On p. 21 of Forster's Logic, Induction and Sets the reader is asked to check that the following is an exhaustive list of extensions satisfying modus ponens:
- $\lambda pq.q$
- $\lambda pq.(p\leftrightarrow q)$
- $\lambda pq.\neg p$
- $\lambda pq.(\neg p \vee q)$
- $\lambda pq. \bot$
Now clearly we cannot have $(\lambda pq.f)(\top,\bot) = \top$, since whenever a candidate $f$ evaluates to $\top$ and $ p $ is $\top$, then q must also be $\top$. As far as I can see this is just what it means to satisfy modus ponens, and certainly this holds for the above list. However, shouldn't we then expect $2^3$ possible extensions? In particular, why does say, $\lambda pq.(\neg p \land \neg q)$ not satisfy?