**Proof.** **(Proposition 1)** First, we derive the best responses of the players at each information set. The best responses of

${S}_{1}$ after tables

A and

B have been observed are given as:

On the other hand, the best responses of

${S}_{2}$ after tables

A and

B have been observed are as follows:

The receiver’s best responses after observing a message

$ij\in \{AA,AB,BA,BB\}$ depend on the beliefs formed at that information set:

Furthermore, the beliefs, calculated by Bayes’ rule (whenever possible), are as follows:

We want to show that the senders use the same strategy at the two information sets. To arrive at a contradiction, we consider the following cases: (1) One of the senders uses different strategies, while the other sender uses the same strategy at the two information sets; (2) Both of the senders use different strategies at the two information sets.

**Case 1:** Suppose that ${S}_{1}$ uses different strategies, i.e., ${p}_{A}\ne {p}_{B}$; and, without loss of generality, let’s assume ${p}_{A}>{p}_{B}$. First, consider the case ${q}_{A}={q}_{B}=q\in (0,1)$. Then, the consistency of beliefs requires ${\mu}_{AA}>\frac{1}{2}$, ${\mu}_{AB}>\frac{1}{2}$, ${\mu}_{BA}<\frac{1}{2}$, and ${\mu}_{BB}<\frac{1}{2}$. The best responses of the receiver at each information set under these beliefs become ${r}_{AA}=0$, ${r}_{AB}=0$, ${r}_{BA}=1$, and ${r}_{BB}=1$. But then, ${S}_{1}$’s best responses are ${p}_{A}=0$ and ${p}_{B}=1$, which contradicts to our hypothesis that ${p}_{A}>{p}_{B}$. Now, suppose that ${q}_{A}={q}_{B}=q=0$. With these strategies, the beliefs become ${\mu}_{AB}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. Having these beliefs, the receiver’s best responses can be found as ${r}_{AB}=0$ and ${r}_{BB}=1$. Then, the best responses of ${S}_{1}$ are ${p}_{A}=0$ and ${p}_{B}=1$, again contradicting to our hypothesis. Next, assume that ${q}_{A}={q}_{B}=q=1$. Under these strategies (given the hypothesis ${p}_{A}>{p}_{B}$), the beliefs can be calculated as ${\mu}_{AA}>\frac{1}{2}$, and ${\mu}_{BA}<\frac{1}{2}$ with the associated best responses for the receiver being ${r}_{AA}=0$ and ${r}_{BA}=1$. This in turn suggests that the best responses of ${S}_{1}$ should be ${p}_{A}=0$ and ${p}_{B}=1$, which provides the desired contradiction to ${p}_{A}>{p}_{B}$.

**Case 2:** Suppose that ${p}_{A}\ne {p}_{B}$ and ${q}_{A}\ne {q}_{B}$. Without loss of generality, we assume that ${p}_{A}>{p}_{B}\ge 0$ and ${q}_{A}>{q}_{B}\ge 0$. Then, the beliefs can be calculated as ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. The best responses of the receiver at these information sets become ${r}_{AA}=0$ and ${r}_{BB}=1$. If ${q}_{A}<1$, for ${p}_{A}>0$ to be the best response of ${S}_{1}$, the best responses of the receiver should satisfy ${r}_{AA}={r}_{BA}=0$ and ${r}_{AB}={r}_{BB}=1$. But if ${r}_{AA}={r}_{BA}=0$ and ${r}_{AB}={r}_{BB}=1$, then ${q}_{A}=0$, which is a contradiction as ${q}_{A}>{q}_{B}\ge 0$, by assumption. If ${q}_{A}=1$, then for having ${p}_{A}>0$ as a part of equilibrium, the best response of the receiver should satisfy ${r}_{AA}={r}_{BA}=0$. In turn, ${q}_{A}=1$ can be a best response to these strategies only if ${r}_{AB}=0$ and ${p}_{A}=1$ (in addition to ${r}_{AA}={r}_{BA}=0,{r}_{BB}=1$). But, then ${p}_{B}$ equals to 1 if ${q}_{B}<1$ and ${q}_{B}$ equals to 1 if ${p}_{B}<1$, which is the desired contradiction (since by assumption ${p}_{B}\ne 1$ and ${q}_{B}\ne 1$ as ${p}_{B}<{p}_{A}$ and ${q}_{B}<{q}_{A}$).

Since the senders are symmetric we exclude the symmetric situations. In all the other cases, we get at least one of the beliefs different than $\frac{1}{2}$. The corresponding best responses of the receiver at such information sets are pure strategies; and, the best responses of the senders against these pure strategies give the desired contradiction unless the senders use the same strategies at each information sets. Also, when ${p}_{A}={p}_{B}\in (0,1)$ and ${q}_{A}={q}_{B}\in (0,1)$, the beliefs can be easily calculated as ${\mu}_{ij}=\frac{1}{2}$ and they can be assigned in a consistent way off the equilibrium path. ☐

**Proof.** **(Proposition 2)** The best response of

${S}_{1}$ after table

A is observed is as follows:

The best response of

${S}_{1}$ after table

B is observed is as follows:

The best response of

${S}_{2}$ when the actual table is

A and the sender 1 has sent message

A is given by:

The best response of

${S}_{2}$ when the actual table is

A and the sender 1 has sent message

B is given by:

The best response of

${S}_{2}$ when the actual table is

B and the sender 1 has sent message

A is given by:

The best response of

${S}_{2}$ when the actual table is

B and the sender 1 has sent message

B is given by:

The receiver’s best response after observing message

$ij\in \{AA,AB,BA,BB\}$ is given by:

where the beliefs calculated by the Bayes’ rule (whenever possible) are as follows:

In the first step, we want to show that ${S}_{1}$ uses the same strategy at the two information sets. To do that, first, we are going to assume that ${S}_{2}$ uses the same strategies at her information sets, then we will allow for the case in which ${S}_{2}$ may use different strategies.

**Case 1:** Suppose that ${S}_{2}$ uses the same strategy ${q}_{A}\left(A\right)={q}_{B}\left(A\right)$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)$. For a contradiction, without loss of generality, we assume that ${p}_{A}>{p}_{B}$.

Case 1.a: Assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)\in (0,1)$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)\in (0,1)$. Then, the beliefs can be calculated as ${\mu}_{AA}>\frac{1}{2}$, ${\mu}_{AB}>\frac{1}{2}$, ${\mu}_{BA}<\frac{1}{2}$, and ${\mu}_{BB}<\frac{1}{2}$. The associated best responses of the receiver are ${r}_{AA}=0$, ${r}_{AB}=0$, ${r}_{BA}=1$, and ${r}_{BB}=1$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which contradicts to our hypothesis.

Case 1.b: Now, assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=0$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)=0$. Then, the beliefs become ${\mu}_{BB}<\frac{1}{2}$ and ${\mu}_{AB}>\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{BB}=1$ and ${r}_{AB}=0$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which again contradicts to our hypothesis ${p}_{A}>{p}_{B}$.

Case 1.c: Now, assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=1$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)=1$. Then, the beliefs become ${\mu}_{BA}<\frac{1}{2}$ and ${\mu}_{AA}>\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{BA}=1$ and ${r}_{AA}=0$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which gives the desired contradiction.

Case 1.d: We next assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=0$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)\in (0,1)$. Then, the beliefs become ${\mu}_{BB}<\frac{1}{2}$, ${\mu}_{BA}<\frac{1}{2}$, and ${\mu}_{AB}>\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{BB}=1$, ${r}_{BA}=1$, and ${r}_{AB}=0$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which contradicts to our hypothesis.

Case 1.e: We next assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=0$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)=1$. Then, the beliefs become ${\mu}_{BA}<\frac{1}{2}$ and ${\mu}_{AB}>\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{BA}=1$ and ${r}_{AB}=0$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which contradicts to our hypothesis.

Case 1.f: We next assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=1$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)\in (0,1)$. Then, the beliefs become ${\mu}_{BB}<\frac{1}{2}$, ${\mu}_{BA}<\frac{1}{2}$, and ${\mu}_{AA}>\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{BB}=1$, ${r}_{BA}=1$, and ${r}_{AA}=0$. The best responses of ${S}_{1}$ in turn become ${p}_{A}=0$ and ${p}_{B}=1$, which provides a contradiction.

Case 1.g: We next assume that ${q}_{A}\left(A\right)={q}_{B}\left(A\right)=1$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)=0$. Then, the beliefs become ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. The best responses of the receiver corresponding to these beliefs are ${r}_{AA}=0$ and ${r}_{BB}=1$. The best responses of ${S}_{1}$ again become ${p}_{A}=0$ and ${p}_{B}=1$, which is a contradiction.

The other symmetric cases give the similar contradictions, and thus, are omitted.

**Case 2:** We now consider that case where ${S}_{2}$ uses different strategies. Without loss of generality, we assume that ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)\ge 0$ and ${q}_{A}\left(B\right)>{q}_{B}\left(B\right)\ge 0$. Again, suppose for a contradiction that ${p}_{A}>{p}_{B}\ge 0$. Then, the beliefs are calculated as ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. The corresponding best responses of the receiver at these information sets become ${r}_{AA}=0$ and ${r}_{BB}=1$. Note that for ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)$ and ${q}_{A}\left(B\right)>{q}_{B}\left(B\right)$ to be part of an equilibrium, the receiver’s strategies should satisfy ${r}_{AA}\ge {r}_{AB}$ and ${r}_{BA}\ge {r}_{BB}$. As ${r}_{AA}=0$ and ${r}_{BB}=1$, we get ${r}_{AB}=0$ and ${r}_{BB}=1$. Then, the best response of ${S}_{1}$ against the receiver’s strategies become ${p}_{A}=0$ and ${p}_{B}=1$, which is the desired contradiction.

In the next step, we show that ${S}_{2}$ must use the same strategy in equilibrium, i.e., ${q}_{A}\left(A\right)={q}_{B}\left(A\right)={q}_{1}$ and ${q}_{A}\left(B\right)={q}_{B}\left(B\right)={q}_{2}$.

**Case 1:** We first assume that ${S}_{1}$ uses the same strategy.

Case 1.a: Suppose that ${p}_{A}={p}_{B}\in (0,1)$. Assume for a contradiction that ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)$. This implies ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{AB}<\frac{1}{2}$. The best responses of the receiver in turn becomes ${r}_{AA}=0$ and ${r}_{AB}=1$. The best responses of ${S}_{2}$ against the receiver’s strategy is ${q}_{A}\left(A\right)=0$ and ${q}_{B}\left(A\right)=1$, which contradicts to our hypothesis.

Case 1.b: Suppose that ${p}_{A}={p}_{B}=0$. Assume for a contradiction, ${q}_{A}\left(B\right)>{q}_{B}\left(B\right)$. This implies ${\mu}_{BA}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. The best responses of the receiver in turn becomes ${r}_{BB}=1$ and ${r}_{BA}=0$. The best responses of ${S}_{2}$ against the receiver’s strategy is ${q}_{A}\left(B\right)=0$ and ${q}_{B}\left(B\right)=1$, which contradicts to our hypothesis.

Case 1.c: Suppose that ${p}_{A}={p}_{B}=1$. Assume for a contradiction, ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)$. This implies ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{AB}<\frac{1}{2}$. The best responses of the receiver in turn becomes ${r}_{AA}=0$ and ${r}_{AB}=1$. The best responses of ${S}_{2}$ against the receiver’s strategy is ${q}_{B}\left(A\right)=1$ and ${q}_{A}\left(A\right)=0$, which contradicts to our hypothesis.

All other symmetric cases give the desired results.

**Case 2:** We now assume that ${S}_{1}$ uses different strategies; and without loss of generality assume ${p}_{A}>{p}_{B}\ge 0$. To arrive at a contradiction, without loss of generality, we assume that ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)$ and ${q}_{A}\left(B\right)>{q}_{B}\left(B\right)$. The beliefs can be calculated as ${\mu}_{AA}>\frac{1}{2}$ and ${\mu}_{BB}<\frac{1}{2}$. The best responses of the receiver in turn becomes ${r}_{AA}=0$ and ${r}_{BB}=1$. For ${q}_{A}\left(A\right)>{q}_{B}\left(A\right)$ and ${q}_{A}\left(B\right)>{q}_{B}\left(B\right)$ to be a part of equilibrium, the receiver’s equilibrium strategies should satisfy ${r}_{AA}\ge {r}_{AB}$ and ${r}_{BA}\ge {r}_{BB}$. As ${r}_{AA}=0$ and ${r}_{BB}=1$, we get ${r}_{AB}=0$ and ${r}_{BA}=1$. Against these strategies of the receiver, the best responses of ${S}_{1}$ satisfy ${p}_{A}=0$ and ${p}_{B}=1$, which contradicts to the hypothesis. ☐