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 (X is being_I &for x,y,z being Element of X holds x\(y\z)=z\(y\x) & x\0.X=x
  ))
proof
  let X be non empty BCIStr_0;
  thus X is p-Semisimple BCI-algebra implies (X is being_I & for x,y,z being
  Element of X holds x\(y\z)=z\(y\x) & x\0.X=x) by Th2,Th57;
  assume that
A1: X is being_I and
A2: for x,y,z being Element of X holds x\(y\z)=z\(y\x) & x\0.X=x;
A3: now
    let x,y,z be Element of X;
    thus x\x=0.X by A1;
    ((x\y)\(x\z))\(z\y)=(z\(x\(x\y)))\(z\y) by A2
      .=(z\(y\(x\x)))\(z\y) by A2
      .=(z\(y\0.X))\(z\y) by A1
      .=(z\y)\(z\y) by A2;
    hence ((x\y)\(x\z))\(z\y)=0.X by A1;
    (x\(x\y))\y=(y\(x\x))\y by A2
      .=(y\0.X)\y by A1
      .=y\y by A2;
    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)=y\(x\x) by A2;
      then x\(x\y)= y\0.X by A1;
      hence thesis by A2;
    end;
  end;
  now
    let x,y be Element of X;
    assume that
A4: x\y = 0.X and
    y\x = 0.X;
    x=x\0.X by A2
      .=y\(x\x) by A2,A4
      .= y\0.X by A1;
    hence x=y by A2;
  end;
  then X is being_BCI-4;
  hence thesis by A1,A3,Def26,Th1;
end;
