let X be commutative bounded BCK-algebra; :: thesis: for a being Element of X st a is being_greatest holds

( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x )

let a be Element of X; :: thesis: ( a is being_greatest implies ( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x ) )

assume A1: a is being_greatest ; :: thesis: ( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x )

thus ( X is BCK-implicative implies for x being Element of X holds x \ (a \ x) = x ) ; :: thesis: ( ( for x being Element of X holds x \ (a \ x) = x ) implies X is BCK-implicative )

assume A2: for x being Element of X holds x \ (a \ x) = x ; :: thesis: X is BCK-implicative

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

( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x )

let a be Element of X; :: thesis: ( a is being_greatest implies ( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x ) )

assume A1: a is being_greatest ; :: thesis: ( X is BCK-implicative iff for x being Element of X holds x \ (a \ x) = x )

thus ( X is BCK-implicative implies for x being Element of X holds x \ (a \ x) = x ) ; :: thesis: ( ( for x being Element of X holds x \ (a \ x) = x ) implies X is BCK-implicative )

assume A2: for x being Element of X holds x \ (a \ x) = x ; :: thesis: X is BCK-implicative

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

proof

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

A3: (x \ (a \ x)) \ x = (x \ x) \ (a \ x) by BCIALG_1:7

.= (a \ x) ` by BCIALG_1:def 5

.= 0. X by BCIALG_1:def 8 ;

y \ a = 0. X by A1;

then y <= a ;

then y \ x <= a \ x by BCIALG_1:5;

then A4: x \ (a \ x) <= x \ (y \ x) by BCIALG_1:5;

x \ (x \ (a \ x)) = x \ x by A2

.= 0. X by BCIALG_1:def 5 ;

then x \ (a \ x) = x by A3, BCIALG_1:def 7;

then A5: x \ (x \ (y \ x)) = 0. X by A4;

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

.= (y \ x) ` by BCIALG_1:def 5

.= 0. X by BCIALG_1:def 8 ;

hence x \ (y \ x) = x by A5, BCIALG_1:def 7; :: thesis: verum

end;A3: (x \ (a \ x)) \ x = (x \ x) \ (a \ x) by BCIALG_1:7

.= (a \ x) ` by BCIALG_1:def 5

.= 0. X by BCIALG_1:def 8 ;

y \ a = 0. X by A1;

then y <= a ;

then y \ x <= a \ x by BCIALG_1:5;

then A4: x \ (a \ x) <= x \ (y \ x) by BCIALG_1:5;

x \ (x \ (a \ x)) = x \ x by A2

.= 0. X by BCIALG_1:def 5 ;

then x \ (a \ x) = x by A3, BCIALG_1:def 7;

then A5: x \ (x \ (y \ x)) = 0. X by A4;

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

.= (y \ x) ` by BCIALG_1:def 5

.= 0. X by BCIALG_1:def 8 ;

hence x \ (y \ x) = x by A5, BCIALG_1:def 7; :: thesis: verum