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

proof

assume A1: X is BCK-implicative BCK-algebra ; :: thesis: X is BCK-algebra of 1, 0 , 0 , 0
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; :: thesis: Polynom (1,0,x,y) = Polynom (0,0,y,x)
A2: (x \ (x \ y)) \ (x \ y) = y \ (y \ x) by A1, BCIALG_3:35;
(((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 A2, 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) ; :: thesis: verum

end;

hence X is BCK-algebra of 1, 0 , 0 , 0 by A1, Def3; :: thesis: verum

end;

assume A3: X is BCK-algebra of 1, 0 , 0 , 0 ; :: thesis: X is BCK-implicative BCK-algebra
for x, y being Element of X holds (x \ (x \ y)) \ (x \ y) = y \ (y \ x)

proof

let x, y be Element of X; :: thesis: (x \ (x \ y)) \ (x \ y) = y \ (y \ x)
A4: Polynom (1,0,x,y) = Polynom (0,0,y,x) by A3, Def3;
(x \ (x \ y)) \ (x \ y) = (x,(x \ y)) to_power 2 by BCIALG_2:3
.= (((x,(x \ y)) to_power (1 + 1)),(y \ x)) to_power 0 by BCIALG_2:1
.= (y,(y \ x)) to_power 1 by A4, BCIALG_2:1
.= y \ (y \ x) by BCIALG_2:2 ;
hence (x \ (x \ y)) \ (x \ y) = y \ (y \ x) ; :: thesis: verum

end;

hence X is BCK-implicative BCK-algebra by A3, BCIALG_3:35; :: thesis: verum