Metamath Proof Explorer


Theorem fmtno4

Description: The 4 th Fermat number, see remark in ApostolNT p. 7. (Contributed by AV, 13-Jun-2021)

Ref Expression
Assertion fmtno4 ⊢ FermatNo ⁡ 4 = 65537

Proof

Step Hyp Ref Expression
1 4nn0 ⊢ 4 ∈ ℕ 0
2 fmtno ⊢ 4 ∈ ℕ 0 → FermatNo ⁡ 4 = 2 2 4 + 1
3 1 2 ax-mp ⊢ FermatNo ⁡ 4 = 2 2 4 + 1
4 2exp4 ⊢ 2 4 = 16
5 4 oveq2i ⊢ 2 2 4 = 2 16
6 5 oveq1i ⊢ 2 2 4 + 1 = 2 16 + 1
7 2exp16 ⊢ 2 16 = 65536
8 7 oveq1i ⊢ 2 16 + 1 = 65536 + 1
9 6nn0 ⊢ 6 ∈ ℕ 0
10 5nn0 ⊢ 5 ∈ ℕ 0
11 9 10 deccl ⊢ 65 ∈ ℕ 0
12 11 10 deccl ⊢ 655 ∈ ℕ 0
13 3nn0 ⊢ 3 ∈ ℕ 0
14 12 13 deccl ⊢ 6553 ∈ ℕ 0
15 6p1e7 ⊢ 6 + 1 = 7
16 eqid ⊢ 65536 = 65536
17 14 9 15 16 decsuc ⊢ 65536 + 1 = 65537
18 6 8 17 3eqtri ⊢ 2 2 4 + 1 = 65537
19 3 18 eqtri ⊢ FermatNo ⁡ 4 = 65537