let X be BCI-algebra; :: thesis: ( ( for X being BCI-algebra
for x, y being Element of X holds x \ (x \ y) = y \ (y \ x) ) implies X is BCK-algebra )
assume A1: for X being BCI-algebra
for x, y being Element of X holds x \ (x \ y) = y \ (y \ x) ; :: thesis: X is BCK-algebra
for z being Element of X holds z ` = 0. X

proof

let z be Element of X; :: thesis: z ` = 0. X
(z `) ` = z \ (z \ (0. X)) by A1;
then (z `) ` = z \ z by Th2;
then ((z `) `) \ z = z ` by Def5;
hence z ` = 0. X by Th1; :: thesis: verum

end;

hence X is BCK-algebra by Def8; :: thesis: verum