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

proof

assume A1: X is commutative BCK-algebra ; :: thesis: for x, y being Element of X holds x \ (x \ y) = y \ (y \ (x \ (x \ y)))
let x, y be Element of X; :: thesis: x \ (x \ y) = y \ (y \ (x \ (x \ y)))
y \ (y \ x) = y \ (y \ (x \ (x \ y))) by A1, Th3;
hence x \ (x \ y) = y \ (y \ (x \ (x \ y))) by A1, Def1; :: thesis: verum

end;

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

proof

let x, y be Element of X; :: thesis: x \ (x \ y) <= y \ (y \ x)
x \ (x \ y) <= x by Th2;
then A3: y \ x <= y \ (x \ (x \ y)) by BCIALG_1:5;
x \ (x \ y) = y \ (y \ (x \ (x \ y))) by A2;
hence x \ (x \ y) <= y \ (y \ x) by A3, BCIALG_1:5; :: thesis: verum

end;

hence X is commutative BCK-algebra by Th1; :: thesis: verum