Most of the operators extend to sets, in a way which is consistent with the interpretation of sets as independent nondeterministic choices. Generally, a unary operator f obeys the law
f{x,y} = {f(x),f(y)}
Thus, for example,
-(2..3) = (-2..-3)and
~{0,1} = {1,0}
For a binary operator *, we have
{x,y} * z = {x * z, y * z}
x * {y,z} = {x * y, x * z}
For example,
3 + {4,5} = {7,8}
and
0 & {0,1} = 0
1 & {0,1} = {0,1}
(which are actually special cases of the laws given above for ``and'').
This behavior of sets is somewhat counterintuitive when the equality operator is applied to sets. For example, the result of
{a,b} = {a,b}
is not equal to 1 (true). The way to understand this is to think of
each set as representing an arbitrary choice among its elements. Thus,
the result of the above expression is the set 
, since we may
choose equal elements or we may choose unequal elements.
The exception to the above rule is the ``in'' operator, which compares a value and a set of values. In this case, only the left represents a nondeterministic choice. That is:
{x,y} in z = {x in z, y in z}
However, as stated previously,
x in {y,z} = (x in y) | (x in z)
[N.B. This makes ``in'' the only operator in the language which is not monotonic with respect to set containment. The ``in'' operator is only monotonic in its left argument. A formal verification system that relies on monotonicity (such as ternary symbolic simulation) should allow only constant sets on the right hand side of ``in''.]