reserve X for BCI-algebra;
reserve X1 for non empty Subset of X;
reserve A,I for Ideal of X;
reserve x,y,z for Element of X;
reserve a for Element of A;
reserve X for BCK-algebra;
reserve X for BCI-algebra;
reserve X for BCK-algebra;
reserve I for Ideal of X;
reserve I for Ideal of X;
reserve X for BCK-algebra;
reserve I for Ideal of X;

theorem
  I is implicative-ideal of X implies I is commutative Ideal of X & I is
  positive-implicative-ideal of X
proof
  assume
A1: I is implicative-ideal of X;
A2: for x,y being Element of X st x\y in I holds x\(y\(y\x)) in I
  proof
    let x,y be Element of X;
    (x\(y\(y\x)))\x = (x\x)\(y\(y\x)) by BCIALG_1:7
      .=(y\(y\x))` by BCIALG_1:def 5
      .=0.X by BCIALG_1:def 8;
    then (x\(y\(y\x)))<=x;
    then y\x<=y\(x\(y\(y\x))) by BCIALG_1:5;
    then (x\(y\(y\x)))\(y\(x\(y\(y\x))))<=(x\(y\(y\x)))\(y\x) by BCIALG_1:5;
    then
A3: (x\(y\(y\x)))\(y\(x\(y\(y\x))))\(x\y) <=(x\(y\(y\x)))\(y\x)\(x\y) by
BCIALG_1:5;
    (x\(y\(y\x)))\(y\x)=(x\(y\x))\(y\(y\x)) by BCIALG_1:7;
    then (x\(y\(y\x)))\(y\(x\(y\(y\x))))\(x\y)<=0.X by A3,BCIALG_1:def 3;
    then (x\(y\(y\x)))\(y\(x\(y\(y\x))))\(x\y)\0.X=0.X;
    then (x\(y\(y\x)))\(y\(x\(y\(y\x))))\(x\y)=0.X by BCIALG_1:2;
    then
A4: (x\(y\(y\x)))\(y\(x\(y\(y\x))))<=x\y;
    assume x\y in I;
    hence thesis by A1,A4,Th5,Th50;
  end;
  for x,y being Element of X st (x\y)\y in I holds x\y in I
  proof
    let x,y be Element of X;
    (x\y)\(x\(x\y))\((x\y)\y)=0.X by BCIALG_1:1;
    then
A5: (x\y)\(x\(x\y))<=((x\y)\y);
    assume (x\y)\y in I;
    hence thesis by A1,A5,Th5,Th50;
  end;
  hence thesis by A2,Th33,Th51;
end;
