set X1 = BCK-part X;
take BCK-part X ; :: thesis: ( BCK-part X is Ideal of X & BCK-part X is commutative )
A1: for x, y being Element of X st x \ y in BCK-part X & y in BCK-part X holds
x in BCK-part X

proof

let x, y be Element of X; :: thesis: ( x \ y in BCK-part X & y in BCK-part X implies x in BCK-part X )
assume that
A2: x \ y in BCK-part X and
A3: y in BCK-part X ; :: thesis: x in BCK-part X
ex x2 being Element of X st
( y = x2 & 0. X <= x2 ) by A3;
then A4: y ` = 0. X ;
ex x1 being Element of X st
( x \ y = x1 & 0. X <= x1 ) by A2;
then (x \ y) ` = 0. X ;
then (x `) \ (0. X) = 0. X by A4, BCIALG_1:9;
then (0. X) \ x = 0. X by BCIALG_1:2;
then 0. X <= x ;
hence x in BCK-part X ; :: thesis: verum

end;

A5: for x, y, z being Element of X st (x \ y) \ z in BCK-part X & z in BCK-part X holds
x \ (y \ (y \ x)) in BCK-part X

proof

let x, y, z be Element of X; :: thesis: ( (x \ y) \ z in BCK-part X & z in BCK-part X implies x \ (y \ (y \ x)) in BCK-part X )
assume that
(x \ y) \ z in BCK-part X and
z in BCK-part X ; :: thesis: x \ (y \ (y \ x)) in BCK-part X
(0. X) \ (x \ (y \ (y \ x))) = (x \ (y \ (y \ x))) `
.= 0. X by BCIALG_1:def 8 ;
then 0. X <= x \ (y \ (y \ x)) ;
hence x \ (y \ (y \ x)) in BCK-part X ; :: thesis: verum

end;

0. X in BCK-part X by BCIALG_1:19;
then BCK-part X is Ideal of X by A1, BCIALG_1:def 18;
hence ( BCK-part X is Ideal of X & BCK-part X is commutative ) by A5, Def6; :: thesis: verum