let X be commutative bounded BCK-algebra; :: thesis: for a being Element of X st a is being_greatest holds
( X is BCK-implicative iff for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y )
let a be Element of X; :: thesis: ( a is being_greatest implies ( X is BCK-implicative iff for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y ) )
assume A1: a is being_greatest ; :: thesis: ( X is BCK-implicative iff for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y )
thus ( X is BCK-implicative implies for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y ) :: thesis: ( ( for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y ) implies X is BCK-implicative )

proof

assume A2: X is BCK-implicative ; :: thesis: for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y
then A3: ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra ) by Th34;
let x, y be Element of X; :: thesis: (a \ y) \ ((a \ y) \ x) = x \ y
a \ y = (a \ y) \ y by A3, Th28;
then A4: ((a \ y) \ ((a \ y) \ x)) \ (x \ y) = 0. X by BCIALG_1:1;
A5: X is involutory by A1, A2, Th37;
A6: (a \ y) \ ((a \ y) \ x) = x \ (x \ (a \ y)) by Def1;
A7: x \ (a \ y) = y \ (a \ x) by A1, A5, Th23;
(y \ (a \ x)) \ y = (y \ y) \ (a \ x) by BCIALG_1:7
.= (a \ x) ` by BCIALG_1:def 5
.= 0. X by BCIALG_1:def 8 ;
then x \ (a \ y) <= y by A7, BCIALG_1:def 11;
then x \ y <= (a \ y) \ ((a \ y) \ x) by A6, BCIALG_1:5;
then (x \ y) \ ((a \ y) \ ((a \ y) \ x)) = 0. X by BCIALG_1:def 11;
hence (a \ y) \ ((a \ y) \ x) = x \ y by A4, BCIALG_1:def 7; :: thesis: verum

end;

assume A8: for x, y being Element of X holds (a \ y) \ ((a \ y) \ x) = x \ y ; :: thesis: X is BCK-implicative
for x being Element of X holds (a \ x) \ ((a \ x) \ x) = 0. X

proof

let x be Element of X; :: thesis: (a \ x) \ ((a \ x) \ x) = 0. X
(a \ x) \ ((a \ x) \ x) = x \ x by A8;
hence (a \ x) \ ((a \ x) \ x) = 0. X by BCIALG_1:def 5; :: thesis: verum

end;

hence X is BCK-implicative by A1, Th43; :: thesis: verum