[proofplan] We first normalize the finite set by translating and dividing by its additive gcd, because these operations preserve cardinalities of sumsets and turn arithmetic progressions back into arithmetic progressions. The only structural input is the Freiman interval bound: a normalized set $B \subset \{0,\dots,n\}$ with $0,n \in B$, $\gcd(B)=1$, and $|B+B| \le 3|B|-4$ satisfies $n \le |B+B|-|B|$. Once this bound is available, $B$ lies in the interval $\{0,\dots,n\}$ of length at most $|B+B|-|B|+1$, and undoing the normalization gives the required arithmetic progression for $A$. [/proofplan]

[step:Normalize the set by translation and dilation]Let $a_{\min} := \min A$ and $a_{\max} := \max A$. Define the translated set $A_0 \subset \mathbb{Z}$ by \begin{align*} A_0 := A - a_{\min} = \{a - a_{\min} : a \in A\}. \end{align*} Then $0 \in A_0$, $\max A_0 = a_{\max} - a_{\min}$, and $|A_0| = |A| = k$. The map \begin{align*} \tau: A &\to A_0 \\ a &\mapsto a - a_{\min} \end{align*} is a bijection, and the map \begin{align*} \tau_2: A+A &\to A_0 + A_0 \\ s &\mapsto s - 2a_{\min} \end{align*} is also a bijection. Hence \begin{align*} |A_0 + A_0| = |A+A| \le 3k-4. \end{align*} Let $d \in \mathbb{N}$ be the additive gcd of $A_0$, defined by \begin{align*} d := \gcd\{x : x \in A_0\}. \end{align*} Since $0 \in A_0$ and $|A_0|=k\ge 3$, the set $A_0$ contains a positive integer, so $d$ is well-defined and positive. Define the normalized set $B \subset \mathbb{Z}$ by \begin{align*} B := d^{-1}A_0 = \{x/d : x \in A_0\}. \end{align*} Then $|B|=k$, $0 \in B$, $\gcd(B)=1$, and if $n := \max B$, then $B \subset \{0,1,\dots,n\}$ and $n \in B$. The map \begin{align*} \delta_2: A_0 + A_0 &\to B+B \\ s &\mapsto s/d \end{align*} is a bijection, so \begin{align*} |B+B| = |A_0+A_0| = |A+A| \le 3k-4. \end{align*}[/step]

[guided]The purpose of this step is to put the set into the standard form required by the interval theorem. We begin by removing the minimum element. Let $a_{\min} := \min A$ and $a_{\max} := \max A$, and define \begin{align*} A_0 := A - a_{\min} = \{a - a_{\min} : a \in A\}. \end{align*} The translation map \begin{align*} \tau: A &\to A_0 \\ a &\mapsto a - a_{\min} \end{align*} is bijective, so $|A_0|=|A|=k$. It also sends the smallest element of $A$ to $0$, so $0 \in A_0$, and the largest element becomes $a_{\max}-a_{\min}$. We must also check that the sumset size is unchanged. Define \begin{align*} \tau_2: A+A &\to A_0 + A_0 \\ s &\mapsto s - 2a_{\min}. \end{align*} If $s=a_1+a_2$ with $a_1,a_2\in A$, then \begin{align*} \tau_2(s)=(a_1-a_{\min})+(a_2-a_{\min}) \in A_0+A_0. \end{align*} Conversely, every element of $A_0+A_0$ has this form, and the inverse map is $t\mapsto t+2a_{\min}$. Thus $\tau_2$ is bijective and \begin{align*} |A_0 + A_0| = |A+A| \le 3k-4. \end{align*} Next we remove any common additive divisor. Let \begin{align*} d := \gcd\{x : x \in A_0\}. \end{align*} Because $k\ge 3$, the set $A_0$ has a positive element, namely $a_{\max}-a_{\min}$, so $d\in\mathbb N$ is positive. Define \begin{align*} B := d^{-1}A_0 = \{x/d : x \in A_0\}. \end{align*} Every element of $A_0$ is divisible by $d$, so $B\subset\mathbb Z$. Division by $d$ is injective, hence $|B|=|A_0|=k$. By construction $0\in B$, and the gcd of all elements of $B$ is $1$. Let $n:=\max B$. Then $B\subset \{0,1,\dots,n\}$ and $n\in B$. Finally, division by $d$ also preserves the cardinality of the sumset: the map \begin{align*} \delta_2: A_0 + A_0 &\to B+B \\ s &\mapsto s/d \end{align*} is bijective with inverse $u\mapsto du$. Therefore \begin{align*} |B+B| = |A_0+A_0| = |A+A| \le 3k-4. \end{align*}[/guided]

[step:Apply the normalized Freiman interval bound]We use the following structural form of Freiman's interval theorem. [claim:Freiman Lev Smeliansky Stanchescu interval theorem] Let $B\subset\mathbb Z$ be finite with $|B|=k\ge 3$. Suppose $B\subset\{0,1,\dots,n\}$, $0,n\in B$, $\gcd(B)=1$, and \begin{align*} |B+B|\le 3k-4. \end{align*} Then \begin{align*} n\le |B+B|-k. \end{align*} [/claim] [proof] This is the interval form of the Freiman--Lev--Smeliansky--Stanchescu inverse theorem for subsets of the integers, used here as an external structural input rather than as a restatement of the present theorem. In this interval form, the hypotheses are exactly: the set is finite, the endpoints $0$ and $n$ both belong to the set, the additive gcd is $1$, and the doubling bound is at most $3k-4$. Its conclusion is that the ambient interval length $n+1$ is at most $|B+B|-k+1$, equivalently $n\le |B+B|-k$. [/proof] The set $B$ constructed in the previous step satisfies every hypothesis of the Freiman--Lev--Smeliansky--Stanchescu interval theorem: it is finite, $|B|=k\ge3$, $B\subset\{0,1,\dots,n\}$, $0,n\in B$, $\gcd(B)=1$, and $|B+B|\le3k-4$. Therefore \begin{align*} n\le |B+B|-k. \end{align*}[/step]

[guided]At this point all elementary normalization work is complete. The remaining input is the standard normalized interval form of the Freiman--Lev--Smeliansky--Stanchescu $3k-4$ theorem. We record it as a claim: if $B\subset\{0,1,\dots,n\}$ has endpoints $0,n$, has additive gcd $1$, has $|B|=k\ge3$, and satisfies \begin{align*} |B+B|\le 3k-4, \end{align*} then \begin{align*} n\le |B+B|-k. \end{align*} We verify the hypotheses for the present set $B$. The previous step proved that $B$ is finite and $|B|=k\ge3$. It also defined $n:=\max B$, so $B\subset\{0,1,\dots,n\}$ and $n\in B$; since $0\in A_0$, division by $d$ gives $0\in B$. The definition of $d$ as the gcd of the translated set gives $\gcd(B)=1$. Finally, the bijection between $A+A$ and $B+B$ gives \begin{align*} |B+B|=|A+A|\le3k-4. \end{align*} All hypotheses of the Freiman--Lev--Smeliansky--Stanchescu interval theorem are therefore satisfied, so \begin{align*} n\le |B+B|-k. \end{align*}[/guided]

[step:Place the normalized set in a short arithmetic progression] Since $B\subset\{0,1,\dots,n\}$, the set $B$ is contained in the arithmetic progression \begin{align*} P_B := \{0,1,\dots,n\}. \end{align*} This progression has common difference $1$ and length $n+1$. By the bound from the previous step, \begin{align*} |P_B| = n+1 \le |B+B|-k+1. \end{align*} Because $|B+B|=|A+A|$, we obtain \begin{align*} |P_B| \le |A+A|-k+1. \end{align*} If necessary, extend $P_B$ on the right to an arithmetic progression $P_B'\subset\mathbb Z$ with common difference $1$ and exact length $|A+A|-k+1$; since the extension contains $P_B$, it also contains $B$. [/step]

[step:Undo the normalization to obtain the progression containing $A$]Let $L := |A+A|-k+1$. From the previous step there is an arithmetic progression $P_B'\subset\mathbb Z$ of length $L$ and common difference $1$ such that $B\subset P_B'$. Let $r\in\mathbb Z$ be the first term of $P_B'$, so \begin{align*} P_B' = \{r,r+1,\dots,r+L-1\}. \end{align*} Define the first term $a_0\in\mathbb Z$ by $a_0:=a_{\min}+dr$, and define \begin{align*} P_A := a_{\min} + dP_B' = \{a_0,a_0+d,\dots,a_0+(L-1)d\}. \end{align*} Then $P_A$ is an arithmetic progression in $\mathbb Z$ with first term $a_0$, common difference $d$, and length $L$. Since $B=d^{-1}(A-a_{\min})$, every $a\in A$ has the form $a=a_{\min}+db$ for some $b\in B\subset P_B'$. Hence $a\in P_A$, so $A\subset P_A$. Therefore $A$ is contained in an arithmetic progression of length \begin{align*} L=|A+A|-k+1, \end{align*} as required.[/step]

[guided]Let \begin{align*} L := |A+A|-k+1. \end{align*} The previous step produced an arithmetic progression $P_B'\subset\mathbb Z$ of length $L$ and common difference $1$ with $B\subset P_B'$. Let $r\in\mathbb Z$ be its first term, so \begin{align*} P_B'=\{r,r+1,\dots,r+L-1\}. \end{align*} We now reverse the two normalizing operations: first multiply by $d$, then translate by $a_{\min}$. Define $a_0:=a_{\min}+dr$, and define \begin{align*} P_A := a_{\min} + dP_B' = \{a_0,a_0+d,\dots,a_0+(L-1)d\}. \end{align*} Because $P_B'$ is an arithmetic progression with common difference $1$, its image under $p\mapsto a_{\min}+dp$ is an arithmetic progression in $\mathbb Z$ with first term $a_0$ and common difference $d$. This map is injective, so the length remains $L$. It remains to check containment. Since $B=d^{-1}(A-a_{\min})$, for each $a\in A$ there exists $b\in B$ such that \begin{align*} a=a_{\min}+db. \end{align*} Since $B\subset P_B'$, this element $b$ lies in $P_B'$, and hence $a\in P_A$. Thus $A\subset P_A$. The progression $P_A$ has length \begin{align*} L=|A+A|-k+1, \end{align*} which is exactly the length required in the theorem statement.[/guided]