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 Th51:
  I is positive-implicative-ideal of X iff for x,y being Element
  of X st (x\y)\y in I holds x\y in I
proof
  thus I is positive-implicative-ideal of X implies for x,y being Element of X
  st (x\y)\y in I holds x\y in I
  proof
    assume
A1: I is positive-implicative-ideal of X;
    let x,y be Element of X;
    y\y =0.X by BCIALG_1:def 5;
    then
A2: y\y in I by A1,Def8;
    assume (x\y)\y in I;
    hence thesis by A1,A2,Def8;
  end;
  thus (for x,y being Element of X st (x\y)\y in I holds x\y in I) implies I
  is positive-implicative-ideal of X
  proof
    assume
A3: for x,y being Element of X st (x\y)\y in I holds x\y in I;
A4: for x,y,z being Element of X st (x\y)\z in I & y\z in I holds x\z in I
    proof
      let x,y,z be Element of X;
      assume that
A5:   (x\y)\z in I and
A6:   y\z in I;
      ((x\z)\z)\((x\y)\z)\((x\z)\(x\y))=0.X by BCIALG_1:def 3;
      then ((x\z)\z)\((x\y)\z)<=(x\z)\(x\y);
      then ((x\z)\z)\((x\y)\z)\(y\z)<=(x\z)\(x\y)\(y\z) by BCIALG_1:5;
      then ((x\z)\z)\((x\y)\z)\(y\z)<=0.X by BCIALG_1:1;
      then ((x\z)\z)\((x\y)\z)\(y\z)\0.X=0.X;
      then ((x\z)\z)\((x\y)\z)\(y\z)=0.X by BCIALG_1:2;
      then ((x\z)\z)\((x\y)\z)<=y\z;
      then ((x\z)\z)\((x\y)\z) in I by A6,Th5;
      then (x\z)\z in I by A5,BCIALG_1:def 18;
      hence thesis by A3;
    end;
    0.X in I by BCIALG_1:def 18;
    hence thesis by A4,Def8;
  end;
end;
