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 A1: ( X is being_I & ( 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
A2: X is being_BCI-4

proof

now

let x, y be Element of X; :: thesis: ( x \ y = 0. X & y \ x = 0. X implies x = y )
assume A3: ( x \ y = 0. X & y \ x = 0. X ) ; :: thesis: x = y
x = x \ (0. X) by A1
.= y \ (x \ x) by A1, A3
.= y \ (0. X) by A1, Def5 ;
hence x = y by A1; :: thesis: verum

end;

hence X is being_BCI-4 by Def7; :: thesis: verum

end;

now

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, Def5; :: 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 ) )
thus ((x \ y) \ (x \ z)) \ (z \ y) = 0. X :: thesis: ( (x \ (x \ y)) \ y = 0. X & ( for x, y being Element of X holds x \ (x \ y) = y ) )

proof

((x \ y) \ (x \ z)) \ (z \ y) = (z \ (x \ (x \ y))) \ (z \ y) by A1
.= (z \ (y \ (x \ x))) \ (z \ y) by A1
.= (z \ (y \ (0. X))) \ (z \ y) by A1, Def5
.= (z \ y) \ (z \ y) by A1 ;
hence ((x \ y) \ (x \ z)) \ (z \ y) = 0. X by A1, Def5; :: thesis: verum

end;

thus (x \ (x \ y)) \ y = 0. X :: thesis: for x, y being Element of X holds x \ (x \ y) = y

proof

(x \ (x \ y)) \ y = (y \ (x \ x)) \ y by A1
.= (y \ (0. X)) \ y by A1, Def5
.= y \ y by A1 ;
hence (x \ (x \ y)) \ y = 0. X by A1, Def5; :: thesis: verum

end;

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 A1;
then x \ (x \ y) = y \ (0. X) by A1, Def5;
hence x \ (x \ y) = y by A1; :: thesis: verum

end;

end;

hence X is p-Semisimple BCI-algebra by A1, A2, Def26, Th1; :: thesis: verum