give a $3$ dimensional visualization of the agent’s value for each allocation which gives intuition into the proof of \[lem:equal-pay\].

\[lem:fixed-menu\] If $\tilde{x}$ is a deterministic DSIC extension of $\detallocs$, then for $\typei[1] = $ $\frac{1}{3}A + \frac{1}{3}B + \frac{1}{3}D$, and any $V_2 \in \dettypespace$, $\extalloci[1](\typei[1], V_2) \in $ $\{\extalloci[1](A, V_2), \extalloci[1](B, V_2),$ $\extalloci[1](C, V_2),$ $ \extalloci[1](D, V_2)\}$.

For the sake of contradiction, assume $\extallocs$ is a DSIC extension of $\detallocs$, implementable by payment rule $\pays$, and for $\typei[1]$ and $V_2 \in \dettypespace$, $\extalloci[1](\typei[1], V_2) = \detalloci[1](V_1, V_2')$ for some $(V_1, V_2') \in \dettypespaces$ and $V_2' \neq V_2$.

For any $V_2$, one of $\detalloci[1](A, V_2)$ and $\detalloci[1](B, V_2)$ gives an equal positive value to both $A$ and $B$. Let $V_1^*$ be the type that induces this equally valued allocation. (For example, if $V_2 = B$, then $\detalloci[1](A, B) = [2, 0, 2]$ and $\detalloci[1](B, B) = [2, 2, 0]$. Both $A$ and $B$ have the same value for $\detalloci[1](B, B)$ and so type $B$ would be $V_1^*$.)

Observe that, for the allocation $\detalloci[1](V_1, V_2')$, $V_1$ has positive value $\langle V_1, \detalloci[1](V_1, V_2') \rangle$, and no other type has higher value for it. Therefore, type $\typei[1]$ has value at most $\langle 