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;

theorem
  for x,y being Element of X holds x\(x\y) in I implies x\((x\y)\((x\y)\
  x)) in I & (y\(y\x))\x in I & y\(y\x)\(x\y) in I
proof
  let x,y be Element of X;
  assume
A1: x\(x\y) in I;
  (x\y)\((x\y)\x)=(x\y)\((x\x)\y) by BCIALG_1:7
    .=(x\((x\x)\y))\y by BCIALG_1:7
    .=(x\y`)\y by BCIALG_1:def 5
    .=(x\0.X)\y by BCIALG_1:def 8
    .=x\y by BCIALG_1:2;
  hence x\((x\y)\((x\y)\x)) in I by A1;
  (y\0.X)\(y\x)\(x\0.X)=0.X by BCIALG_1:1;
  then (y\0.X)\(y\x)\(x\0.X) in I by BCIALG_1:def 18;
  then y\(y\x)\(x\0.X) in I by BCIALG_1:2;
  hence y\(y\x)\x in I by BCIALG_1:2;
  (y\0.X)\(y\x)\(x\0.X)=0.X by BCIALG_1:1;
  then (y\0.X)\(y\x)<=(x\0.X);
  then (y\0.X)\(y\x)\(x\y)<=(x\0.X)\(x\y) by BCIALG_1:5;
  then y\(y\x)\(x\y)<=(x\0.X)\(x\y) by BCIALG_1:2;
  then y\(y\x)\(x\y)<=x\(x\y) by BCIALG_1:2;
  then y\(y\x)\(x\y)\(x\(x\y)) =0.X;
  then y\(y\x)\(x\y)\(x\(x\y)) in I by BCIALG_1:def 18;
  hence thesis by A1,BCIALG_1:def 18;
end;
