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!