theorem Th22:
  i in dom B1 & i <> len B1 implies Sum lmlt (Line(Jordan_block(L,
  len B1),i),B1)= L*(B1/.i)+B1/.(i+1)
proof
  assume that
A1: i in dom B1 and
A2: i<>len B1;
  set N=len B1;
A3: dom B1=Seg N by FINSEQ_1:def 3;
  i<=N by A1,FINSEQ_3:25;
  then i<N by A2,XXREAL_0:1;
  then 1<=i+1 & i+1<=N by NAT_1:11,13;
  then
A4: i+1 in dom B1 by A3;
  set ONE=1.(K,N);
A5: len Line(ONE,i+1)=width ONE by CARD_1:def 7;
  width ONE=N by MATRIX_0:24;
  then
A6: dom Line(ONE,i+1)= dom B1 by A5,FINSEQ_3:29;
  len (L*Line(ONE,i))=len Line(ONE,i) & len Line(ONE,i)=width ONE by
CARD_1:def 7,MATRIXR1:16;
  then dom (L*Line(ONE,i))=dom Line(ONE,i+1) by A5,FINSEQ_3:29;
  then
A7: dom lmlt(L*Line(ONE,i),B1) =dom B1 by A6,MATRLIN:12;
  dom lmlt(Line(ONE,i+1),B1) =dom B1 by A6,MATRLIN:12;
  then
A8: len lmlt (L*Line(ONE,i),B1)=len lmlt(Line(ONE,i+1),B1) by A7,FINSEQ_3:29;
  thus Sum lmlt (Line(Jordan_block(L,len B1),i),B1) = Sum lmlt (L*Line(ONE,i)+
  Line(ONE,i+1),B1) by A1,A2,A3,Th4
    .= Sum (lmlt (L*Line(ONE,i),B1)+lmlt(Line(ONE,i+1),B1)) by MATRLIN2:7
    .= (Sum lmlt (L*Line(ONE,i),B1))+(Sum lmlt(Line(ONE,i+1),B1)) by A8,
MATRLIN2:10
    .= (L*Sum lmlt (Line(ONE,i),B1))+(Sum lmlt(Line(ONE,i+1),B1)) by
MATRLIN2:13
    .= L*(B1/.i)+(Sum lmlt(Line(ONE,i+1),B1)) by A1,MATRLIN2:16
    .= L*(B1/.i)+B1/.(i+1) by A4,MATRLIN2:16;
end;
