let X be BCI-algebra; :: thesis: ( ( for X being BCI-algebra

for x, y being Element of X holds (x \ y) \ ((x \ y) \ (y \ x)) = 0. X ) implies X is BCK-algebra )

assume A1: for X being BCI-algebra

for x, y being Element of X holds (x \ y) \ ((x \ y) \ (y \ x)) = 0. X ; :: thesis: X is BCK-algebra

for s being Element of X holds s ` = 0. X

for x, y being Element of X holds (x \ y) \ ((x \ y) \ (y \ x)) = 0. X ) implies X is BCK-algebra )

assume A1: for X being BCI-algebra

for x, y being Element of X holds (x \ y) \ ((x \ y) \ (y \ x)) = 0. X ; :: thesis: X is BCK-algebra

for s being Element of X holds s ` = 0. X

proof

hence
X is BCK-algebra
by Def8; :: thesis: verum
let s be Element of X; :: thesis: s ` = 0. X

(s `) \ ((s `) \ (s \ (0. X))) = 0. X by A1;

then ((s `) \ ((s `) \ s)) \ s = s ` by Th2;

hence s ` = 0. X by Th1; :: thesis: verum

end;(s `) \ ((s `) \ (s \ (0. X))) = 0. X by A1;

then ((s `) \ ((s `) \ s)) \ s = s ` by Th2;

hence s ` = 0. X by Th1; :: thesis: verum