let X be BCK-algebra; :: thesis: ( X is BCK-implicative BCK-algebra iff ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra ) )

thus ( X is BCK-implicative BCK-algebra implies ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra ) ) :: thesis: ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra implies X is BCK-implicative BCK-algebra )

A3: X is commutative BCK-algebra and

A4: X is BCK-positive-implicative BCK-algebra ; :: thesis: X is BCK-implicative BCK-algebra

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

thus ( X is BCK-implicative BCK-algebra implies ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra ) ) :: thesis: ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra implies X is BCK-implicative BCK-algebra )

proof

assume that
assume A1:
X is BCK-implicative BCK-algebra
; :: thesis: ( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra )

A2: for x, y being Element of X holds x \ (x \ y) <= y \ (y \ x)

end;A2: for x, y being Element of X holds x \ (x \ y) <= y \ (y \ x)

proof

for x, y being Element of X holds x \ y = (x \ y) \ y
let x, y be Element of X; :: thesis: x \ (x \ y) <= y \ (y \ x)

(x \ (x \ y)) \ y = (x \ y) \ (x \ y) by BCIALG_1:7

.= 0. X by BCIALG_1:def 5 ;

then x \ (x \ y) <= y ;

then (x \ (x \ y)) \ (y \ x) <= y \ (y \ x) by BCIALG_1:5;

then (x \ (y \ x)) \ (x \ y) <= y \ (y \ x) by BCIALG_1:7;

hence x \ (x \ y) <= y \ (y \ x) by A1, Def12; :: thesis: verum

end;(x \ (x \ y)) \ y = (x \ y) \ (x \ y) by BCIALG_1:7

.= 0. X by BCIALG_1:def 5 ;

then x \ (x \ y) <= y ;

then (x \ (x \ y)) \ (y \ x) <= y \ (y \ x) by BCIALG_1:5;

then (x \ (y \ x)) \ (x \ y) <= y \ (y \ x) by BCIALG_1:7;

hence x \ (x \ y) <= y \ (y \ x) by A1, Def12; :: thesis: verum

proof

hence
( X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra )
by A2, Th1, Th28; :: thesis: verum
let x, y be Element of X; :: thesis: x \ y = (x \ y) \ y

(x \ y) \ (y \ (x \ y)) = x \ y by A1, Def12;

hence x \ y = (x \ y) \ y by A1, Def12; :: thesis: verum

end;(x \ y) \ (y \ (x \ y)) = x \ y by A1, Def12;

hence x \ y = (x \ y) \ y by A1, Def12; :: thesis: verum

A3: X is commutative BCK-algebra and

A4: X is BCK-positive-implicative BCK-algebra ; :: thesis: X is BCK-implicative BCK-algebra

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

proof

hence
X is BCK-implicative BCK-algebra
by Def12; :: thesis: verum
let x, y be Element of X; :: thesis: x \ (y \ x) = x

x \ (x \ (x \ (y \ x))) = x \ (y \ x) by BCIALG_1:8;

then A5: x \ (y \ x) = x \ ((y \ x) \ ((y \ x) \ x)) by A3, Def1;

(y \ x) \ ((y \ x) \ x) = (y \ x) \ (y \ x) by A4, Th28

.= 0. X by BCIALG_1:def 5 ;

hence x \ (y \ x) = x by A5, BCIALG_1:2; :: thesis: verum

end;x \ (x \ (x \ (y \ x))) = x \ (y \ x) by BCIALG_1:8;

then A5: x \ (y \ x) = x \ ((y \ x) \ ((y \ x) \ x)) by A3, Def1;

(y \ x) \ ((y \ x) \ x) = (y \ x) \ (y \ x) by A4, Th28

.= 0. X by BCIALG_1:def 5 ;

hence x \ (y \ x) = x by A5, BCIALG_1:2; :: thesis: verum