theorem Th81:
  k in Seg n & c9 = z.k implies (c*z).k = c*c9
proof
  assume that
A1: k in Seg n and
A2: c9 = z.k;
  k in dom(c*z) by A1,Lm4;
  hence (c*z).k = (c multcomplex).c9 by A2,FUNCT_1:12
    .= c*c9 by Lm3;
end;
