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

thus ( X is BCI-commutative implies for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z) ) :: thesis: ( ( for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z) ) implies X is BCI-commutative )
proof
assume A1: X is BCI-commutative ; :: thesis: for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z)

for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z)
proof
let x, y, z be Element of X; :: thesis: ( x <= y & x <= z implies x <= y \ (y \ z) )
assume that
A2: x <= y and
A3: x <= z ; :: thesis: x <= y \ (y \ z)
A4: x \ z = 0. X by A3;
x \ y = 0. X by A2;
then A5: x = y \ (y \ x) by A1;
then x \ (y \ (y \ z)) = (y \ (y \ (y \ z))) \ (y \ x) by BCIALG_1:7
.= (y \ z) \ (y \ x) by BCIALG_1:8
.= 0. X by ;
hence x <= y \ (y \ z) ; :: thesis: verum
end;
hence for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z) ; :: thesis: verum
end;
assume A6: for x, y, z being Element of X st x <= y & x <= z holds
x <= y \ (y \ z) ; :: thesis:
for x, y being Element of X st x \ y = 0. X holds
x = y \ (y \ x)
proof
let x, y be Element of X; :: thesis: ( x \ y = 0. X implies x = y \ (y \ x) )
x \ x = 0. X by BCIALG_1:def 5;
then A7: x <= x ;
assume x \ y = 0. X ; :: thesis: x = y \ (y \ x)
then x <= y ;
then x <= y \ (y \ x) by A6, A7;
then A8: x \ (y \ (y \ x)) = 0. X ;
(y \ (y \ x)) \ x = (y \ x) \ (y \ x) by BCIALG_1:7
.= 0. X by BCIALG_1:def 5 ;
hence x = y \ (y \ x) by ; :: thesis: verum
end;
hence X is BCI-commutative ; :: thesis: verum