Metamath Proof Explorer


Theorem ballotlemfrci

Description: Reverse counting preserves a tie at the first tie. (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 ballotlemfrci COEFRCIC=0

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 1 2 3 4 5 6 7 8 ballotlemiex COEIC1M+NFCIC=0
13 12 simpld COEIC1M+N
14 elfzuz IC1M+NIC1
15 eluzfz2 IC1IC1IC
16 13 14 15 3syl COEIC1IC
17 1 2 3 4 5 6 7 8 9 10 11 ballotlemfrc COEIC1ICFRCIC=C×˙SCICIC
18 16 17 mpdan COEFRCIC=C×˙SCICIC
19 1 2 3 4 5 6 7 8 9 ballotlemsi COESCIC=1
20 19 oveq1d COESCICIC=1IC
21 20 oveq2d COEC×˙SCICIC=C×˙1IC
22 18 21 eqtrd COEFRCIC=C×˙1IC
23 fz1ssfz0 1M+N0M+N
24 23 13 sselid COEIC0M+N
25 1 2 3 4 5 6 7 8 9 10 11 ballotlemfg COEIC0M+NFCIC=C×˙1IC
26 24 25 mpdan COEFCIC=C×˙1IC
27 12 simprd COEFCIC=0
28 22 26 27 3eqtr2d COEFRCIC=0