Metamath Proof Explorer


Theorem ballotlemfg

Description: Express the value of ( FC ) in terms of .^ . (Contributed by Thierry Arnoux, 21-Apr-2017)

Ref Expression
Hypotheses ballotth.m M
ballotth.n N
ballotth.o O=c𝒫1M+N|c=M
ballotth.p P=x𝒫OxO
ballotth.f F=cOi1ic1ic
ballotth.e E=cO|i1M+N0<Fci
ballotth.mgtn N<M
ballotth.i I=cOEsupk1M+N|Fck=0<
ballotth.s S=cOEi1M+NifiIcIc+1-ii
ballotth.r R=cOEScc
ballotlemg ×˙=uFin,vFinvuvu
Assertion ballotlemfg COEJ0M+NFCJ=C×˙1J

Proof

Step Hyp Ref Expression
1 ballotth.m M
2 ballotth.n N
3 ballotth.o O=c𝒫1M+N|c=M
4 ballotth.p P=x𝒫OxO
5 ballotth.f F=cOi1ic1ic
6 ballotth.e E=cO|i1M+N0<Fci
7 ballotth.mgtn N<M
8 ballotth.i I=cOEsupk1M+N|Fck=0<
9 ballotth.s S=cOEi1M+NifiIcIc+1-ii
10 ballotth.r R=cOEScc
11 ballotlemg ×˙=uFin,vFinvuvu
12 eldifi COECO
13 12 adantr COEJ0M+NCO
14 elfzelz J0M+NJ
15 14 adantl COEJ0M+NJ
16 1 2 3 4 5 13 15 ballotlemfval COEJ0M+NFCJ=1JC1JC
17 fzfi 1M+NFin
18 1 2 3 ballotlemelo COC1M+NC=M
19 18 simplbi COC1M+N
20 ssfi 1M+NFinC1M+NCFin
21 17 19 20 sylancr COCFin
22 13 21 syl COEJ0M+NCFin
23 fzfid COEJ0M+N1JFin
24 1 2 3 4 5 6 7 8 9 10 11 ballotlemgval CFin1JFinC×˙1J=1JC1JC
25 22 23 24 syl2anc COEJ0M+NC×˙1J=1JC1JC
26 16 25 eqtr4d COEJ0M+NFCJ=C×˙1J