reserve X for BCK-algebra;
reserve x,y for Element of X;
reserve IT for non empty Subset of X;

theorem Th6:
  for X being non empty BCIStr_0 holds (X is commutative
BCK-algebra iff for x,y,z being Element of X holds x\(0.X\y) = x & (x\z)\(x\y)
  = (y\z)\(y\x) )
proof
  let X be non empty BCIStr_0;
  thus X is commutative BCK-algebra implies for x,y,z being Element of X holds
  x\(0.X\y) = x & (x\z)\(x\y) = (y\z)\(y\x)
  proof
    assume
A1: X is commutative BCK-algebra;
    let x,y,z be Element of X;
    (x\(x\y))\z = (y\(y\x))\z by A1,Def1;
    then
A2: (x\z)\(x\y) = (y\(y\x))\z by A1,BCIALG_1:7
      .= (y\z)\(y\x) by A1,BCIALG_1:7;
    0.X\y = y` .= 0.X by A1,BCIALG_1:def 8;
    hence thesis by A1,A2,BCIALG_1:2;
  end;
  assume
A3: for x,y,z being Element of X holds x\(0.X\y) = x & (x\z)\(x\y) = (y\
  z)\(y\x);
A4: for x,y being Element of X holds x\(x\y) = y\(y\x)
  proof
    let x,y be Element of X;
    x\(x\y) = (x\(0.X\y))\(x\y) by A3
      .= (y\(0.X\y))\(y\x) by A3
      .= y\(y\x) by A3;
    hence thesis;
  end;
A5: for x,y being Element of X holds x\0.X = x
  proof
    let x,y be Element of X;
    0.X\(0.X\0.X) = 0.X by A3;
    hence thesis by A3;
  end;
  for x,y being Element of X holds (x\y=0.X & y\x=0.X implies x = y)
  proof
    let x,y be Element of X;
    assume x\y=0.X & y\x=0.X;
    then (x\0.X)\0.X = (y\0.X)\0.X by A3;
    then x\0.X = (y\0.X)\0.X by A5
      .= (y\0.X) by A5;
    hence x = (y\0.X) by A5
      .= y by A5;
  end;
  then
A6: X is being_BCI-4;
A7: for x being Element of X holds x\x = 0.X
  proof
    let x be Element of X;
    x = (x\0.X) by A5;
    then x\x = (0.X\0.X)\(0.X\x) by A3
      .= 0.X\(0.X\x) by A5
      .= 0.X by A3;
    hence thesis;
  end;
A8: for x being Element of X holds 0.X\x = 0.X
  proof
    let x be Element of X;
    0.X = (0.X\x)\(0.X\x) by A7
      .= 0.X\x by A3;
    hence thesis;
  end;
A9: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X
  proof
    let x,y,z be Element of X;
    ((x\y)\(x\z))\(z\y) = ((z\y)\(z\x))\(z\y) by A3
      .= ((z\y)\(z\x))\((z\y)\0.X) by A5
      .= (0.X\(z\x))\(0.X\(z\y)) by A3
      .= 0.X\(0.X\(z\y)) by A8
      .= 0.X by A8;
    hence thesis;
  end;
A10: for x,y being Element of X holds (x\(x\y))\y = 0.X
  proof
    let x,y be Element of X;
    0.X = ((x\0.X)\(x\y))\(y\0.X) by A9
      .= (x\(x\y))\(y\0.X) by A5
      .= (x\(x\y))\y by A5;
    hence thesis;
  end;
A11: X is being_I by A7;
  for x being Element of X holds x`=0.X by A8;
  hence thesis by A4,A9,A10,A11,A6,Def1,BCIALG_1:1,def 8;
end;
