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

proof

assume A1: X is commutative BCK-algebra ; :: thesis: X is BCI-algebra of 0 , 0 , 0 , 0
for x, y being Element of X holds Polynom 0 ,0 ,x,y = Polynom 0 ,0 ,y,x

proof

let x, y be Element of X; :: thesis: Polynom 0 ,0 ,x,y = Polynom 0 ,0 ,y,x
A2: x \ (x \ y) = y \ (y \ x) by A1, BCIALG_3:def 1;
(x,(x \ y) to_power 1),(y \ x) to_power 0 = x,(x \ y) to_power 1 by BCIALG_2:1
.= y \ (y \ x) by A2, BCIALG_2:2
.= y,(y \ x) to_power 1 by BCIALG_2:2
.= (y,(y \ x) to_power 1),(x \ y) to_power 0 by BCIALG_2:1 ;
hence Polynom 0 ,0 ,x,y = Polynom 0 ,0 ,y,x ; :: thesis: verum

end;

hence X is BCI-algebra of 0 , 0 , 0 , 0 by Def3; :: thesis: verum

end;

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

proof

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

end;

hence X is commutative BCK-algebra by A3, Th37, BCIALG_3:def 1; :: thesis: verum