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) )
thus ( X is commutative BCK-algebra implies for x, y being Element of X holds x \ (x \ y) <= y \ (y \ x) ) :: thesis: ( ( for x, y being Element of X holds x \ (x \ y) <= y \ (y \ x) ) 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)
let x, y be Element of X; :: thesis: x \ (x \ y) <= y \ (y \ x)
x \ (x \ y) = y \ (y \ x) by ;
then (x \ (x \ y)) \ (y \ (y \ x)) = 0. X by BCIALG_1:def 5;
hence x \ (x \ y) <= y \ (y \ x) ; :: thesis: verum
end;
assume A2: for x, y being Element of X holds x \ (x \ y) <= y \ (y \ x) ; :: thesis:
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)
y \ (y \ x) <= x \ (x \ y) by A2;
then A3: (y \ (y \ x)) \ (x \ (x \ y)) = 0. X ;
x \ (x \ y) <= y \ (y \ x) by A2;
then (x \ (x \ y)) \ (y \ (y \ x)) = 0. X ;
hence x \ (x \ y) = y \ (y \ x) by ; :: thesis: verum
end;
hence X is commutative BCK-algebra by Def1; :: thesis: verum