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I am trying to get an intuition (ohh, the irony) about the logical truth tables. In particular, I am looking at the basic conditional P->Q with the following truth table:

P Q | P -> Q T T T F T T T F F F F T

How does one obtain this truth table? If it is an axiom, what is the motivation behind this particular form?

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    $\begingroup$ It is a definition. See Conditional. $\endgroup$
    – Mauro ALLEGRANZA
    Mar 29, 2017 at 13:37
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    $\begingroup$ See many many posts; e.g. defining-material-conditional $\endgroup$
    – Mauro ALLEGRANZA
    Mar 29, 2017 at 13:43
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    $\begingroup$ And couldnt-we-have-defined-the-material-conditional-differently. $\endgroup$
    – Mauro ALLEGRANZA
    Mar 29, 2017 at 13:44
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    $\begingroup$ We have some "natural" expectations: $T \to T$ must be $T$ and $T \to F$ must be $F$. But also: $p \to q$ must be equivalent to $\lnot q \to \lnot p$, and this, with $T \to T$, needs that $F \to F$ must be $T$. $\endgroup$
    – Mauro ALLEGRANZA
    Mar 29, 2017 at 14:16
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    $\begingroup$ The idea in @MauroALLEGRANZA's last comment can be used more widely. Once you've decided (as in his previous comment) the correct values for $T\to T$ and $T\to F$, there are only 4 possible truth tables (2 options for each of $F\to T$ and $F\to F$. One of those options makes $p\to q$ equivalent to $p\iff q$ (as Mauro said); another makes it equivalent to $p\land q$, and a third makes it equivalent to $q$. None of those makes much sense, so only one option, the standard truth table for $p\to q$, remains. $\endgroup$
    – Andreas Blass
    Mar 29, 2017 at 14:29

1 Answer 1

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There is a long-standing debate whether or not the conditional is truth-functional in the first place (that is: is the truth-value of $P \to Q$ a function of the truth-values of $P$ and $Q$?).

But if we treat it as such (that is: if we had to pick one of the truth-tables), then here is an argument for setting the truth-values as we do.

Consider Modus Ponens:

$$P \rightarrow Q$$

$$P$$

$$\therefore Q$$

Now suppose $P = T$ and $Q = F$. If $T \rightarrow F$ were set to $T$, then this argument would be invalid! Clearly that's not what we want. So, we should set $T \rightarrow F = F$

Now let's consider:

$$P \rightarrow P$$

OK, clearly we want this to be a tautology, no matter what $P$ is saying, and no matter whether $P$ is true or false ( Indeed, even if $P$ is a contradiction, it should still hold that ' If P then P'!). OK, but this means that we can't set $T \rightarrow T$ to $F$, for then $P \rightarrow P$ would not be a tautology, so we set $T \rightarrow T = T$. Likewise, we can't set $F \rightarrow F$ to $F$, so we set $F \rightarrow F = T$.

Finally, we want $\rightarrow$ to be 'asymmetrical' or non-commutative: clearly 'if P then Q' is completely different from 'if Q then P'. But given the other truth-values already set as they are, if we set $F \rightarrow T$ to $F$, then it would become commutative! So, we set $F \rightarrow T =T$.

In short, setting the truth-values as we do is the only way to ensure:

  1. Modus Ponens is valid

  2. $P \rightarrow P$ is a tautology

  3. $\to$ is non-commutative

And, just to have some more arguments for setting the truth-values as we do, consider:

$$P \rightarrow Q$$

$$Q$$

$$\therefore P$$

This should clearly be an invalid argument, with the counterexample of $P = F$ and $Q = T$. But if we were to set $F \rightarrow T$ to $F$, this would not be a counterexample at all! So, we better set $F \rightarrow T = T$.

Finally, let's note that we want:

$$P \rightarrow Q \Leftrightarrow \neg Q \rightarrow \neg P$$

This means that $T \rightarrow T$ and $F \rightarrow F$ better have the same truth-value. So, once you are convinced that one of them should be $T$, then this contraposition equivalence should convince you that the other should be $T$ as well.

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  • $\begingroup$ I reversed my downvote, but I think the links above, and any search on MSE using logic, material conditional, logical implication has long before beaten this topic to death. It's not wrong, but it's not entirely correct, either. $\endgroup$
    – amWhy
    Sep 24, 2017 at 19:10

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