is $A$. Since $\typei[1]$ could report any type in $\dettypespace$, she has as options $\detalloci[1](A, A), \detalloci[1](B, A), \detalloci[1](C, A)$ and $\detalloci[1](D, A) \}$, all at the same price by Lemma \[lem:general-equal-pay\]. By Lemma \[lem:fixed-menu\], these are also all the allocations she could possibly get.

Since $[1, 1, 0]$ is the only allocation for which both types $A$ and $B$ have positive value, it is $\typei[1]$’s preferred allocation, i.e., $\extalloci[1](\typei[1], A)$ must be $[1, 1, 0]$. This in turn implies $\extalloci[2](\typei[1], A) = [1, 1, 0]$. (Recall that ${\mathscr{F}}$ is not a product space, and the only allocation in which agent $1$ gets $[1, 1, 0]$ is $\detallocs(A, A) = [1, 1, 0, 1, 1, 0]$.)

Similarly, one can show $\extalloci[1](\typei[1], D) = [4, 4, 0]$, which implies $\extalloci[2](\typei[1], D) = [2, 2, 0]$ or $[4, 4, 0]$. But in either case, weak monotonicity is violated for agent $2$’s types $A$ and $D$. For example, if $\extalloci[2](\typei[1], D) = [2, 2, 0]$, we have $$\begin{aligned}
    \langle A, [1, 1, 0] \rangle + \langle D, [2, 2, 0] \rangle < \langle A, [2, 2, 0] \rangle + \langle D, [1, 1, 0] \rangle.\end{aligned}$$ Therefore, no deterministic DSIC extension of $\detallocs$ is 