let X be BCI-algebra; :: thesis: ( ( for x being Element of X holds x ` = x ) implies for x, y being Element of X holds x \ y = y \ x )
assume A1: for x being Element of X holds x ` = x ; :: thesis: for x, y being Element of X holds x \ y = y \ x
let x, y be Element of X; :: thesis: x \ y = y \ x
A2: (x \ y) \ (y \ x) = ((x ` ) \ y) \ (y \ x) by A1
.= ((x ` ) \ (y ` )) \ (y \ x) by A1
.= 0. X by Th1 ;
(y \ x) \ (x \ y) = ((y ` ) \ x) \ (x \ y) by A1
.= ((y ` ) \ (x ` )) \ (x \ y) by A1
.= 0. X by Th1 ;
hence x \ y = y \ x by A2, Def7; :: thesis: verum