reserve X for BCI-algebra;
reserve x,y,z,u,a,b for Element of X;
reserve IT for non empty Subset of X;

theorem
  X is alternative & a\x=b\x implies a=b
proof
  assume that
A1: X is alternative and
A2: a\x=b\x;
  a\(x\x)=(b\x)\x by A1,A2;
  then a\(x\x)=b\(x\x) by A1;
  then a\0.X=b\(x\x) by Def5;
  then a\0.X=b\0.X by Def5;
  then a=b\0.X by Th2;
  hence thesis by Th2;
end;
