= Restricted sumset bound for unequal subsets of a prime field
{title2=$|\{a+b:a\in A,b\in B,a\ne b\}|\geq\min(p,|A|+|B|-2)$}
Suppose $|A|>|B|>0$ in $\mathbb F_p$. Put $a=|A|$ and $b'=\min(|B|,p+2-a)$, and choose $B'\subseteq B$ of size $b'$. Let $r=a+b'-2$. If the restricted sumset fits in a set $D$ of size $r-1$, then $(x-y)\prod_{d\in D}(x+y-d)$ vanishes on $A\times B'$. Its <coefficient> of $x^{a-1}y^{b'-1}$ is
$$
(a-b')\frac{(r-1)!}{(a-1)!(b'-1)!}\ne0\quad\text{in }\mathbb F_p.
$$
All factorial arguments are below $p$ and $0<a-b'<p$. The <Combinatorial Nullstellensatz> gives a contradiction. Since $r=\min(p,|A|+|B|-2)$, this includes the saturated case as well as the small-set case.
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