let X be BCI-algebra; :: thesis: ( X is p-Semisimple iff for x, y being Element of X st x \ y = 0. X holds
x = y )
thus ( X is p-Semisimple implies for x, y being Element of X st x \ y = 0. X holds
x = y ) :: thesis: ( ( for x, y being Element of X st x \ y = 0. X holds
x = y ) implies X is p-Semisimple )

proof

assume A1: X is p-Semisimple ; :: thesis: for x, y being Element of X st x \ y = 0. X holds
x = y
for x, y being Element of X st x \ y = 0. X holds
x = y

proof

let x, y be Element of X; :: thesis: ( x \ y = 0. X implies x = y )
assume A2: x \ y = 0. X ; :: thesis: x = y
(0. X) \ (x \ y) = (y \ x) \ ((0. X) `) by A1, BCIALG_1:66
.= (y \ x) \ (0. X) by BCIALG_1:def 5
.= y \ x by BCIALG_1:2 ;
then y \ x = 0. X by A2, BCIALG_1:def 5;
hence x = y by A2, BCIALG_1:def 7; :: thesis: verum

end;

hence for x, y being Element of X st x \ y = 0. X holds
x = y ; :: thesis: verum

end;

assume A3: for x, y being Element of X st x \ y = 0. X holds
x = y ; :: thesis: X is p-Semisimple
for x, y being Element of X holds x \ (x \ y) = y

proof

let x, y be Element of X; :: thesis: x \ (x \ y) = y
(x \ (x \ y)) \ y = (x \ y) \ (x \ y) by BCIALG_1:7
.= 0. X by BCIALG_1:def 5 ;
hence x \ (x \ y) = y by A3; :: thesis: verum

end;

hence X is p-Semisimple ; :: thesis: verum