let X be BCI-algebra; :: thesis: ( X is implicative BCI-algebra implies X is BCI-algebra of 0 ,1, 0 , 0 )

assume A1: X is implicative BCI-algebra ; :: thesis: X is BCI-algebra of 0 ,1, 0 , 0

for x, y being Element of X holds Polynom (0,1,x,y) = Polynom (0,0,y,x)

assume A1: X is implicative BCI-algebra ; :: thesis: X is BCI-algebra of 0 ,1, 0 , 0

for x, y being Element of X holds Polynom (0,1,x,y) = Polynom (0,0,y,x)

proof

hence
X is BCI-algebra of 0 ,1, 0 , 0
by Def3; :: thesis: verum
let x, y be Element of X; :: thesis: Polynom (0,1,x,y) = Polynom (0,0,y,x)

A2: (x \ (x \ y)) \ (y \ x) = y \ (y \ x) by A1, BCIALG_1:def 24;

(((x,(x \ y)) to_power 1),(y \ x)) to_power 1 = ((x \ (x \ y)),(y \ x)) to_power 1 by BCIALG_2:2

.= (x \ (x \ y)) \ (y \ x) by BCIALG_2:2

.= (y,(y \ x)) to_power 1 by A2, BCIALG_2:2

.= (((y,(y \ x)) to_power 1),(x \ y)) to_power 0 by BCIALG_2:1 ;

hence Polynom (0,1,x,y) = Polynom (0,0,y,x) ; :: thesis: verum

end;A2: (x \ (x \ y)) \ (y \ x) = y \ (y \ x) by A1, BCIALG_1:def 24;

(((x,(x \ y)) to_power 1),(y \ x)) to_power 1 = ((x \ (x \ y)),(y \ x)) to_power 1 by BCIALG_2:2

.= (x \ (x \ y)) \ (y \ x) by BCIALG_2:2

.= (y,(y \ x)) to_power 1 by A2, BCIALG_2:2

.= (((y,(y \ x)) to_power 1),(x \ y)) to_power 0 by BCIALG_2:1 ;

hence Polynom (0,1,x,y) = Polynom (0,0,y,x) ; :: thesis: verum