theorem Th15:
  x1 <> x2 implies f*((x1,x2)-->(p1,p2)) = (x1,x2)-->(f(*)p1,f(*)p2)
proof
  set F = (x1,x2)-->(p1,p2), F9 = (x1,x2)-->(f(*)p1,f(*)p2);
  assume
A1: x1 <> x2;
  now
    let x;
    assume
A2: x in {x1,x2};
    then x = x1 or x = x2 by TARSKI:def 2;
    then F/.x = p1 & F9/.x = f(*)p1 or F/.x = p2 & F9/.x = f(*)p2 by A1,Th3;
    hence (f*F)/.x = F9/.x by A2,Def6;
  end;
  hence thesis by Th1;
end;
