let X be non empty BCIStr_0 ; :: thesis: ( X is positive-implicative BCI-algebra iff for x, y, z being Element of X holds
( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & x \ (0. X) = x & x \ y = ((x \ y) \ y) \ (y `) & (x \ (x \ y)) \ (y \ x) = ((y \ (y \ x)) \ (y \ x)) \ (x \ y) ) )
thus ( X is positive-implicative BCI-algebra implies for x, y, z being Element of X holds
( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & x \ (0. X) = x & x \ y = ((x \ y) \ y) \ (y `) & (x \ (x \ y)) \ (y \ x) = ((y \ (y \ x)) \ (y \ x)) \ (x \ y) ) ) by Lm24, Th1, Th2, Th84; :: thesis: ( ( for x, y, z being Element of X holds
( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & x \ (0. X) = x & x \ y = ((x \ y) \ y) \ (y `) & (x \ (x \ y)) \ (y \ x) = ((y \ (y \ x)) \ (y \ x)) \ (x \ y) ) ) implies X is positive-implicative BCI-algebra )
assume A1: for x, y, z being Element of X holds
( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & x \ (0. X) = x & x \ y = ((x \ y) \ y) \ (y `) & (x \ (x \ y)) \ (y \ x) = ((y \ (y \ x)) \ (y \ x)) \ (x \ y) ) ; :: thesis: X is positive-implicative BCI-algebra
now :: thesis: for x being Element of X holds x \ x = 0. X
let x be Element of X; :: thesis: x \ x = 0. X
((x \ (0. X)) \ (x \ (0. X))) \ ((0. X) `) = 0. X by A1;
then ((x \ (0. X)) \ (x \ (0. X))) \ (0. X) = 0. X by A1;
then ((x \ (0. X)) \ x) \ (0. X) = 0. X by A1;
then (x \ x) \ (0. X) = 0. X by A1;
hence x \ x = 0. X by A1; :: thesis: verum
end;
then A2: X is being_I ;
now :: thesis: for x, y being Element of X st x \ y = 0. X & y \ x = 0. X holds
x = y
let x, y be Element of X; :: thesis: ( x \ y = 0. X & y \ x = 0. X implies x = y )
assume ( x \ y = 0. X & y \ x = 0. X ) ; :: thesis: x = y
then (x \ (0. X)) \ (0. X) = ((y \ (0. X)) \ (0. X)) \ (0. X) by A1;
then x \ (0. X) = ((y \ (0. X)) \ (0. X)) \ (0. X) by A1;
then x = ((y \ (0. X)) \ (0. X)) \ (0. X) by A1;
then x = (y \ (0. X)) \ (0. X) by A1;
then x = y \ (0. X) by A1;
hence x = y by A1; :: thesis: verum
end;
then A3: X is being_BCI-4 ;
now :: thesis: for x, y, z being Element of X holds
( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & (x \ (x \ y)) \ y = 0. X )
let x, y, z be Element of X; :: thesis: ( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & (x \ (x \ y)) \ y = 0. X )
thus ((x \ y) \ (x \ z)) \ (z \ y) = 0. X by A1; :: thesis: (x \ (x \ y)) \ y = 0. X
((x \ (0. X)) \ (x \ y)) \ (y \ (0. X)) = 0. X by A1;
then (x \ (x \ y)) \ (y \ (0. X)) = 0. X by A1;
hence (x \ (x \ y)) \ y = 0. X by A1; :: thesis: verum
end;
then X is weakly-positive-implicative BCI-algebra by A1, A2, A3, Th1, Th84;
hence X is positive-implicative BCI-algebra by Lm25; :: thesis: verum