Metamath Proof Explorer


Theorem ballotlemro

Description: Range of R is included in O . (Contributed by Thierry Arnoux, 17-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 sup k 1 M + N | F c k = 0 <
ballotth.s S = c O E i 1 M + N if i I c I c + 1 - i i
ballotth.r R = c O E S c c
Assertion ballotlemro C O E R C O

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 ballotth.s S = c O E i 1 M + N if i I c I c + 1 - i i
10 ballotth.r R = c O E S c c
11 1 2 3 4 5 6 7 8 9 10 ballotlemrval C O E R C = S C C
12 imassrn S C C ran S C
13 1 2 3 4 5 6 7 8 9 ballotlemsf1o C O E S C : 1 M + N 1-1 onto 1 M + N S C -1 = S C
14 13 simpld C O E S C : 1 M + N 1-1 onto 1 M + N
15 f1ofo S C : 1 M + N 1-1 onto 1 M + N S C : 1 M + N onto 1 M + N
16 forn S C : 1 M + N onto 1 M + N ran S C = 1 M + N
17 14 15 16 3syl C O E ran S C = 1 M + N
18 12 17 sseqtrid C O E S C C 1 M + N
19 11 18 eqsstrd C O E R C 1 M + N
20 f1of1 S C : 1 M + N 1-1 onto 1 M + N S C : 1 M + N 1-1 1 M + N
21 14 20 syl C O E S C : 1 M + N 1-1 1 M + N
22 eldifi C O E C O
23 1 2 3 ballotlemelo C O C 1 M + N C = M
24 22 23 sylib C O E C 1 M + N C = M
25 24 simpld C O E C 1 M + N
26 id C O E C O E
27 f1imaeng S C : 1 M + N 1-1 1 M + N C 1 M + N C O E S C C C
28 21 25 26 27 syl3anc C O E S C C C
29 11 28 eqbrtrd C O E R C C
30 hasheni R C C R C = C
31 29 30 syl C O E R C = C
32 24 simprd C O E C = M
33 31 32 eqtrd C O E R C = M
34 1 2 3 ballotlemelo R C O R C 1 M + N R C = M
35 19 33 34 sylanbrc C O E R C O