then for any $V \in \{A, B, C, D\}$, $\payi[1](A, V) = \payi[1](B, V) = \payi[1](C, V) = \payi[1](D, V)$, and $\payi[2](V, A) = \payi[2](V, B) = \payi[2](V, C) = \payi[2](V, D)$.

We prove the lemma for agent $1$, and the statement for agent $2$ follows by symmetry.

By DSIC, for any $V \in \dettypespace$, we have $$\begin{aligned}
    \langle A , \detalloci[1](A,V)\rangle & -\payi[1](A,V)  \\
     & \geq  \langle A , \detalloci[1](C,V)\rangle -\payi[1](C,V); \\
    \langle B , \detalloci[1](B,V)\rangle & -\payi[1](B,V) \\
    & \geq \langle B , \detalloci[1](A,V)\rangle -\payi[1](A,V);
     \\
     \langle C, \detalloci[1](C,V)\rangle & -\payi[1](C,V) \\
     & \geq \langle C , \detalloci[1](B,V)\rangle -\payi[1](B,V).\end{aligned}$$ Note that $$\begin{aligned}
    \langle A , \detalloci[1](A,V)\rangle = \langle A , \detalloci[1](C,V)\rangle; \\
    \langle C , \detalloci[1](C,V)\rangle = \langle C , \detalloci[1](B,V)\rangle; \\
    \langle B , \detalloci[1](B,V)\rangle = \langle B , \detalloci[1](A,V)\rangle.
    \end{aligned}$$ Therefore $\payi[1](A,V) \leq  \payi[1](C,V) \leq \payi[1](B, V) \leq \payi[1](A, V)$. Hence all inequalities are tight and we have $\payi[1](C,V) = \payi[1](A,V) =  \payi[1](B,V)$.

For type $D$’s payment, note that $\detalloci[1](B, V)= \detalloci[1](D, V)$. If $\payi[1](B, V) \neq \payi[1](D, V)$, one of $B$ and $D$ must be incentivized to misreport the other type. Therefore $\payi[1](B, V) = \payi[1](D, V)$.

In figure 2 in the supplementary materials we 