[proofplan]
We reduce the symmetric restricted sumset bound $|A \mathbin{\dot{+}} A| \geq 2|A| - 3$ to the [Asymmetric Restricted Sumset Lower Bound](/theorems/2620) by removing a single element from $A$. Setting $A' = A \setminus \{a\}$ for any $a \in A$, we have $A' \mathbin{\dot{+}} A \subseteq A \mathbin{\dot{+}} A$ and $|A'| < |A|$, so the asymmetric bound applies to the pair $(A', A)$ and yields the result.
[/proofplan]
[step:Dispatch the trivial case $|A| = 1$]
If $|A| = 1$, then $A \mathbin{\dot{+}} A = \{a + b : a, b \in A, a \neq b\} = \varnothing$ since $A$ has only one element and no pair with $a \neq b$ exists. The bound $|A \mathbin{\dot{+}} A| \geq 2|A| - 3 = -1$ is vacuously satisfied because $|\varnothing| = 0 \geq -1$.
[/step]
[step:Reduce to the asymmetric bound by removing one element]
Assume $|A| \geq 2$. Fix any $a \in A$ and set $A' = A \setminus \{a\}$, so $|A'| = |A| - 1 \geq 1$.
We claim $A' \mathbin{\dot{+}} A \subseteq A \mathbin{\dot{+}} A$. Take any element $s \in A' \mathbin{\dot{+}} A$, so $s = a' + b$ where $a' \in A'$, $b \in A$, and $a' \neq b$. Since $A' \subseteq A$, we have $a' \in A$, so $s = a' + b$ with $a', b \in A$ and $a' \neq b$, giving $s \in A \mathbin{\dot{+}} A$.
Therefore $|A \mathbin{\dot{+}} A| \geq |A' \mathbin{\dot{+}} A|$.
[guided]
Why remove just one element? The asymmetric bound requires $|A'| < |B|$, i.e., strict inequality between the two set sizes. Since we start with a single set $A$ and want a bound on $A \mathbin{\dot{+}} A$, we need to create an asymmetry. Removing one element from $A$ is the minimal perturbation: it produces a pair $(A', A)$ with $|A'| = |A| - 1 < |A|$, and the containment $A' \mathbin{\dot{+}} A \subseteq A \mathbin{\dot{+}} A$ ensures the asymmetric bound transfers to the symmetric one.
Note that the direction of the containment matters: every restricted sum from the pair $(A', A)$ is also a restricted sum from the pair $(A, A)$, because $A' \subseteq A$ and the distinctness condition $a' \neq b$ is preserved. The converse fails in general -- sums $a + b$ with both $a, b \in A \setminus A'$ might not appear in $A' \mathbin{\dot{+}} A$ -- but we only need the one-sided containment.
[/guided]
[/step]
[step:Verify the hypotheses of the asymmetric bound and conclude]
We check the conditions of the [Asymmetric Restricted Sumset Lower Bound](/theorems/2620) for the pair $(A', A)$:
- $|A'| = |A| - 1 \geq 1$ and $|A| \geq 2$, so $2 \leq |A'| + 1 = |A|$. We need $|A'| \geq 2$, i.e., $|A| \geq 3$. If $|A| = 2$, then $|A'| = 1$, and the asymmetric bound requires $|A'| \geq 2$, so we handle this case directly: $A = \{a_1, a_2\}$ with $a_1 \neq a_2$, so $A \mathbin{\dot{+}} A = \{a_1 + a_2\}$ and $2|A| - 3 = 1 = |A \mathbin{\dot{+}} A|$.
- For $|A| \geq 3$: $|A'| = |A| - 1 \geq 2$, $|A'| < |A|$ (since $|A| - 1 < |A|$), and $|A'| + |A| = 2|A| - 1 \leq p + 2$ (since $2|A| \leq p + 3$ by hypothesis). All conditions are satisfied.
Applying the [Asymmetric Restricted Sumset Lower Bound](/theorems/2620):
\begin{align*}
|A' \mathbin{\dot{+}} A| \geq |A'| + |A| - 2 = (|A| - 1) + |A| - 2 = 2|A| - 3.
\end{align*}
Combining with the containment:
\begin{align*}
|A \mathbin{\dot{+}} A| \geq |A' \mathbin{\dot{+}} A| \geq 2|A| - 3.
\end{align*}
[guided]
Let us trace the size condition carefully. The asymmetric bound requires $|A'| + |A| \leq p + 2$, which is $2|A| - 1 \leq p + 2$, i.e., $2|A| \leq p + 3$. This is exactly the hypothesis of the Erdos-Heilbronn conjecture, so the condition propagates perfectly.
For the edge case $|A| = 2$: with $A = \{a_1, a_2\} \subseteq \mathbb{Z}_p$ and $a_1 \neq a_2$ (since $A$ is a set), the restricted sumset is $A \mathbin{\dot{+}} A = \{a_1 + a_2\}$, a single element. The bound gives $|A \mathbin{\dot{+}} A| \geq 2 \cdot 2 - 3 = 1$, which is tight. This case must be handled separately because $|A'| = 1 < 2$ violates the hypothesis of the asymmetric bound.
[/guided]
[/step]
