theorem
  X is alternative & x\a=x\b implies a=b
proof
  assume that
A1: X is alternative and
A2: x\a=x\b;
  (x\x)\a=x\(x\b) by A1,A2;
  then (x\x)\a=(x\x)\b by A1;
  then a`=(x\x)\b by Def5;
  then a`=b` by Def5;
  then a=b` by A1,Th76;
  hence thesis by A1,Th76;
end;
