theorem Th22:
  x0,x1,x2 are_mutually_distinct implies [!AffineMap(a,b),x0,x1, x2!]=0
proof
  assume
A1: x0,x1,x2 are_mutually_distinct;
  then
A2: x1<>x2 by ZFMISC_1:def 5;
  x0<>x1 by A1,ZFMISC_1:def 5;
  then
  [!AffineMap(a,b),x0,x1,x2!] = (a-[!AffineMap(a,b),x1,x2!])/(x0-x2) by Th21
    .= (a-a)/(x0-x2) by A2,Th21;
  hence thesis;
end;
