let X be BCI-algebra; :: thesis: ( X is BCK-positive-implicative BCK-algebra iff X is BCK-algebra of 0 ,1, 0 ,1 )
thus ( X is BCK-positive-implicative BCK-algebra implies X is BCK-algebra of 0 ,1, 0 ,1 ) :: thesis: ( X is BCK-algebra of 0 ,1, 0 ,1 implies X is BCK-positive-implicative BCK-algebra )
proof
assume A1: X is BCK-positive-implicative BCK-algebra ; :: thesis: X is BCK-algebra of 0 ,1, 0 ,1
for x, y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
proof
let x, y be Element of X; :: thesis: Polynom (0,1,x,y) = Polynom (0,1,y,x)
(x \ y) \ (x \ y) = 0. X by BCIALG_1:def 5;
then (x \ (x \ y)) \ y = 0. X by BCIALG_1:7;
then A2: x \ (x \ y) <= y ;
x \ (x \ y) = (x \ (x \ y)) \ (x \ y) by ;
then x \ (x \ y) <= y \ (x \ y) by ;
then (x \ (x \ y)) \ (y \ x) <= (y \ (x \ y)) \ (y \ x) by BCIALG_1:5;
then A3: (x \ (x \ y)) \ (y \ x) <= (y \ (y \ x)) \ (x \ y) by BCIALG_1:7;
(y \ x) \ (y \ x) = 0. X by BCIALG_1:def 5;
then (y \ (y \ x)) \ x = 0. X by BCIALG_1:7;
then y \ (y \ x) <= x ;
then A4: (y \ (y \ x)) \ (y \ x) <= x \ (y \ x) by BCIALG_1:5;
y \ (y \ x) = (y \ (y \ x)) \ (y \ x) by ;
then (y \ (y \ x)) \ (x \ y) <= (x \ (y \ x)) \ (x \ y) by ;
then (y \ (y \ x)) \ (x \ y) <= (x \ (x \ y)) \ (y \ x) by BCIALG_1:7;
then A5: (x \ (x \ y)) \ (y \ x) = (y \ (y \ x)) \ (x \ y) by ;
(((x,(x \ y)) to_power 1),(y \ x)) to_power 1 = ((x,(x \ y)) to_power 1) \ (y \ x) by BCIALG_2:2
.= (x \ (x \ y)) \ (y \ x) by BCIALG_2:2
.= ((y \ (y \ x)),(x \ y)) to_power 1 by
.= (((y,(y \ x)) to_power 1),(x \ y)) to_power 1 by BCIALG_2:2 ;
hence Polynom (0,1,x,y) = Polynom (0,1,y,x) ; :: thesis: verum
end;
hence X is BCK-algebra of 0 ,1, 0 ,1 by ; :: thesis: verum
end;
assume A6: X is BCK-algebra of 0 ,1, 0 ,1 ; :: thesis:
for x, y being Element of X holds x \ y = (x \ y) \ y
proof
let x, y be Element of X; :: thesis: x \ y = (x \ y) \ y
A7: Polynom (0,1,x,(x \ y)) = Polynom (0,1,(x \ y),x) by ;
A8: (x \ y) \ x = (x \ x) \ y by BCIALG_1:7
.= y ` by BCIALG_1:def 5
.= 0. X by ;
then A9: ((x \ y) \ ((x \ y) \ x)) \ (x \ (x \ y)) = (x \ y) \ (x \ (x \ y)) by BCIALG_1:2
.= (x \ (x \ (x \ y))) \ y by BCIALG_1:7
.= (x \ y) \ y by BCIALG_1:8 ;
A10: (x \ (x \ (x \ y))) \ ((x \ y) \ x) = (x \ y) \ ((x \ y) \ x) by BCIALG_1:8
.= x \ y by ;
(x \ (x \ (x \ y))) \ ((x \ y) \ x) = ((x,(x \ (x \ y))) to_power 1) \ ((x \ y) \ x) by BCIALG_2:2
.= ((((x \ y),((x \ y) \ x)) to_power 1),(x \ (x \ y))) to_power 1 by
.= (((x \ y) \ ((x \ y) \ x)),(x \ (x \ y))) to_power 1 by BCIALG_2:2
.= ((x \ y) \ ((x \ y) \ x)) \ (x \ (x \ y)) by BCIALG_2:2 ;
hence x \ y = (x \ y) \ y by ; :: thesis: verum
end;
hence X is BCK-positive-implicative BCK-algebra by ; :: thesis: verum