= Solution
Write $a=|A|$ and $b=|B|$, so $p\geq a>b\geq1$. To cover both the unsaturated and saturated cases with the same <coefficient> calculation, set
$$
b'=\min\{b,p+2-a\},\qquad r=a+b'-2=\min\{p,a+b-2\}.
$$
Choose $B'\subseteq B$ of size $b'$. These parameters satisfy $1\leq b'<a\leq p$ and $1\leq r\leq p$.
Assume, for a contradiction, that the restricted sumset $C$ has fewer than $r$ elements. Extend it to a <subset> $D\subseteq\mathbb F_p$ of size $r-1$, and form
$$
F(x,y)=(x-y)\prod_{z\in D}(x+y-z).
$$
On $A\times B'$, this <polynomial> vanishes: when $x=y$ the first factor vanishes, and when $x\ne y$ the sum belongs to $C\subseteq D$. Its total degree is $r=(a-1)+(b'-1)$. Only the leading homogeneous part $(x-y)(x+y)^{r-1}$ contributes to the <coefficient> of $x^{a-1}y^{b'-1}$. That <coefficient> is
$$
\binom{r-1}{a-2}-\binom{r-1}{a-1}
=(a-b')\frac{(r-1)!}{(a-1)!(b'-1)!}.
$$
The equality includes $b'=1$, with an out-of-range binomial <coefficient> interpreted as zero. All factorial arguments are between zero and $p-1$, and $0<a-b'<p$, so this <coefficient> is nonzero in $\mathbb F_p$. The <Combinatorial Nullstellensatz> applied to $A\times B'$ forces a nonzero value, contradicting the vanishing just proved. Hence the <restricted sumset bound for unequal subsets of a prime field> is
$$
\boxed{|C|\geq\min\{p,|A|+|B|-2\}.}
$$
The strict size inequality is used precisely in the nonzero factor $a-b'$. The argument includes $r=p$ without dividing by $p!$, so no separate saturation assumption is hidden.
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