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

then reconsider X1 = BCK-part X as Ideal of X by A1, Def18;

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

0. X in BCK-part X
by Th19;
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;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

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

hence
BCK-part X is closed Ideal of X
by Def19; :: thesis: verumlet 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;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