let X be BCK-algebra; :: thesis: ( X is BCK-implicative BCK-algebra iff for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x) )
thus ( X is BCK-implicative BCK-algebra implies for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x) ) :: thesis: ( ( for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x) ) implies X is BCK-implicative BCK-algebra )

proof

assume A1: X is BCK-implicative BCK-algebra ; :: thesis: for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x)
let x, y be Element of X; :: thesis: (x \ (x \ y)) \ (x \ y) = y \ (y \ x)
X is commutative BCK-algebra by A1, Th34;
then (x \ (x \ y)) \ (x \ y) = (y \ (y \ x)) \ (x \ y) by Def1
.= (y \ (x \ y)) \ (y \ x) by BCIALG_1:7 ;
hence (x \ (x \ y)) \ (x \ y) = y \ (y \ x) by A1, Def12; :: thesis: verum

end;

assume A2: for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x) ; :: thesis: X is BCK-implicative BCK-algebra
for x, y being Element of X holds (x \ y) \ y = x \ y

proof

let x, y be Element of X; :: thesis: (x \ y) \ y = x \ y
A3: (x \ (x \ (x \ y))) \ (x \ (x \ y)) = (x \ y) \ (x \ (x \ y)) by BCIALG_1:8
.= (x \ (x \ (x \ y))) \ y by BCIALG_1:7
.= (x \ y) \ y by BCIALG_1:8 ;
A4: (x \ y) \ x = (x \ x) \ y by BCIALG_1:7
.= y ` by BCIALG_1:def 5
.= 0. X by BCIALG_1:def 8 ;
(x \ (x \ (x \ y))) \ (x \ (x \ y)) = (x \ y) \ ((x \ y) \ x) by A2;
hence (x \ y) \ y = x \ y by A3, A4, BCIALG_1:2; :: thesis: verum

end;

then A5: X is BCK-positive-implicative BCK-algebra by Th28;
for x, y being Element of X st x <= y holds
x = y \ (y \ x)

proof

let x, y be Element of X; :: thesis: ( x <= y implies x = y \ (y \ x) )
assume x <= y ; :: thesis: x = y \ (y \ x)
then A6: x \ y = 0. X ;
then y \ (y \ x) = (x \ (x \ y)) \ (0. X) by A2
.= x \ (0. X) by A6, BCIALG_1:2 ;
hence x = y \ (y \ x) by BCIALG_1:2; :: thesis: verum

end;

then X is commutative BCK-algebra by Th5;
hence X is BCK-implicative BCK-algebra by A5, Th34; :: thesis: verum