Metamath Proof Explorer


Theorem ballotlemfval

Description: The value of F . (Contributed by Thierry Arnoux, 23-Nov-2016)

Ref Expression
Hypotheses ballotth.m M
ballotth.n N
ballotth.o O=c𝒫1M+N|c=M
ballotth.p P=x𝒫OxO
ballotth.f F=cOi1ic1ic
ballotlemfval.c φCO
ballotlemfval.j φJ
Assertion ballotlemfval φFCJ=1JC1JC

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 ballotlemfval.c φCO
7 ballotlemfval.j φJ
8 simpl b=Cib=C
9 8 ineq2d b=Ci1ib=1iC
10 9 fveq2d b=Ci1ib=1iC
11 8 difeq2d b=Ci1ib=1iC
12 11 fveq2d b=Ci1ib=1iC
13 10 12 oveq12d b=Ci1ib1ib=1iC1iC
14 13 mpteq2dva b=Ci1ib1ib=i1iC1iC
15 ineq2 b=c1ib=1ic
16 15 fveq2d b=c1ib=1ic
17 difeq2 b=c1ib=1ic
18 17 fveq2d b=c1ib=1ic
19 16 18 oveq12d b=c1ib1ib=1ic1ic
20 19 mpteq2dv b=ci1ib1ib=i1ic1ic
21 20 cbvmptv bOi1ib1ib=cOi1ic1ic
22 5 21 eqtr4i F=bOi1ib1ib
23 zex V
24 23 mptex i1iC1iCV
25 14 22 24 fvmpt COFC=i1iC1iC
26 6 25 syl φFC=i1iC1iC
27 oveq2 i=J1i=1J
28 27 ineq1d i=J1iC=1JC
29 28 fveq2d i=J1iC=1JC
30 27 difeq1d i=J1iC=1JC
31 30 fveq2d i=J1iC=1JC
32 29 31 oveq12d i=J1iC1iC=1JC1JC
33 32 adantl φi=J1iC1iC=1JC1JC
34 ovexd φ1JC1JCV
35 26 33 7 34 fvmptd φFCJ=1JC1JC