Metamath Proof Explorer


Theorem ballotlem1c

Description: If the first vote is for A, the vote on the first tie is for B. (Contributed by Thierry Arnoux, 4-Apr-2017)

Ref Expression
Hypotheses ballotth.m M
ballotth.n N
ballotth.o O = c 𝒫 1 M + N | c = M
ballotth.p P = x 𝒫 O x O
ballotth.f F = c O i 1 i c 1 i c
ballotth.e E = c O | i 1 M + N 0 < F c i
ballotth.mgtn N < M
ballotth.i I = c O E inf k 1 M + N | F c k = 0 <
Assertion ballotlem1c C O E 1 C ¬ I C C

Proof

Step Hyp Ref Expression
1 ballotth.m M
2 ballotth.n N
3 ballotth.o O = c 𝒫 1 M + N | c = M
4 ballotth.p P = x 𝒫 O x O
5 ballotth.f F = c O i 1 i c 1 i c
6 ballotth.e E = c O | i 1 M + N 0 < F c i
7 ballotth.mgtn N < M
8 ballotth.i I = c O E inf k 1 M + N | F c k = 0 <
9 eldifi C O E C O
10 9 ad2antrr C O E 1 C I C C C O
11 1 2 3 4 5 6 7 8 ballotlemiex C O E I C 1 M + N F C I C = 0
12 11 simpld C O E I C 1 M + N
13 elfznn I C 1 M + N I C
14 12 13 syl C O E I C
15 14 adantr C O E 1 C I C
16 1 2 3 4 5 6 7 8 ballotlemii C O E 1 C I C 1
17 eluz2b3 I C 2 I C I C 1
18 15 16 17 sylanbrc C O E 1 C I C 2
19 uz2m1nn I C 2 I C 1
20 18 19 syl C O E 1 C I C 1
21 20 adantr C O E 1 C I C C I C 1
22 fveq2 i = 1 F C i = F C 1
23 22 breq2d i = 1 0 F C i 0 F C 1
24 elnnuz I C 1 I C 1 1
25 24 biimpi I C 1 I C 1 1
26 eluzfz1 I C 1 1 1 1 I C 1
27 20 25 26 3syl C O E 1 C 1 1 I C 1
28 27 adantr C O E 1 C I C C 1 1 I C 1
29 0le1 0 1
30 1e0p1 1 = 0 + 1
31 29 30 breqtri 0 0 + 1
32 1nn 1
33 32 a1i C O E 1
34 1 2 3 4 5 9 33 ballotlemfp1 C O E ¬ 1 C F C 1 = F C 1 1 1 1 C F C 1 = F C 1 1 + 1
35 34 simprd C O E 1 C F C 1 = F C 1 1 + 1
36 35 imp C O E 1 C F C 1 = F C 1 1 + 1
37 1m1e0 1 1 = 0
38 37 fveq2i F C 1 1 = F C 0
39 38 oveq1i F C 1 1 + 1 = F C 0 + 1
40 39 a1i C O E 1 C F C 1 1 + 1 = F C 0 + 1
41 1 2 3 4 5 ballotlemfval0 C O F C 0 = 0
42 9 41 syl C O E F C 0 = 0
43 42 adantr C O E 1 C F C 0 = 0
44 43 oveq1d C O E 1 C F C 0 + 1 = 0 + 1
45 36 40 44 3eqtrrd C O E 1 C 0 + 1 = F C 1
46 31 45 breqtrid C O E 1 C 0 F C 1
47 46 adantr C O E 1 C I C C 0 F C 1
48 23 28 47 rspcedvdw C O E 1 C I C C i 1 I C 1 0 F C i
49 df-neg 1 = 0 1
50 1 2 3 4 5 9 14 ballotlemfp1 C O E ¬ I C C F C I C = F C I C 1 1 I C C F C I C = F C I C 1 + 1
51 50 simprd C O E I C C F C I C = F C I C 1 + 1
52 51 imp C O E I C C F C I C = F C I C 1 + 1
53 11 simprd C O E F C I C = 0
54 53 adantr C O E I C C F C I C = 0
55 52 54 eqtr3d C O E I C C F C I C 1 + 1 = 0
56 0cnd C O E I C C 0
57 1cnd C O E I C C 1
58 9 adantr C O E I C C C O
59 14 nnzd C O E I C
60 59 adantr C O E I C C I C
61 1zzd C O E I C C 1
62 60 61 zsubcld C O E I C C I C 1
63 1 2 3 4 5 58 62 ballotlemfelz C O E I C C F C I C 1
64 63 zcnd C O E I C C F C I C 1
65 56 57 64 subadd2d C O E I C C 0 1 = F C I C 1 F C I C 1 + 1 = 0
66 55 65 mpbird C O E I C C 0 1 = F C I C 1
67 49 66 eqtrid C O E I C C 1 = F C I C 1
68 neg1lt0 1 < 0
69 67 68 eqbrtrrdi C O E I C C F C I C 1 < 0
70 69 adantlr C O E 1 C I C C F C I C 1 < 0
71 1 2 3 4 5 10 21 48 70 ballotlemfcc C O E 1 C I C C k 1 I C 1 F C k = 0
72 1 2 3 4 5 6 7 8 ballotlemimin C O E ¬ k 1 I C 1 F C k = 0
73 72 ad2antrr C O E 1 C I C C ¬ k 1 I C 1 F C k = 0
74 71 73 pm2.65da C O E 1 C ¬ I C C