[proofplan] We prove the containment by taking an arbitrary element of $A_t+B$ and showing that it lies in both sets on the right-hand side. Membership in $A+B$ follows from the part of the definition $A_t \subset A$. Membership in $B+B+t$ follows from the part of the definition $A_t \subset B+t$. The special case is obtained by substituting $B=A$. [/proofplan] [step:Show that every element of $A_t+B$ lies in $A+B$] Let $x \in A_t+B$. By definition of the sumset $A_t+B$, there exist $a \in A_t$ and $b \in B$ such that \begin{align*} x = a+b. \end{align*} Since $A_t = A \cap (B+t)$, we have $a \in A$. Therefore $a \in A$ and $b \in B$, so the same representation gives \begin{align*} x = a+b \in A+B. \end{align*} [/step] [step:Use the translate condition defining $A_t$ to place the same element in $B+B+t$] Again let $x \in A_t+B$, and write $x=a+b$ with $a \in A_t$ and $b \in B$. Since $A_t = A \cap (B+t)$, we also have $a \in B+t$. By definition of the translate $B+t$, there exists $b' \in B$ such that \begin{align*} a = b' + t. \end{align*} Substituting this expression for $a$ into $x=a+b$ gives \begin{align*} x = a+b = (b'+t)+b. \end{align*} Because $G$ is abelian, we may reorder the terms: \begin{align*} x = b'+b+t. \end{align*} Here $b',b \in B$, so $b'+b \in B+B$, and hence \begin{align*} x \in B+B+t. \end{align*} [/step] [step:Intersect the two memberships and specialize to $B=A$] The previous two steps show that every $x \in A_t+B$ satisfies both $x \in A+B$ and $x \in B+B+t$. Hence \begin{align*} x \in (A+B)\cap(B+B+t). \end{align*} Since $x \in A_t+B$ was arbitrary, this proves \begin{align*} A_t+B \subset (A+B)\cap(B+B+t). \end{align*} For the stated special case, set $B=A$. Then for $t \in A-A$ the set $A_t$ becomes \begin{align*} A_t = A \cap (A+t), \end{align*} and the containment just proved becomes \begin{align*} (A\cap(A+t))+A \subset (A+A)\cap(A+A+t). \end{align*} This is exactly the claimed special case. [/step]