Fixpoints of a transformation of an extended WHILE program
Suppose $f$ is a fixpoint of $A$. First,
$$
f(0) = f(1) = f(2) = f(3) = 3f(3),
$$
and so $f(0) = f(1) = f(2) = f(3) = 0$. Second,
$$
f(5) = 5f(4) = 5f(5),
$$
and so $f(4) = f(5) = 0$. Third, if $n = 4m+3 \geq 7$ then $$f(n) = n f(2m+3),$$ while if $n = 4m+1 \geq 9$ then $$f(n) = n f(2m+2) = nf(2m+3).$$ In both cases, $2m+3 &lt; n$, and so you can prove by induction that $f(n) = 0$ for all odd $n$. That implies that $f(n) = 0$ also for all even $n$, and we conclude that $f(n) = 0$ for all $n$.

Edit: This is the maximal fixpoint. The minimal fixpoint is $\bot$. There are two other fixpoints: $f(n) = 0$ for $n \leq 3$ and undefined otherwise, and $f(n) = 0$ for $n &gt; 3$ and undefined otherwise.