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I understand the usual motivation behind the truth table for the logical connective $\to$.

However, I would like to know if there is a more fundamental reason for that truth table. Something that would have to do with arguments and validity.

A.G.Hamilton writes in Logic for Mathematicians that "the significance of the conditional statement $A\to B$ is that its truth enables the truth of $B$ to be inferred from the truth of $A$, and nothing in particular to be inferred from the falsity of $A$".

Question: What does Hamilton mean precisely? Something like $\to$ is the only binary truth function such that

  • The argument form $(p\to q),p;\therefore q$ is valid

  • The argument form $(p\to q),\sim p;\mathcal{A}$ is invalid unless $\mathcal{A}$ is a statement form logically equivalent to $\sim p$ ?

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    The key point is the link with *modus ponens*, i.e. the fundamental rule of inference: we want to use "safely" MP to make inference and we make infernec from premises known (or assumed) to be *true*. In such case we have that $A$ and $A \to B$ are *true* and **we want** that this is enough to conclude with the truth of $B$. If $A$ is not *true*, then... it does not really matter, because in such case we are not "interested" to infer anything from it **and** $A \to B$. Thisis the "reason why" we use the stipulation: $A \to B$ is true when $A$ is *false*. – 2017-02-02

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First of all, the latter requirement in your Question is too strong: clearly you want to be able to infer $P \to Q$ from $P \to Q$ and $\neg P$. And saying that $\phi$ cannot be inferred unless it is equivalent to either $P \to Q$ or $\neg P$ doesn't work either: we still want to be able to infer $\neg P \lor R$.

Still, while your requirement as stated is too strong, we are indeed left wondering: how exactly are we to understand these 'things in particular' that we should not be able to infer from $P \to Q$ and $\neg P$?

Well, I think it is pretty clear that we should not be able to infer either $Q$ or $\neg Q$. That is, the following argument should be invalid:

$P \to Q$

$\neg P$

$\therefore Q$

And to do that, we need a counterexample with $P = F$ and $Q = F$, and thus we should set $F \to F = T$

Likewise the following argument should be invalid:

$P \to Q$

$\neg P$

$\therefore \neg Q$

and that means that we have to set $F \to T = T$.

OK, but what about $T \to T$? Would setting it to $F$ mean that suddenly the truth of $P \to Q$ does enable us to infer 'things in particular' from the falsity of P? Well, clearly not!

Consider: Suppose we set the three truth-values as already established. Then whatever statement $\phi$ we can infer from $P \to Q$ and $\neg P$ with $T \to T = T$, we can of course also infer from $P \to Q$ and $\neg P$ with $T \to T = F$, because any counterexample to any inference will require $P$ to be false, and for that case, the two possible definitions work exactly the same, meaning they have the exact same set of counterexamples, and thus have the exact same set of sentences that are logical consequences.

Thus: it is not the case that not being able to infer 'things in particular' (however they might be defined) from $P\ to Q$ and $\neg $ will force the truth-table for the material conditional to be what they are!

OK, so is there an independent argument for setting the truth-values as we do? Here is one:

$T \rightarrow F = F$. OK, this makes immediate sense intuitively, but if you want a more technical argument, you can indeed just point to Modus Ponens:

$$P \rightarrow Q$$

$$P$$

$$\therefore Q$$

Now suppose $P = T$ and $Q = F$. If $T \rightarrow F$ were $T$, 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-cummutative

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 $Qt$, then this contraposition equivalence should convince you that the other should be $T$ as well.

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    Nice answer (+1)! You showed that $\to$ is the only binary truth function that: 1. satisfies the modus ponens 2. $P\to P$ is a tautology 3. is non-commutative. Before accepting your answer I have to ask if the truth values for $T\to T$, $F\to T$ and $F\to F$ could also be forced (with proof) by a requirement as in my question, i.e. by imposing that we can't infer anything new from $p\to q$ and $\sim p$ or that the argument form $(p\to q),\sim p;\therefore\mathcal{A}$ be invalid whenever $\mathcal{A}$ is a statement form which is not a tautology or logically equivalent to $p\to q$ or $\sim p$ ? – 2017-02-03
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    @NeedForHelp Hmm, I'll need some time to think about that ... – 2017-02-03
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    @NeedForHelp OK, the $F\to F$ and $F \to T$ cases follow by considering $p \to q, \neg p \therefore q$ and $p \to q, \neg p \therefore \neg q$ respectively. Still working on the $T \to T$ case ... – 2017-02-03
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    @NeedForHelp OK, so that requirement does *not* force $T \to T$ to be $T$, because if it were false (and with the other three values set as they are), then you still couldn't validly infer anything from $p \to q, \neg p \therefore \phi$ that you couldn't from that if it were true, for if it is invalid in the latter case, then it is possible for $p \to q$ to be true, $\neg p$ to be true, and $\phi$ to be false, meaning that any such counterexample needs $p$ to be false, and for a false $p$, the two different definitions would work the same, meaning that the same counterexample still works. – 2017-02-03
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    @NeedForHelp .... So .... Either Hamilton did not mean what you thought he meant (but he surely wasn't clear either what he meant!!)... Or Hamilton had it wrong! – 2017-02-03
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    Hmm. Nice argument! You're very bright. The litteral quotation is: "The difficulty arises with the truth value $T$ assigned to $A\to B$ in the case where $A$ is false. Consideration of examples of conditional statements in which the antecedent is false might perhaps lead one to the conclusion that such statements do not have a truth value at all. One might also gain the impression that such statements are not useful or meaningful. For example, the statement "If grass is red then the moon is made of green cheese" could fairly be said to be meaningless... – 2017-02-03
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    ... However, we shall be interested in deduction and methods of proof, principally in mathematics. In this context the significance of a condition statement $A\to B$ is that its truth enables the truth of $B$ to be inferred from the truth of $A$, and nothing in particular to be inferred from the falsity of $A$." Then he goes on to give the example $\forall n:n>2\to n^2>4$ to justify the truth table. – 2017-02-03
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    Then I guess he just meant that we can't infer $q$ or $\sim q$ from $p\to q$ and $\sim p$. In any case, I'm a bit surprised that modus ponens + invalidity of $(p\to q),\sim p\therefore\mathcal{A}$ doesn't characterize $\to$ ! Naively I thought it was true (but hey, I'm no logician!). Anyway, thank you for the lesson and for your dedication! – 2017-02-03
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    @NeedForHelp Well, I am really glad you asked this question!! And you were right to want some clarity, and for questioning whether his demand would force the truth-values to be set as they are... and on my analysis, his characterization really *doesn;t* force all values, *no matter* how 'things in particular' is being interpreted! So good catch, I say! – 2017-02-03
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The latter condition is too strong - for example, $p \implies q$ can be validly inferred from $p \implies q, \neg p$, as can any tautology. The idea is simply that the argument $(p \implies q),\neg p; \neg q$ is not valid. That is, from $A \implies B$ and $\neg A$, we may be able to deduce many things, but we will not be able to deduce whether $B$ holds.

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    So $\to$ is the only binary truth function such that $(p\to q),p;\therefore q$ is valid and both $(p\to q),\sim p;\therefore q$ and $(p\to q),\sim p;\therefore\sim q$ are invalid? – 2017-02-02
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    Yes, @NeedForHelp, the only things that can be inferred from $p{\to}q$ and $\neg p$ are $p{\to}q$ or $\neg p$; nothing new. – 2017-02-02
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From the statements of $\{A{\to}B, A\}$ we can infer something new; that $B$ is true.

However, from the statements of $\{A{\to}B, \neg A\}$ we cannot infer anything new.

That is all.

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    More precisely, if $S$ is a set of statement forms and $\operatorname{Inf}S$ denotes the set of statements forms that can be inferred from the ones in $S$ as conclusions of valid argument forms, do you mean $\operatorname{Inf}\{p\to q,\neg p\}=\operatorname{Inf}\{p\to q\}\cup\operatorname{Inf}\{\neg p\}=\operatorname{Inf}\{\neg p\}$ (which is quite easy to prove)? Unfortunately 1. $p\to q,p\therefore q$ and 2. $\operatorname{Inf}\{p\to q,\neg p\}=\operatorname{Inf}\{p\to q\}\cup\operatorname{Inf}\{\neg p\}$ don't characterize $\to$ (even if we require it to not to be a contradiction) I think. – 2017-02-03