let X be BCI-algebra; :: thesis: BCK-part X is closed Ideal of X
set X1 = BCK-part X;
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 x1 being Element of X st
( x \ y = x1 & 0. X <= x1 ) by A2;
then (x \ y) ` = 0. X ;
then A4: (x `) \ (y `) = 0. X by Th9;
ex x2 being Element of X st
( y = x2 & 0. X <= x2 ) by A3;
then (x `) \ (0. X) = 0. X by A4;
then x ` = 0. X by Th2;
then 0. X <= x ;
hence x in BCK-part X ; :: thesis: verum
end;
0. X in BCK-part X by Th19;
then reconsider X1 = BCK-part X as Ideal of X by A1, Def18;
now :: thesis: for x being Element of X1 holds x ` in X1
let x be Element of X1; :: thesis: x ` in X1
x in X1 ;
then ex x1 being Element of X st
( x = x1 & 0. X <= x1 ) ;
then x ` = 0. X ;
then (x `) ` = 0. X by Def5;
then 0. X <= x ` ;
hence x ` in X1 ; :: thesis: verum
end;
hence BCK-part X is closed Ideal of X by Def19; :: thesis: verum