type $D$ has the same value for all allocations given a fixed type of the opponent, and that $\payi[1](D, V_2) = \payi[1](V_1, V_2)$ by Lemma \[lem:equal-pay\]. Crucially, and  cannot both be tight, because by construction, for any $V_2 \neq V_2'$, $$\begin{gathered}
\langle V_1, \detalloci[1](V_1, V_2) - \detalloci[1](V_1, V_2') \rangle \\
= \frac{3}{2} \langle D, \detalloci[1](D, V_2) - \detalloci[1](V_1, V_2') \rangle \neq 0.  \end{gathered}$$ Therefore, $\frac{2}{3} \cdot$ $+ \frac{1}{3} \cdot$ gives $$\begin{gathered}
\langle \frac{2}{3} V_1^* + \frac{1}{3} D, \detalloci[1](V_1^*, V_2) \rangle - \payi[1](V_1^*, V_2)
\\ 
> \langle \frac{2}{3} V_1 + \frac{1}{3} D, \detalloci[1](V_1, V_2') \rangle - \payi[1](\typei[1], V_2),\end{gathered}$$ which contradicts .

By the same reasoning as for Lemma \[lem:equal-pay\], the following lemma follows from Lemma \[lem:fixed-menu\].

\[lem:general-equal-pay\] If $\extallocs$ is a deterministic DSIC extension of $\detallocs$, implementable by payment rule $\pays$, then for any $\typei[1]$ in the interior of $\operatorname{Conv}(\dettypespace)$ and any $V_2 \in \dettypespace$, $\payi[1](\typei[1], V_2) = \payi[1](A, V_2)$.

We are now ready to prove Theorem \[thm:det\].

Suppose $\extallocs$ is a deterministic DSIC extension of $\detallocs$. We show a contradiction by showing that $\extallocs$ must violate weak monotonicity.

Consider $\typei[1] = \tfrac 1 3 A + \tfrac 1 3 B + \tfrac 1 3 D$ and when agent $2$’s type 