let X be BCI-algebra; :: thesis: ( ( for x, y being Element of X holds (x `) \ y = (x \ y) ` ) implies for x, y being Element of X holds (x \ y) \ (y \ x) in BCK-part X )
assume A1: for x, y being Element of X holds (x `) \ y = (x \ y) ` ; :: thesis: for x, y being Element of X holds (x \ y) \ (y \ x) in BCK-part X
let x, y be Element of X; :: thesis: (x \ y) \ (y \ x) in BCK-part X
((x \ y) \ (y \ x)) ` = ((x \ y) `) \ (y \ x) by A1
.= ((x `) \ (y `)) \ (y \ x) by Th9
.= (((y `) `) \ x) \ (y \ x) by Th7
.= (((y `) \ x) `) \ (y \ x) by A1
.= (((y \ x) `) `) \ (y \ x) by A1
.= 0. X by Th1 ;
then 0. X <= (x \ y) \ (y \ x) ;
hence (x \ y) \ (y \ x) in BCK-part X ; :: thesis: verum