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 0
4 fmtnorec1 4 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 0
8 3 7 deccl 42 0
9 9nn0 9 0
10 8 9 deccl 429 0
11 10 3 deccl 4294 0
12 11 9 deccl 42949 0
13 6nn0 6 0
14 12 13 deccl 429496 0
15 7nn0 7 0
16 14 15 deccl 4294967 0
17 16 7 deccl 42949672 0
18 17 9 deccl 429496729 0
19 6p1e7 6 + 1 = 7
20 5nn0 5 0
21 13 20 deccl 65 0
22 21 20 deccl 655 0
23 3nn0 3 0
24 22 23 deccl 6553 0
25 1nn0 1 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
31 ax-1cn 1
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