Reactive Systems: Modelling, Specification and Verification - Cs.ioc.ee

4.2. TARSKI’S FIXED POINT THEOREM 93
of elements of D. As D is finite, the sequence must be eventually constant, i.e.,
there is an m such that f k (⊥) = f m (⊥) for all k ≥ m. In particular,
f(f m (⊥)) = f m+1 (⊥) = f m (⊥) ,
which is the same as saying that f m (⊥) is a fixed point for f.
To prove that f m (⊥) is the least fixed point for f, assume that d is another
fixed point for f. Then we have that ⊥ ⊑ d and therefore, as f is monotonic,
that f(⊥) ⊑ f(d) = d. By repeating this reasoning m−1 more times we get that
f m (⊥) ⊑ d. We can therefore conclude that f m (⊥) is the least fixed point for f.
The proof of the statement that characterizes largest fixed points is similar, and
is left as an exercise for the reader. ✷
Exercise 4.8 (For the theoretically minded) Fill in the details in the proof of the
above theorem.
Example 4.3 Consider the function f : 2 {0,1} → 2 {0,1} defined by
f(X) = X ∪ {0} .
This function is monotonic, and 2 {0,1} is a complete lattice, when ordered using set
inclusion, with the empty set as least element and {0, 1} as largest element. The
above theorem gives an algorithm for computing the least and largest fixed point
of f. To compute the least fixed point, we begin by applying f to the empty set.
The result is {0}. Since, we have added 0 to the input of f, we have not found our
least fixed point yet. Therefore we proceed by applying f to {0}. We have that
f({0}) = {0} ∪ {0} = {0} .
It follows that, not surprisingly, {0} is the least fixed point of the function f.
To compute the largest fixed point of f, we begin by applying f to the top
element in our lattice, namely the set {0, 1}. Observe that
f({0, 1}) = {0, 1} ∪ {0} = {0, 1} .
Therefore {0, 1} is the largest fixed point of the function f.
Exercise 4.9 Consider the function g : 2 {0,1,2} → 2 {0,1,2} defined by
g(X) = (X ∩ {1}) ∪ {2} .
Use Theorem 4.2 to compute the least and largest fixed point of g.