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 & (x\a)\b = 0.X implies a=x\b & b=x\a
proof
  assume that
A1: X is alternative and
A2: (x\a)\b = 0.X;
  (x\a)\(b\b) = b` by A1,A2;
  then (x\a)\0.X = b` by Def5;
  then x\a = b` by Th2;
  then x\(x\a) = x\b by A1,Th76;
  then (x\x)\a = x\b by A1;
  then a` = x\b by Def5;
  then a=x\b by A1,Th76;
  hence thesis by A1,Th76;
end;
