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 ) = 6 5 5 3 7

Proof

Step Hyp Ref Expression
1 4nn0 4 ∈ ℕ0
2 fmtno ( 4 ∈ ℕ0 → ( FermatNo ‘ 4 ) = ( ( 2 ↑ ( 2 ↑ 4 ) ) + 1 ) )
3 1 2 ax-mp ( FermatNo ‘ 4 ) = ( ( 2 ↑ ( 2 ↑ 4 ) ) + 1 )
4 2exp4 ( 2 ↑ 4 ) = 1 6
5 4 oveq2i ( 2 ↑ ( 2 ↑ 4 ) ) = ( 2 ↑ 1 6 )
6 5 oveq1i ( ( 2 ↑ ( 2 ↑ 4 ) ) + 1 ) = ( ( 2 ↑ 1 6 ) + 1 )
7 2exp16 ( 2 ↑ 1 6 ) = 6 5 5 3 6
8 7 oveq1i ( ( 2 ↑ 1 6 ) + 1 ) = ( 6 5 5 3 6 + 1 )
9 6nn0 6 ∈ ℕ0
10 5nn0 5 ∈ ℕ0
11 9 10 deccl 6 5 ∈ ℕ0
12 11 10 deccl 6 5 5 ∈ ℕ0
13 3nn0 3 ∈ ℕ0
14 12 13 deccl 6 5 5 3 ∈ ℕ0
15 6p1e7 ( 6 + 1 ) = 7
16 eqid 6 5 5 3 6 = 6 5 5 3 6
17 14 9 15 16 decsuc ( 6 5 5 3 6 + 1 ) = 6 5 5 3 7
18 6 8 17 3eqtri ( ( 2 ↑ ( 2 ↑ 4 ) ) + 1 ) = 6 5 5 3 7
19 3 18 eqtri ( FermatNo ‘ 4 ) = 6 5 5 3 7