
theorem Th1:
  for X being non empty BCIStr_0 holds (X is BCI-algebra iff (X is
being_I & X is being_BCI-4 &for x,y,z being Element of X holds ((x\y)\(x\z))\(z
  \y)=0.X & (x\(x\y))\y = 0.X ))
proof
  let X be non empty BCIStr_0;
  thus X is BCI-algebra implies (X is being_I & X is being_BCI-4 &for x,y,z
  being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & (x\(x\y))\y = 0.X )
  proof
    assume
A1: X is BCI-algebra;
    now
      let x,y,z be Element of X;
A2:   ((x\y)\(z\y))\(x\z)=0.X by A1,Def3;
A3:   for x,y,z being Element of X holds (x\y)\z = (x\z)\y
      proof
        let x,y,z be Element of X;
        ((x\y)\z)\((x\z)\y)=0.X & ((x\z)\y)\((x\y)\z)=0.X by A1,Def4;
        hence thesis by A1,Def7;
      end;
      then (x\(x\y))\y=(x\y)\(x\y);
      hence
      ((x\y)\(x\z))\(z\y)=0.X & (x\(x\y))\y = 0.X & x\x = 0.X & (x\y = 0.
      X & y\x = 0.X implies x = y) by A1,A2,A3,Def5,Def7;
    end;
    hence thesis;
  end;
  assume that
A4: X is being_I and
A5: X is being_BCI-4 and
A6: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & (x\(x\y
  ))\y = 0.X;
A7: for x being Element of X st x\0.X=0.X holds x=0.X
  proof
    let x be Element of X;
    assume
A8: x\0.X =0.X;
    then x` = (x \ (x\x)) \ x by A4
      .= 0.X by A6;
    hence thesis by A5,A8;
  end;
A9: for x being Element of X holds x \ 0.X = x
  proof
    let x be Element of X;
    (x\(x\0.X))\0.X = 0.X by A6;
    then
A10: x\(x\0.X) = 0.X by A7;
    0.X = (x\(x\x))\x by A6
      .= (x\0.X)\x by A4;
    hence thesis by A5,A10;
  end;
A11: for x,y,z being Element of X st x\y=0.X & y\z=0.X holds x\z=0.X
  proof
    let x,y,z be Element of X;
    assume that
A12: x\y=0.X and
A13: y\z=0.X;
    ((x\z)\(x\y))\(y\z)=0.X by A6;
    then (x\z)\(x\y)=0.X by A9,A13;
    hence thesis by A9,A12;
  end;
A14: for x,y,z being Element of X holds ((x\y)\z)\((x\z)\y) = 0.X
  proof
    let x,y,z be Element of X;
    (((x\y)\z)\((x\y)\(x\(x\z))))\((x\(x\z))\z)=0.X by A6;
    then (((x\y)\z)\((x\y)\(x\(x\z))))\0.X=0.X by A6;
    then
A15: ((x\y)\z)\((x\y)\(x\(x\z)))=0.X by A7;
    ((x\y)\(x\(x\z)))\((x\z)\y) = 0.X by A6;
    hence thesis by A11,A15;
  end;
A16: for x,y,z being Element of X st x\y=0.X holds (x\z)\(y\z)=0.X&(z\y)\(z\
  x)=0.X
  proof
    let x,y,z be Element of X;
    assume
A17: x\y=0.X;
    ((z\y)\(z\x))\(x\y)=0.X & ((x\z)\(x\y))\(y\z)=0.X by A6;
    hence thesis by A9,A17;
  end;
  for x,y,z being Element of X holds ((x\y)\(z\y))\(x\z) = 0.X
  proof
    let x,y,z be Element of X;
    ((x\y)\(x\z))\(z\y) = 0.X by A6;
    then ((x\y)\(z\y))\((x\y)\((x\y)\(x\z))) = 0.X by A16;
    then (((x\y)\(z\y))\(x\z))\(((x\y)\((x\y)\(x\z)))\(x\z)) = 0.X by A16;
    then (((x\y)\(z\y))\(x\z))\ 0.X = 0.X by A6;
    hence thesis by A7;
  end;
  hence thesis by A4,A5,A14,Def3,Def4;
end;
