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

theorem
  for X being non empty BCIStr_0 holds (X is BCK-implicative 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))\(x\y) )
proof
  let X be non empty BCIStr_0;
  thus X is BCK-implicative 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))\(x\y)
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    then
A2: X is commutative BCK-algebra by Th34;
    let x,y,z be Element of X;
A3: x\(0.X\y) = x\y` .= x\0.X by A1,BCIALG_1:def 8
      .= x by A1,BCIALG_1:2;
    ((y\z)\(y\x))\(x\y) = ((y\z)\(x\y))\(y\x) by A1,BCIALG_1:7
      .= ((y\(x\y))\z)\(y\x) by A1,BCIALG_1:7
      .= (y\z)\(y\x) by A1,Def12
      .= (y\(y\x))\z by A1,BCIALG_1:7
      .= (x\(x\y))\z by A2,Def1
      .= (x\z)\(x\y) by A1,BCIALG_1:7;
    hence thesis by A3;
  end;
  assume
A4: for x,y,z being Element of X holds x\(0.X\y) = x & (x\z)\(x\y) = ((y
  \z)\(y\x))\(x\y);
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 A4;
    hence thesis by A4;
  end;
A6: for x,y being Element of X st x\y=0.X & y\x=0.X holds 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)\0.X by A4
      .= (y\0.X)\0.X by A5;
    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;
A7: for x,y being Element of X holds x\(x\y) = (y\(y\x))\(x\y)
  proof
    let x,y be Element of X;
    x\(x\y) = (x\0.X)\(x\y) by A5
      .= ((y\0.X)\(y\x))\(x\y) by A4;
    hence thesis by A5;
  end;
A8: for y being Element of X holds y\y = 0.X
  proof
    let y be Element of X;
    0.X\(0.X\y) = (y\(y\0.X))\(0.X\y) by A7
      .= (y\y)\(0.X\y) by A5
      .= y\y by A4;
    hence thesis by A4;
  end;
A9: for x being Element of X holds 0.X\x = 0.X
  proof
    let x be Element of X;
    (0.X\x)\(0.X\x) = 0.X\x by A4;
    hence thesis by A8;
  end;
A10: for x,y being Element of X holds (x\y)\x = 0.X
  proof
    let x,y be Element of X;
    (x\y)\x = (x\y)\(x\0.X) by A5
      .= ((0.X\y)\(0.X\x))\(x\0.X) by A4
      .= ((0.X)\(0.X\x))\(x\0.X) by A9
      .= 0.X\(x\0.X) by A9;
    hence thesis by A9;
  end;
A11: for x,y,z being Element of X holds ((x\z)\(x\y))\((y\z)\(y\x)) = 0.X
  proof
    let x,y,z be Element of X;
    (((y\z)\(y\x))\(x\y))\((y\z)\(y\x)) = 0.X by A10;
    hence thesis by A4;
  end;
A12: for x,y,z being Element of X holds ((x\z)\(x\y)) = ((y\z)\(y\x))
  proof
    let x,y,z be Element of X;
    ((y\z)\(y\x))\((x\z)\(x\y)) = 0.X & ((x\z)\(x\y))\((y\z)\(y\x)) = 0.X
    by A11;
    hence thesis by A6;
  end;
  then
A13: X is commutative BCK-algebra by A4,Th6;
  for x,y being Element of X holds (x\(x\y))\(x\y) = (y\(y\x))
  proof
    let x,y be Element of X;
    x\(x\y) = (y\(y\x)) by A13,Def1;
    hence thesis by A7;
  end;
  hence thesis by A4,A12,Th6,Th35;
end;
