reserve x,y for set,
  i,j,k,l,m,n for Nat,
  K for Field,
  N for without_zero finite Subset of NAT,
  a,b for Element of K,
  A,B,B1,B2,X,X1,X2 for (Matrix of K),
  A9 for (Matrix of m,n,K),
  B9 for (Matrix of m,k,K);
reserve D for non empty set,
  bD for FinSequence of D,
  b,f,g for FinSequence of K,
  MD for Matrix of D;

theorem
  X in Solutions_of(A,B) & i in Seg width X & Col(X,i) = len X |-> 0.K
  implies Col(B,i) = len B |-> 0.K
proof
  assume that
A1: X in Solutions_of(A,B) and
A2: i in Seg width X and
A3: Col(X,i) = len X |-> 0.K;
  set LB0=len B |-> 0.K;
  consider X1 such that
A4: X = X1 and
A5: len X1 = width A and
A6: width X1 = width B and
A7: A * X1 = B by A1;
A8: now
    let k such that
A9: 1 <=k & k <= len B;
A10: k in Seg len B by A9;
    Indices B=[:Seg len B,Seg width B:] by FINSEQ_1:def 3;
    then [k,i] in Indices B by A2,A4,A6,A10,ZFMISC_1:87;
    then
A11: B*(k,i) = Line(A,k) "*" Col(X1,i) by A5,A7,MATRIX_3:def 4
      .= Sum(0.K*Line(A,k)) by A3,A4,A5,FVSUM_1:66
      .= 0.K* Sum(Line(A,k)) by FVSUM_1:73
      .= 0.K
      .= LB0.k by A10,FINSEQ_2:57;
    k in dom B by A10,FINSEQ_1:def 3;
    hence (Col(B,i)).k=LB0.k by A11,MATRIX_0:def 8;
  end;
  len Col(B,i) = len B & len LB0 = len B by CARD_1:def 7;
  hence thesis by A8;
end;
