let X be BCK-algebra; :: thesis: for I being Ideal of X holds
( 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 )
let I be Ideal of X; :: thesis: ( 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 )
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 ) :: thesis: ( ( 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 A1: I is positive-implicative-ideal of X ; :: thesis: for x, y being Element of X st (x \ y) \ y in I holds
x \ y in I
let x, y be Element of X; :: thesis: ( (x \ y) \ y in I implies x \ y in I )
y \ y = 0. X by BCIALG_1:def 5;
then A2: y \ y in I by A1, Def8;
assume (x \ y) \ y in I ; :: thesis: x \ y in I
hence x \ y in I by A1, A2, Def8; :: thesis: verum

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 ) :: thesis: verum

proof

assume A3: for x, y being Element of X st (x \ y) \ y in I holds
x \ y in I ; :: thesis: I is positive-implicative-ideal of X
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; :: thesis: ( (x \ y) \ z in I & y \ z in I implies x \ z in I )
assume that
A5: (x \ y) \ z in I and
A6: y \ z in I ; :: thesis: x \ 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 x \ z in I by A3; :: thesis: verum

end;

0. X in I by BCIALG_1:def 18;
hence I is positive-implicative-ideal of X by A4, Def8; :: thesis: verum

end;