SetDirectory["/home/travis/Documents/Workbook/Mathematica/Braidors"];
<< ../WKOIV/FreeLie2.m;
Off[General::stop]
$SeriesShowDegree=4;
Degr=8;

σ∘dρ=0 Implies the Pentagon?

As[2,1]=As[3,1]=As[3,2]=0;
As[3,1,2]=0;
A=DKS[3,As];

Bs[2,1]=Bs[3,1]=Bs[3,2]=0;
Bs[3,1,2]=0;
B=DKS[3,Bs];

Solving just  σ∘dρ=0

SeriesSolve[B,(B^(σ[3,1,2]))-(B^(σ[2,1,3]))+B≡DKS[0]];
B@{Degr} // Timing

Solving  σ∘dρ=0 and the pentagon

SeriesSolve[A,((A^(σ[3,1,2]))-(A^(σ[2,1,3]))+A≡DKS[0] )&& (-(A^(σ[1,2,34]))-(A^(σ[12,3,4]))+(A^(σ[2,3,4]))+(A^(σ[1,23,4]))+(A^(σ[1,2,3]))≡DKS[0])];
A@{Degr} // Timing

(A≡B) @{Degr}

Pentagon Implies σ∘dρ=0?

Fs[2,1]=Fs[3,1]=Fs[3,2]=0;
Fs[3,1,2]=0;
F=DKS[3,Fs];

Solving just the pentagon

SeriesSolve[F,(-(F^(σ[1,2,34]))-(F^(σ[12,3,4]))+(F^(σ[2,3,4]))+(F^(σ[1,23,4]))+(F^(σ[1,2,3]))≡DKS[0])];

F@{Degr} // Timing

(A≡F) @{Degr}

Comparing with the solution of the pentagon and  σ∘dρ=0 from above,  this gives evidence the pentagon implies the  σ∘dρ=0.