Matthew: okay, thanks for going through all that - I think you're right! But I think this particular feasibility relation \\( \Phi : \mathbf{Bool} \nrightarrow [0,\infty) \\) is also the conjoint of the monotone function

\[ G : [0,\infty) \to \mathbf{Bool} \]

given by

\[ G(y) = \begin{cases} \texttt{true} &\mbox{if } y \ge 500 \\\\

\texttt{false} & \mbox{otherwise.}\end{cases} \]

Let's see: the conjoint of \\(G\\) is defined by

\[ \check{G}(x,y) = \mathbf{Bool}(x, G(y)) .\]

so it evaluates as \\(\texttt{true}\\) iff \\(x \le G(y)\\).

Two cases here:

* \\(y \ge 500\\). In this case \\(G(y) = \texttt{true}\\) so we have \\(x \le G(y) \\) for all \\(x \in \mathbf{Bool}\\). Thus, in this case \\(\check{G}(x,y) = \texttt{true}\\) for all \\(x\\).

* \\(y \lt 500 \\). In this case \\(G(y) = \texttt{false}\\) so we have \\(x \le G(y) \\) iff \\(x = \texttt{false}\\). Thus, in this case \\(\check{G}(x,y) = \texttt{true}\\) only for \\(x = \texttt{false}\\).

In short, \\(\check{G}(x,y) = \texttt{true}\\) iff either \\(y \ge 500\\) or \\(x = \texttt{false}\\).

Or in the English: the feasible situations are precisely those where you either have more than $500, or you don't buy a ticket!

So yes, I think we were both right. I think this feasibility relation is both a companion and a conjoint!

Somehow the conjoint seems more intuitive to me: from the amount of money you have, I can determine whether or not you can buy a ticket.

The companion approach is more mysterious to me: from whether or not you can buy a ticket, I cannot determine how much money you have!

Nonetheless you were able to dream up a monotone function from truth values to amounts of money, whose companion gives the desired feasibility relation. Interesting!

\[ G : [0,\infty) \to \mathbf{Bool} \]

given by

\[ G(y) = \begin{cases} \texttt{true} &\mbox{if } y \ge 500 \\\\

\texttt{false} & \mbox{otherwise.}\end{cases} \]

Let's see: the conjoint of \\(G\\) is defined by

\[ \check{G}(x,y) = \mathbf{Bool}(x, G(y)) .\]

so it evaluates as \\(\texttt{true}\\) iff \\(x \le G(y)\\).

Two cases here:

* \\(y \ge 500\\). In this case \\(G(y) = \texttt{true}\\) so we have \\(x \le G(y) \\) for all \\(x \in \mathbf{Bool}\\). Thus, in this case \\(\check{G}(x,y) = \texttt{true}\\) for all \\(x\\).

* \\(y \lt 500 \\). In this case \\(G(y) = \texttt{false}\\) so we have \\(x \le G(y) \\) iff \\(x = \texttt{false}\\). Thus, in this case \\(\check{G}(x,y) = \texttt{true}\\) only for \\(x = \texttt{false}\\).

In short, \\(\check{G}(x,y) = \texttt{true}\\) iff either \\(y \ge 500\\) or \\(x = \texttt{false}\\).

Or in the English: the feasible situations are precisely those where you either have more than $500, or you don't buy a ticket!

So yes, I think we were both right. I think this feasibility relation is both a companion and a conjoint!

Somehow the conjoint seems more intuitive to me: from the amount of money you have, I can determine whether or not you can buy a ticket.

The companion approach is more mysterious to me: from whether or not you can buy a ticket, I cannot determine how much money you have!

Nonetheless you were able to dream up a monotone function from truth values to amounts of money, whose companion gives the desired feasibility relation. Interesting!