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 0
2 fmtno 4 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 0
10 5nn0 5 0
11 9 10 deccl 65 0
12 11 10 deccl 655 0
13 3nn0 3 0
14 12 13 deccl 6553 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