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 (a \ x) \ x = a \ 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 (a \ x) \ x = a \ x ) )
assume A1: a is being_greatest ; :: thesis: ( X is BCK-implicative iff for x being Element of X holds (a \ x) \ x = a \ x )
thus ( X is BCK-implicative implies for x being Element of X holds (a \ x) \ x = a \ x ) :: thesis: ( ( for x being Element of X holds (a \ x) \ x = a \ x ) implies X is BCK-implicative )
proof
assume A2: X is BCK-implicative ; :: thesis: for x being Element of X holds (a \ x) \ x = a \ x
let x be Element of X; :: thesis: (a \ x) \ x = a \ x
A3: ((a \ x) \ x) \ (a \ x) = ((a \ x) \ (a \ x)) \ x by BCIALG_1:7
.= x ` by BCIALG_1:def 5
.= 0. X by BCIALG_1:def 8 ;
(a \ x) \ ((a \ x) \ x) = 0. X by A1, A2, Th43;
hence (a \ x) \ x = a \ x by ; :: thesis: verum
end;
assume A4: for x being Element of X holds (a \ x) \ x = a \ x ; :: thesis:
let x, y be Element of X; :: according to BCIALG_3:def 12 :: thesis: x \ (y \ x) = x
(a \ x) \ ((a \ x) \ x) = (a \ x) \ (a \ x) by A4
.= 0. X by BCIALG_1:def 5 ;
then A5: x \ (x \ (a \ x)) = 0. X by Def1;
y \ a = 0. X by A1;
then y <= a ;
then y \ x <= a \ x by BCIALG_1:5;
then A6: x \ (a \ x) <= x \ (y \ x) by BCIALG_1:5;
(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 ;
then x \ (a \ x) = x by ;
then A7: x \ (x \ (y \ x)) = 0. X by A6;
(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 ; :: thesis: verum