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
  for X being non empty BCIStr_0 holds (X is p-Semisimple BCI-algebra
  iff for x,y,z being Element of X holds (x\y)\(x\z)=z\y & x\0.X=x )
proof
  let X be non empty BCIStr_0;
  thus X is p-Semisimple BCI-algebra implies for x,y,z being Element of X
  holds (x\y)\(x\z)=z\y & x\0.X=x by Th2,Th56;
  assume
A1: for x,y,z being Element of X holds (x\y)\(x\z)=z\y & x\0.X=x;
A2: now
    let x,y,z be Element of X;
    ((x\y)\(x\z))\(z\y)=(z\y)\(z\y) by A1
      .=((z\y)\0.X)\(z\y) by A1
      .=((z\y)\0.X)\((z\y)\0.X) by A1
      .=(0.X)` by A1;
    hence ((x\y)\(x\z))\(z\y)=0.X by A1;
    (x\(x\y))\y=((x\0.X)\(x\y))\y by A1
      .=(y\0.X)\y by A1
      .=(y\0.X)\(y\0.X) by A1
      .=(0.X)` by A1;
    hence (x\(x\y))\y = 0.X by A1;
    thus for x,y being Element of X holds x\(x\y) = y
    proof
      let x,y be Element of X;
      x\(x\y)=(x\0.X)\(x\y) by A1;
      then x\(x\y)= y\0.X by A1;
      hence thesis by A1;
    end;
  end;
  now
    let x,y be Element of X;
    assume that
A3: x\y = 0.X and
    y\x = 0.X;
    x=x\0.X by A1
      .=(x\0.X)\(x\y) by A1,A3
      .= y\0.X by A1;
    hence x=y by A1;
  end;
  then
A4: X is being_BCI-4;
  now
    let x be Element of X;
    x\x=(x\0.X)\x by A1
      .=(x\0.X)\(x\0.X) by A1
      .=(0.X)` by A1;
    hence x\x=0.X by A1;
  end;
  then X is being_I;
  hence thesis by A4,A2,Def26,Th1;
end;
