let X be BCI-algebra; :: thesis: for I being Ideal of X holds
( I is p-ideal of X iff for x, y being Element of X st x in I & x <= y holds
y in I )

let I be Ideal of X; :: thesis: ( I is p-ideal of X iff for x, y being Element of X st x in I & x <= y holds
y in I )

thus ( I is p-ideal of X implies for x, y being Element of X st x in I & x <= y holds
y in I ) :: thesis: ( ( for x, y being Element of X st x in I & x <= y holds
y in I ) implies I is p-ideal of X )
proof
assume I is p-ideal of X ; :: thesis: for x, y being Element of X st x in I & x <= y holds
y in I

then A1: BCK-part X c= I by Th19;
let x, y be Element of X; :: thesis: ( x in I & x <= y implies y in I )
assume that
A2: x in I and
A3: x <= y ; :: thesis: y in I
A4: x \ y = 0. X by A3;
then (y \ x) ` = x \ y by BCIALG_1:6;
then 0. X <= y \ x by A4;
then y \ x in BCK-part X ;
hence y in I by ; :: thesis: verum
end;
assume A5: for x, y being Element of X st x in I & x <= y holds
y in I ; :: thesis: I is p-ideal of X
A6: for x, y, z being Element of X st (x \ z) \ (y \ z) in I & y in I holds
x in I
proof
let x, y, z be Element of X; :: thesis: ( (x \ z) \ (y \ z) in I & y in I implies x in I )
assume that
A7: (x \ z) \ (y \ z) in I and
A8: y in I ; :: thesis: x in I
((x \ z) \ (y \ z)) \ (x \ y) = 0. X by BCIALG_1:def 3;
then (x \ z) \ (y \ z) <= x \ y ;
then x \ y in I by A5, A7;
hence x in I by ; :: thesis: verum
end;
0. X in I by BCIALG_1:def 18;
hence I is p-ideal of X by ; :: thesis: verum