Let $p$ be a prime and let all variables be in $\Bbb{Z}_p$.
Then you can write the result of
if(i > 0)
k = (a + b)c;
(C code)
as a polynomial $k := i^{p-1} (a+b)c + (1 - i^{p-1}) k$ (notice $:=$ and not $=$). But what about
if (i > j)
k = (a + b)c;
?
If you try converting $i \gt j$ when your ordering on $\Bbb{Z}_p$ is $0, 1 \lt 2 \lt \dots \lt p-1 $, then we have that $i - j \gt 0 \not\implies i \gt j$, but the converse holds.
Then how can we construct a polynomial $F$ in $i,j,k,a,b,c$ such that $k := F(i,j,k,a,b,c)$ yields the same computation?
I bet there is one!!!