theorem Th41:
  [:rng nt,rng mt:] c= Indices M implies a * Segm(M,nt,mt) = Segm( a * M,nt,mt)
proof
  set Sa=Segm(a * M,nt,mt);
  set S=Segm(M,nt,mt);
  set aS=a*S;
A1: Indices (a*M)=Indices M by MATRIXR1:18;
A2: Indices Sa=Indices S by MATRIX_0:26;
  assume
A3: [:rng nt,rng mt:] c= Indices M;
  now
    let i,j such that
A4: [i,j] in Indices Sa;
A5: aS*(i,j)=a * (S*(i,j)) by A2,A4,MATRIX_3:def 5;
    reconsider i9=i,j9=j as Element of NAT by ORDINAL1:def 12;
A6: Sa*(i9,j9)=(a*M)*(nt.i,mt.j) by A4,Def1;
A7: S*(i9,j9)=M*(nt.i,mt.j) by A2,A4,Def1;
    [nt.i9,mt.j9] in Indices M by A3,A1,A4,Th17;
    hence Sa*(i,j)=aS*(i,j) by A6,A5,A7,MATRIX_3:def 5;
  end;
  hence thesis by MATRIX_0:27;
end;
