Metamath Proof Explorer


Theorem ballotlemsup

Description: The set of zeroes of F satisfies the conditions to have a supremum. (Contributed by Thierry Arnoux, 1-Dec-2016) (Revised by AV, 6-Oct-2020)

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 sup k 1 M + N | F c k = 0 <
Assertion ballotlemsup C O E z w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w

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 sup k 1 M + N | F c k = 0 <
9 fzfi 1 M + N Fin
10 ssrab2 k 1 M + N | F C k = 0 1 M + N
11 ssfi 1 M + N Fin k 1 M + N | F C k = 0 1 M + N k 1 M + N | F C k = 0 Fin
12 9 10 11 mp2an k 1 M + N | F C k = 0 Fin
13 12 a1i C O E k 1 M + N | F C k = 0 Fin
14 1 2 3 4 5 6 7 ballotlem5 C O E k 1 M + N F C k = 0
15 rabn0 k 1 M + N | F C k = 0 k 1 M + N F C k = 0
16 14 15 sylibr C O E k 1 M + N | F C k = 0
17 fz1ssnn 1 M + N
18 nnssre
19 17 18 sstri 1 M + N
20 10 19 sstri k 1 M + N | F C k = 0
21 20 a1i C O E k 1 M + N | F C k = 0
22 13 16 21 3jca C O E k 1 M + N | F C k = 0 Fin k 1 M + N | F C k = 0 k 1 M + N | F C k = 0
23 ltso < Or
24 22 23 jctil C O E < Or k 1 M + N | F C k = 0 Fin k 1 M + N | F C k = 0 k 1 M + N | F C k = 0
25 fiinf2g < Or k 1 M + N | F C k = 0 Fin k 1 M + N | F C k = 0 k 1 M + N | F C k = 0 z k 1 M + N | F C k = 0 w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w
26 20 sseli z k 1 M + N | F C k = 0 z
27 26 anim1i z k 1 M + N | F C k = 0 w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w z w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w
28 27 reximi2 z k 1 M + N | F C k = 0 w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w z w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w
29 24 25 28 3syl C O E z w k 1 M + N | F C k = 0 ¬ w < z w z < w y k 1 M + N | F C k = 0 y < w