let X be non empty BCIStr_0 ; :: thesis: ( X is p-Semisimple BCI-algebra iff for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) )

thus ( X is p-Semisimple BCI-algebra implies for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ) by Th2, Th56; :: thesis: ( ( for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ) implies X is p-Semisimple BCI-algebra )

assume A1: for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ; :: thesis: X is p-Semisimple BCI-algebra

hence X is p-Semisimple BCI-algebra by A4, A2, Def26, Th1; :: thesis: verum

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) )

thus ( X is p-Semisimple BCI-algebra implies for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ) by Th2, Th56; :: thesis: ( ( for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ) implies X is p-Semisimple BCI-algebra )

assume A1: for x, y, z being Element of X holds

( (x \ y) \ (x \ z) = z \ y & x \ (0. X) = x ) ; :: thesis: X is p-Semisimple BCI-algebra

A2: 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 & ( for x, y being Element of X holds x \ (x \ y) = y ) )

( ((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 \ 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 \ y) \ (z \ y) by A1

.= ((z \ y) \ (0. X)) \ (z \ y) by A1

.= ((z \ y) \ (0. X)) \ ((z \ y) \ (0. X)) by A1

.= (0. X) ` by A1 ;

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 = ((x \ (0. X)) \ (x \ y)) \ y by A1

.= (y \ (0. X)) \ y by A1

.= (y \ (0. X)) \ (y \ (0. X)) by A1

.= (0. X) ` by A1 ;

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

end;((x \ y) \ (x \ z)) \ (z \ y) = (z \ y) \ (z \ y) by A1

.= ((z \ y) \ (0. X)) \ (z \ y) by A1

.= ((z \ y) \ (0. X)) \ ((z \ y) \ (0. X)) by A1

.= (0. X) ` by A1 ;

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 = ((x \ (0. X)) \ (x \ y)) \ y by A1

.= (y \ (0. X)) \ y by A1

.= (y \ (0. X)) \ (y \ (0. X)) by A1

.= (0. X) ` by A1 ;

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

now :: thesis: for x, y being Element of X st x \ y = 0. X & y \ x = 0. X holds

x = y

then A4:
X is being_BCI-4
;x = y

let x, y be Element of X; :: thesis: ( x \ y = 0. X & y \ x = 0. X implies x = y )

assume that

A3: x \ y = 0. X and

y \ x = 0. X ; :: thesis: x = y

x = x \ (0. X) by A1

.= (x \ (0. X)) \ (x \ y) by A1, A3

.= y \ (0. X) by A1 ;

hence x = y by A1; :: thesis: verum

end;assume that

A3: x \ y = 0. X and

y \ x = 0. X ; :: thesis: x = y

x = x \ (0. X) by A1

.= (x \ (0. X)) \ (x \ y) by A1, A3

.= y \ (0. X) by A1 ;

hence x = y by A1; :: thesis: verum

now :: thesis: for x being Element of X holds x \ x = 0. X

then
X is being_I
;let x be Element of X; :: thesis: x \ x = 0. X

x \ x = (x \ (0. X)) \ x by A1

.= (x \ (0. X)) \ (x \ (0. X)) by A1

.= (0. X) ` by A1 ;

hence x \ x = 0. X by A1; :: thesis: verum

end;x \ x = (x \ (0. X)) \ x by A1

.= (x \ (0. X)) \ (x \ (0. X)) by A1

.= (0. X) ` by A1 ;

hence x \ x = 0. X by A1; :: thesis: verum

hence X is p-Semisimple BCI-algebra by A4, A2, Def26, Th1; :: thesis: verum