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 ::T1.1.6
  for X being non empty BCIStr_0 holds (X is BCI-algebra iff (X is
being_BCI-4 & for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.
  X = x ))
proof
  let X be non empty BCIStr_0;
  thus X is BCI-algebra implies (X is being_BCI-4 & for x,y,z being Element of
  X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X = x ) by Th1,Th2;
  thus (X is being_BCI-4 & for x,y,z being Element of X holds ((x\y)\(x\z))\(z
  \y)=0.X & x\0.X = x )implies X is BCI-algebra
  proof
    assume that
A1: X is being_BCI-4 and
A2: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X = x;
A3: X is being_I
    proof
      let x be Element of X;
      x\x=(x\x)\0.X by A2;
      then x\x=((x\0.X)\x)\0.X by A2;
      then x\x=((x\0.X)\(x\0.X))\0.X by A2;
      then x\x=((x\0.X)\(x\0.X))\(0.X)` by A2;
      hence thesis by A2;
    end;
    now
      let x,y,z be Element of X;
      ((x\0.X)\(x\y))\(y\0.X)=0.X by A2;
      then (x\(x\y))\(y\0.X)=0.X by A2;
      hence ((x\y)\(x\z))\(z\y)=0.X & (x\(x\y))\y = 0.X by A2;
    end;
    hence thesis by A1,A3,Th1;
  end;
end;
