈ ℕ 0
8 3 7 deccl ⊢ 42 ∈ ℕ 0
9 9nn0 ⊢ 9 ∈ ℕ 0
10 8 9 deccl ⊢ 429 ∈ ℕ 0
11 10 3 deccl ⊢ 4294 ∈ ℕ 0
12 11 9 deccl ⊢ 42949 ∈ ℕ 0
13 6nn0 ⊢ 6 ∈ ℕ 0
14 12 13 deccl ⊢ 429496 ∈ ℕ 0
15 7nn0 ⊢ 7 ∈ ℕ 0
16 14 15 deccl ⊢ 4294967 ∈ ℕ 0
17 16 7 deccl ⊢ 42949672 ∈ ℕ 0
18 17 9 deccl ⊢ 429496729 ∈ ℕ 0
19 6p1e7 ⊢ 6 + 1 = 7
20 5nn0 ⊢ 5 ∈ ℕ 0
21 13 20 deccl ⊢ 65 ∈ ℕ 0
22 21 20 deccl ⊢ 655 ∈ ℕ 0
23 3nn0 ⊢ 3 ∈ ℕ 0
24 22 23 deccl ⊢ 6553 ∈ ℕ 0
25 1nn0 ⊢ 1 ∈ ℕ 0
26 fmtno4 ⊢ FermatNo ⁡ 4 = 65537
27 3p1e4 ⊢ 3 + 1 = 4
28 eqid ⊢ 6553 = 6553
29 22 23 27 28 decsuc ⊢ 6553 + 1 = 6554
30 6cn ⊢ 6 ∈ ℂ
31 ax-1cn ⊢ 1 ∈ ℂ
32 df-7 ⊢ 7 = 6 + 1
33 30 31 32 mvrraddi ⊢ 7 − 1 = 6
34 24 15 25 26 29 33 decsubi ⊢ FermatNo ⁡ 4 − 1 = 65536
35 34 oveq1i ⊢ FermatNo ⁡ 4 − 1 2 = 65536 2
36 fmtno5lem4 ⊢ 65536 2 = 4294967296
37 35 36 eqtri ⊢ FermatNo ⁡ 4 − 1 2 = 4294967296
38 18 13 19 37 decsuc ⊢ FermatNo ⁡ 4 − 1 2 + 1 = 4294967297
39 6 38 eqtri ⊢ FermatNo ⁡ 5 = 4294967297