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 φA
fltltc.b φB
fltltc.c φC
fltltc.n φN3
fltltc.1 φAN+BN=CN
fltnlta.1 φA<B
Assertion fltnlta φN<A

Proof

Step Hyp Ref Expression
1 fltltc.a φA
2 fltltc.b φB
3 fltltc.c φC
4 fltltc.n φN3
5 fltltc.1 φAN+BN=CN
6 fltnlta.1 φA<B
7 eluzge3nn N3N
8 4 7 syl φN
9 8 nnred φN
10 3 nnred φC
11 2 nnred φB
12 10 11 resubcld φCB
13 uzuzle23 N3N2
14 uz2m1nn N2N1
15 4 13 14 3syl φN1
16 15 nnnn0d φN10
17 10 16 reexpcld φCN1
18 15 nnred φN1
19 11 16 reexpcld φBN1
20 18 19 remulcld φN1BN1
21 17 20 readdcld φCN1+N1BN1
22 12 21 remulcld φCBCN1+N1BN1
23 1 nnrpd φA+
24 15 nnzd φN1
25 23 24 rpexpcld φAN1+
26 22 25 rerpdivcld φCBCN1+N1BN1AN1
27 1 nnred φA
28 19 20 readdcld φBN1+N1BN1
29 12 28 remulcld φCBBN1+N1BN1
30 29 25 rerpdivcld φCBBN1+N1BN1AN1
31 12 9 remulcld φCB N
32 1cnd φ1
33 15 nncnd φN1
34 19 recnd φBN1
35 32 33 34 adddird φ1+N-1BN1=1BN1+N1BN1
36 8 nncnd φN
37 32 36 pncan3d φ1+N-1=N
38 37 oveq1d φ1+N-1BN1=NBN1
39 34 mullidd φ1BN1=BN1
40 39 oveq1d φ1BN1+N1BN1=BN1+N1BN1
41 35 38 40 3eqtr3rd φBN1+N1BN1=NBN1
42 41 oveq2d φCBBN1+N1BN1=CBNBN1
43 42 oveq1d φCBBN1+N1BN1AN1=CBNBN1AN1
44 43 30 eqeltrrd φCBNBN1AN1
45 8 nnnn0d φN0
46 45 nn0ge0d φ0N
47 1red φ1
48 1 2 3 4 5 fltltc φB<C
49 nnltp1le BCB<CB+1C
50 2 3 49 syl2anc φB<CB+1C
51 48 50 mpbid φB+1C
52 11 leidd φBB
53 10 11 47 11 51 52 lesub3d φ1CB
54 9 12 46 53 lemulge12d φNCB N
55 12 recnd φCB
56 25 rpred φAN1
57 56 recnd φAN1
58 55 36 57 mulassd φCB NAN1=CBNAN1
59 58 oveq1d φCB NAN1AN1=CBNAN1AN1
60 55 36 mulcld φCB N
61 1 nncnd φA
62 1 nnne0d φA0
63 61 62 24 expne0d φAN10
64 60 57 63 divcan4d φCB NAN1AN1=CB N
65 59 64 eqtr3d φCBNAN1AN1=CB N
66 9 56 remulcld φNAN1
67 12 66 remulcld φCBNAN1
68 42 29 eqeltrrd φCBNBN1
69 41 28 eqeltrrd φNBN1
70 difrp BCB<CCB+
71 11 10 70 syl2anc φB<CCB+
72 48 71 mpbid φCB+
73 8 nnrpd φN+
74 2 nnrpd φB+
75 23 74 15 6 ltexp1dd φAN1<BN1
76 56 19 73 75 ltmul2dd φNAN1<NBN1
77 66 69 72 76 ltmul2dd φCBNAN1<CBNBN1
78 67 68 25 77 ltdiv1dd φCBNAN1AN1<CBNBN1AN1
79 65 78 eqbrtrrd φCB N<CBNBN1AN1
80 9 31 44 54 79 lelttrd φN<CBNBN1AN1
81 80 43 breqtrrd φN<CBBN1+N1BN1AN1
82 3 nnrpd φC+
83 74 82 15 48 ltexp1dd φBN1<CN1
84 19 17 20 83 ltadd1dd φBN1+N1BN1<CN1+N1BN1
85 28 21 72 84 ltmul2dd φCBBN1+N1BN1<CBCN1+N1BN1
86 29 22 25 85 ltdiv1dd φCBBN1+N1BN1AN1<CBCN1+N1BN1AN1
87 9 30 26 81 86 lttrd φN<CBCN1+N1BN1AN1
88 27 45 reexpcld φAN
89 1 2 3 4 5 fltnltalem φCBCN1+N1BN1<AN
90 22 88 25 89 ltdiv1dd φCBCN1+N1BN1AN1<ANAN1
91 36 32 nncand φNN1=1
92 91 oveq2d φANN1=A1
93 8 nnzd φN
94 61 62 24 93 expsubd φANN1=ANAN1
95 61 exp1d φA1=A
96 92 94 95 3eqtr3d φANAN1=A
97 90 96 breqtrd φCBCN1+N1BN1AN1<A
98 9 26 27 87 97 lttrd φN<A