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;

theorem
  for X being BCK-algebra,A,I being Ideal of X holds (A/\I={0.X} iff for
  x being Element of A,y being Element of I holds x\y =x )
proof
  let X be BCK-algebra;
  let A,I be Ideal of X;
  thus A/\I={0.X} implies for x being Element of A,y being Element of I holds
  x\y =x
  proof
    assume
A1: A/\I={0.X};
    let x be Element of A;
    let y be Element of I;
    x\(x\y)\y=0.X by BCIALG_1:1;
    then x\(x\y)<=y;
    then
A2: x\(x\y) in I by Th5;
    (x\(x\y))\x = x\x\(x\y) by BCIALG_1:7
      .=(x\y)` by BCIALG_1:def 5
      .=0.X by BCIALG_1:def 8;
    then (x\(x\y))<=x;
    then x\(x\y) in A by Th5;
    then x\(x\y) in {0.X} by A1,A2,XBOOLE_0:def 4;
    then
A3: x\(x\y)=0.X by TARSKI:def 1;
    (x\y)\x = x\x\y by BCIALG_1:7
      .=y` by BCIALG_1:def 5
      .=0.X by BCIALG_1:def 8;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  thus (for x being Element of A,y being Element of I holds x\y =x) implies A
  /\I={0.X}
  proof
    assume
A4: for x being Element of A,y being Element of I holds x\y =x;
    thus A/\I c= {0.X}
    proof
      let x be object;
      assume
A5:   x in A/\I;
      then reconsider xxx=x as Element of A by XBOOLE_0:def 4;
      reconsider xx=x as Element of I by A5,XBOOLE_0:def 4;
      xxx\xx=xxx by A4;
      then x=0.X by BCIALG_1:def 5;
      hence thesis by TARSKI:def 1;
    end;
    let x be object;
    assume x in {0.X};
    then
A6: x =0.X by TARSKI:def 1;
    0.X in A & 0.X in I by BCIALG_1:def 18;
    hence thesis by A6,XBOOLE_0:def 4;
  end;
end;
