theorem Th18:
  y<>0.SF & z<>0.SF implies x/y=(x*z)/(y*z)
proof
  assume
A1: y<>0.SF;
  assume
A2: z<>0.SF;
  thus x/y=x*1_SF*y"
    .=x*(z*z")*y" by A2,Th9
    .=(x*z)*z"*y" by GROUP_1:def 3
    .=(x*z)*(z"*y") by GROUP_1:def 3
    .=(x*z)/(y*z) by A1,A2,Th11;
end;
