let X be BCI-algebra; :: thesis: ( X is BCI-commutative iff for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y )

A1: ( ( for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y ) implies X is BCI-commutative )
proof
assume A2: for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y ; :: thesis:
for x, z being Element of X st x \ z = 0. X holds
x = z \ (z \ x)
proof
let x, z be Element of X; :: thesis: ( x \ z = 0. X implies x = z \ (z \ x) )
set y = z \ (z \ x);
(z \ (z \ (z \ x))) \ (z \ x) = (z \ x) \ (z \ x) by BCIALG_1:8
.= 0. X by BCIALG_1:def 5 ;
then A3: z \ (z \ (z \ x)) <= z \ x ;
assume x \ z = 0. X ; :: thesis: x = z \ (z \ x)
then x <= z ;
then x <= z \ (z \ x) by A2, A3;
then A4: x \ (z \ (z \ x)) = 0. X ;
(z \ (z \ x)) \ x = (z \ x) \ (z \ x) by BCIALG_1:7
.= 0. X by BCIALG_1:def 5 ;
hence x = z \ (z \ x) by ; :: thesis: verum
end;
hence X is BCI-commutative ; :: thesis: verum
end;
( X is BCI-commutative implies for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y )
proof
assume A5: X is BCI-commutative ; :: thesis: for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y

for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y
proof
let x, y, z be Element of X; :: thesis: ( x <= z & z \ y <= z \ x implies x <= y )
assume that
A6: x <= z and
A7: z \ y <= z \ x ; :: thesis: x <= y
x \ z = 0. X by A6;
then A8: x = z \ (z \ x) by A5;
(z \ y) \ (z \ x) = 0. X by A7;
then 0. X = x \ y by ;
hence x <= y ; :: thesis: verum
end;
hence for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y ; :: thesis: verum
end;
hence ( X is BCI-commutative iff for x, y, z being Element of X st x <= z & z \ y <= z \ x holds
x <= y ) by A1; :: thesis: verum