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

theorem
  X is BCK-implicative BCK-algebra iff for x,y being Element of X holds
  x\(x\(y\x)) = 0.X
proof
  thus X is BCK-implicative BCK-algebra implies for x,y being Element of X
  holds x\(x\(y\x)) = 0.X
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    let x,y be Element of X;
    x\(x\(y\x)) = x\x by A1,Def12;
    hence thesis by BCIALG_1:def 5;
  end;
  assume
A2: for x,y being Element of X holds x\(x\(y\x)) = 0.X;
  for x,y being Element of X holds x\(y\x)=x
  proof
    let x,y be Element of X;
A3: (x\(y\x))\x = (x\x)\(y\x) by BCIALG_1:7
      .= (y\x)` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    x\(x\(y\x)) = 0.X by A2;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  hence thesis by Def12;
end;
