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 )

for x, y being Element of X holds x \ (x \ y) = y \ (y \ x)

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 A3:
X is BCI-algebra of 0 , 0 , 0 , 0
; :: thesis: X is commutative BCK-algebra
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)

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

proof

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

for x, y being Element of X holds x \ (x \ y) = y \ (y \ x)

proof

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