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Velleman's logic in sentence 3 under figure 4 is confusing me. He is using lines two and four of the truth table to infer what Q of line 1 should be. But lines two and four assume P --> Q are true while line 1 assumes P --> Q to be false. So the latter doesn't follow the former in the way Velleman is reasoning here.

Furthermore it is already given in the first two columns of P and Q that they are both already F, so how he reasons that there is even a possible "inference" of Q being T is very strange.

Help anyone.


  • 1
    $\begingroup$ Can't help you with Velleman, but you may find my blog posting at dcproof.com/IfPigsCanFly.html to be useful. There, in the first half, I formally derive each line of the truth table for implication as a theorem of propositional logic, thus providing your requested justification, e.g. $A \land B \implies (A\implies B)$ and $A \land \neg B \implies \neg (A\implies B)$. $\endgroup$ – Dan Christensen Apr 25 at 12:32

Attmittedly it took me some time too to figure out what the author meant; here's what I think is intended:

An inference is valid iff in all the rows where all of the premises are true, the conclusion is true as well, i.o.w., there is no row where all premises are true but the conclusion is false. To check whether the entailment holds, we only need to consider the rows which have true in all premise columns; rows where at least one of the premises is false may simply be ignored.

In figure 4, we need the additional premise P to ensure that the entailment holds: If it weren't for P, then with row 1 we would have a row where all the premises (now only consisting of $P \to Q$) are true, but in this row the conclusion $Q$ is false. So from $P \to Q$ alone without $P$ as a second premise we can't deduce $Q$, because the validity of the argument would fail in the first row.

This would be different if we changed the first row for $P \to Q$ to false: Then, this row would be unproblematic for the truth of the conclusion, since we only need to consider rows where all the premises are true, and now that we changed the only premise $P \to Q$ to false in this row, that row passes the test anyway. So we no longer need the falsity of $P$ to ensure that we don't have a case where the premises are true but the conclusion is false, because we already have the falisity of $P \to Q$.

So in the hypothetical scenario layed out where $P \to Q$ were false in the first row, we wouldn't need the second premise $P$ to conclude $Q$, because then the statement "There are no rows such that all premises are true and the conclusion is false" is already guaranteed by $P \to Q$ alone.

But indeed it is a bit strange to assume that the column for $P \to Q$ is alternated in some way, because the truth table for $P \to Q$ as given in 4 is just how implication is defined, so the confusion you uttereted in the second part of your question is understandable.

  • $\begingroup$ Really thorough explanation here! Thanks! $\endgroup$ – user152810 Apr 24 at 22:58

Velleman's logic is correct. Here is what he is saying. Suppose we change the first line of the truth-table for $P \to Q$, so it becomes:

\begin{array}{cc|c} P&Q&P \to Q\\ \hline F&F&F\\ F&T&T\\ T&F&F\\ T&T&T\\ \end{array}

OK, so now look at the following argument:

$$P \to Q$$

$$\therefore Q$$

Is this valid? With the truth-table as defined above, it would be, since in every case where the premise $P \to Q$ is true (lines 2 and 4 of the truth-table), $Q$ is also true.

But, Velleman says, this would be very unintuitive (he gives the example with the eggs). Therefore, we better not use the truth-table above.

  • $\begingroup$ Very concise explanation! Thanks! $\endgroup$ – user152810 Apr 24 at 22:59

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