let X be BCI-algebra; ( X is associative BCI-algebra implies ( X is BCI-algebra of 0 ,1, 0 , 0 & X is BCI-algebra of 1, 0 , 0 , 0 ) )
assume A1:
X is associative BCI-algebra
; ( X is BCI-algebra of 0 ,1, 0 , 0 & X is BCI-algebra of 1, 0 , 0 , 0 )
for x being Element of X holds (x `) ` = x
then A2:
X is p-Semisimple
by BCIALG_1:54;
for x, y being Element of X holds Polynom (1,0,x,y) = Polynom (0,0,y,x)
proof
let x,
y be
Element of
X;
Polynom (1,0,x,y) = Polynom (0,0,y,x)
x \ (x \ y) = y
by A2;
then A3:
(x \ (x \ y)) \ (x \ y) = y \ (y \ x)
by A1, BCIALG_1:48;
(
((x,(x \ y)) to_power (1 + 1)),
(y \ x))
to_power 0 =
(
x,
(x \ y))
to_power 2
by BCIALG_2:1
.=
(x \ (x \ y)) \ (x \ y)
by BCIALG_2:3
.=
(
y,
(y \ x))
to_power 1
by A3, BCIALG_2:2
.=
(
((y,(y \ x)) to_power 1),
(x \ y))
to_power 0
by BCIALG_2:1
;
hence
Polynom (1,
0,
x,
y)
= Polynom (
0,
0,
y,
x)
;
verum
end;
hence
( X is BCI-algebra of 0 ,1, 0 , 0 & X is BCI-algebra of 1, 0 , 0 , 0 )
by A2, Def3, Th42; verum