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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?

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    $\begingroup$ The way I read the exercise is: for each of the proposed truth-functions (this is the meaning of extensions) see if they satisfy "mp" (with the truth-function in place of $\to$). Thus, for the fourth case: if $p$ is true and $\lambda p q . (\lnot p \lor q)$ is true, also $q$ must be. But with $p$ true and $\lambda p q . (\lnot p \land \lnot q)$ this cannot do, because with $p$ true we have $(\lnot p \land \lnot q)$ false. $\endgroup$ – Mauro ALLEGRANZA Sep 24 '17 at 17:26
  • $\begingroup$ @MauroALLEGRANZA well in the third case we also have $\neg p$ false with $p$ true, but this is apparently acceptable. to be precise, the way I read the exercise is: for each proposed truth function $f$, whenever $(f(p,q) = \top) \land (p=\top)$ then we must have also $q=\top$. I take my example $\lambda pq.(\neg p \land \neg q)$ to satisfy this because the condition $(f(p,q) = \top) \land (p=\top)$ is never met. $\endgroup$ – Matt E. Sep 24 '17 at 17:54
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    $\begingroup$ New try; let use Forster's same words: "f $p$ implies $q$, what does this tell us about what $p$ and $q$ evaluate to ? Well, at the very last, it tells us that $p$ can't evaluate to true when $q$ evaluates to false". $\endgroup$ – Mauro ALLEGRANZA Sep 25 '17 at 7:55
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    $\begingroup$ Apparently, this can be read: consider the truth table for the proposed truth functions and check that they have false in the row: $p$ true and $q$ false. If so, you are right: also $(\lnot p \land \lnot q)$ satisfies it. $\endgroup$ – Mauro ALLEGRANZA Sep 25 '17 at 8:28
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Thanks @MauroALLEGRANZA for your comments. I should have gone in search of errata first: there are indeed 8 possibilities.

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