Metamath Proof Explorer


Theorem fltnlta

Description: In a Fermat counterexample, the exponent N is less than all three numbers ( A , B , and C ). Note that A < B (hypothesis) and B < C ( fltltc ). See https://youtu.be/EymVXkPWxyc for an outline. (Contributed by SN, 24-Aug-2023)

Ref Expression
Hypotheses fltltc.a φ A
fltltc.b φ B
fltltc.c φ C
fltltc.n φ N 3
fltltc.1 φ A N + B N = C N
fltnlta.1 φ A < B
Assertion fltnlta φ N < A

Proof

Step Hyp Ref Expression
1 fltltc.a φ A
2 fltltc.b φ B
3 fltltc.c φ C
4 fltltc.n φ N 3
5 fltltc.1 φ A N + B N = C N
6 fltnlta.1 φ A < B
7 eluzge3nn N 3 N
8 4 7 syl φ N
9 8 nnred φ N
10 3 nnred φ C
11 2 nnred φ B
12 10 11 resubcld φ C B
13 uzuzle23 N 3 N 2
14 uz2m1nn N 2 N 1
15 4 13 14 3syl φ N 1
16 15 nnnn0d φ N 1 0
17 10 16 reexpcld φ C N 1
18 15 nnred φ N 1
19 11 16 reexpcld φ B N 1
20 18 19 remulcld φ N 1 B N 1
21 17 20 readdcld φ C N 1 + N 1 B N 1
22 12 21 remulcld φ C B C N 1 + N 1 B N 1
23 1 nnrpd φ A +
24 15 nnzd φ N 1
25 23 24 rpexpcld φ A N 1 +
26 22 25 rerpdivcld φ C B C N 1 + N 1 B N 1 A N 1
27 1 nnred φ A
28 19 20 readdcld φ B N 1 + N 1 B N 1
29 12 28 remulcld φ C B B N 1 + N 1 B N 1
30 29 25 rerpdivcld φ C B B N 1 + N 1 B N 1 A N 1
31 12 9 remulcld φ C B N
32 1cnd φ 1
33 15 nncnd φ N 1
34 19 recnd φ B N 1
35 32 33 34 adddird φ 1 + N - 1 B N 1 = 1 B N 1 + N 1 B N 1
36 8 nncnd φ N
37 32 36 pncan3d φ 1 + N - 1 = N
38 37 oveq1d φ 1 + N - 1 B N 1 = N B N 1
39 34 mulid2d φ 1 B N 1 = B N 1
40 39 oveq1d φ 1 B N 1 + N 1 B N 1 = B N 1 + N 1 B N 1
41 35 38 40 3eqtr3rd φ B N 1 + N 1 B N 1 = N B N 1
42 41 oveq2d φ C B B N 1 + N 1 B N 1 = C B N B N 1
43 42 oveq1d φ C B B N 1 + N 1 B N 1 A N 1 = C B N B N 1 A N 1
44 43 30 eqeltrrd φ C B N B N 1 A N 1
45 8 nnnn0d φ N 0
46 45 nn0ge0d φ 0 N
47 1red φ 1
48 1 2 3 4 5 fltltc φ B < C
49 nnltp1le B C B < C B + 1 C
50 2 3 49 syl2anc φ B < C B + 1 C
51 48 50 mpbid φ B + 1 C
52 11 leidd φ B B
53 10 11 47 11 51 52 lesub3d φ 1 C B
54 9 12 46 53 lemulge12d φ N C B N
55 12 recnd φ C B
56 25 rpred φ A N 1
57 56 recnd φ A N 1
58 55 36 57 mulassd φ C B N A N 1 = C B N A N 1
59 58 oveq1d φ C B N A N 1 A N 1 = C B N A N 1 A N 1
60 55 36 mulcld φ C B N
61 1 nncnd φ A
62 1 nnne0d φ A 0
63 61 62 24 expne0d φ A N 1 0
64 60 57 63 divcan4d φ C B N A N 1 A N 1 = C B N
65 59 64 eqtr3d φ C B N A N 1 A N 1 = C B N
66 9 56 remulcld φ N A N 1
67 12 66 remulcld φ C B N A N 1
68 42 29 eqeltrrd φ C B N B N 1
69 41 28 eqeltrrd φ N B N 1
70 difrp B C B < C C B +
71 11 10 70 syl2anc φ B < C C B +
72 48 71 mpbid φ C B +
73 8 nnrpd φ N +
74 2 nnrpd φ B +
75 23 74 15 6 ltexp1dd φ A N 1 < B N 1
76 56 19 73 75 ltmul2dd φ N A N 1 < N B N 1
77 66 69 72 76 ltmul2dd φ C B N A N 1 < C B N B N 1
78 67 68 25 77 ltdiv1dd φ C B N A N 1 A N 1 < C B N B N 1 A N 1
79 65 78 eqbrtrrd φ C B N < C B N B N 1 A N 1
80 9 31 44 54 79 lelttrd φ N < C B N B N 1 A N 1
81 80 43 breqtrrd φ N < C B B N 1 + N 1 B N 1 A N 1
82 3 nnrpd φ C +
83 74 82 15 48 ltexp1dd φ B N 1 < C N 1
84 19 17 20 83 ltadd1dd φ B N 1 + N 1 B N 1 < C N 1 + N 1 B N 1
85 28 21 72 84 ltmul2dd φ C B B N 1 + N 1 B N 1 < C B C N 1 + N 1 B N 1
86 29 22 25 85 ltdiv1dd φ C B B N 1 + N 1 B N 1 A N 1 < C B C N 1 + N 1 B N 1 A N 1
87 9 30 26 81 86 lttrd φ N < C B C N 1 + N 1 B N 1 A N 1
88 27 45 reexpcld φ A N
89 1 2 3 4 5 fltnltalem φ C B C N 1 + N 1 B N 1 < A N
90 22 88 25 89 ltdiv1dd φ C B C N 1 + N 1 B N 1 A N 1 < A N A N 1
91 36 32 nncand φ N N 1 = 1
92 91 oveq2d φ A N N 1 = A 1
93 8 nnzd φ N
94 61 62 24 93 expsubd φ A N N 1 = A N A N 1
95 61 exp1d φ A 1 = A
96 92 94 95 3eqtr3d φ A N A N 1 = A
97 90 96 breqtrd φ C B C N 1 + N 1 B N 1 A N 1 < A
98 9 26 27 87 97 lttrd φ N < A