let X be non empty BCIStr_0 ; :: thesis: ( X is p-Semisimple BCI-algebra iff ( X is being_I & ( for x, y, z being Element of X holds
( x \ (y \ z) = z \ (y \ x) & x \ (0. X) = x ) ) ) )
thus ( X is p-Semisimple BCI-algebra implies ( X is being_I & ( for x, y, z being Element of X holds
( x \ (y \ z) = z \ (y \ x) & x \ (0. X) = x ) ) ) ) by Th2, Th57; :: thesis: ( X is being_I & ( for x, y, z being Element of X holds
( x \ (y \ z) = z \ (y \ x) & x \ (0. X) = x ) ) implies X is p-Semisimple BCI-algebra )
assume that
A1: X is being_I and
A2: for x, y, z being Element of X holds
( x \ (y \ z) = z \ (y \ x) & x \ (0. X) = x ) ; :: thesis: X is p-Semisimple BCI-algebra
A3: now :: thesis: for x, y, z being Element of X holds
( x \ x = 0. X & ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & (x \ (x \ y)) \ y = 0. X & ( for x, y being Element of X holds x \ (x \ y) = y ) )
let x, y, z be Element of X; :: thesis: ( x \ x = 0. X & ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & (x \ (x \ y)) \ y = 0. X & ( for x, y being Element of X holds x \ (x \ y) = y ) )
thus x \ x = 0. X by A1; :: thesis: ( ((x \ y) \ (x \ z)) \ (z \ y) = 0. X & (x \ (x \ y)) \ y = 0. X & ( for x, y being Element of X holds x \ (x \ y) = y ) )
((x \ y) \ (x \ z)) \ (z \ y) = (z \ (x \ (x \ y))) \ (z \ y) by A2
.= (z \ (y \ (x \ x))) \ (z \ y) by A2
.= (z \ (y \ (0. X))) \ (z \ y) by A1
.= (z \ y) \ (z \ y) by A2 ;
hence ((x \ y) \ (x \ z)) \ (z \ y) = 0. X by A1; :: thesis: ( (x \ (x \ y)) \ y = 0. X & ( for x, y being Element of X holds x \ (x \ y) = y ) )
(x \ (x \ y)) \ y = (y \ (x \ x)) \ y by A2
.= (y \ (0. X)) \ y by A1
.= y \ y by A2 ;
hence (x \ (x \ y)) \ y = 0. X by A1; :: thesis: for x, y being Element of X holds x \ (x \ y) = y
thus for x, y being Element of X holds x \ (x \ y) = y :: thesis: verum
proof
let x, y be Element of X; :: thesis: x \ (x \ y) = y
x \ (x \ y) = y \ (x \ x) by A2;
then x \ (x \ y) = y \ (0. X) by A1;
hence x \ (x \ y) = y by A2; :: thesis: verum
end;
end;
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 that
A4: x \ y = 0. X and
y \ x = 0. X ; :: thesis: x = y
x = x \ (0. X) by A2
.= y \ (x \ x) by A2, A4
.= y \ (0. X) by A1 ;
hence x = y by A2; :: thesis: verum
end;
then X is being_BCI-4 ;
hence X is p-Semisimple BCI-algebra by A1, A3, Def26, Th1; :: thesis: verum