How does one obtain $P\implies Q$ truth table?

Question

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

$$\begin{array}{c|c|c} P & Q & P\implies Q \\ \hline \text{T} & \text{T} & \text{T} \\ \hline \text{F} & \text{T} & \text{T} \\ \hline \text{T} & \text{F} & \text{F} \\ \hline \text{F} & \text{F} & \text{T} \end{array}$$

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

The idea is the only way you could disprove "$p$ implies $q$" is to show that it's possible for have $p$ true without also having $q$ true. If "$p$ implies $q$" can't be shown false, we declare it to be true (innocent unless proven guilty). — quasi, Mar 29 '17 at 13:29
And couldnt-we-have-defined-the-material-conditional-differently. — Mauro ALLEGRANZA, Mar 29 '17 at 13:44
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 $T$. — Mauro ALLEGRANZA, Mar 29 '17 at 14:16
Thus, we are left with the "annoying" case: $F \to T$. Two possibilities for it: either $T$ or $F$. You can easily check that if we choose $F$ for $F \to T$ the result is (material) equivalence. Thus... — Mauro ALLEGRANZA, Mar 29 '17 at 14:19
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. — Andreas Blass, Mar 29 '17 at 14:29
All the above under the basic "assumption" of truth-functionality. If you are still unsatisfied with the way math log maps the natural language use of "if..., then..." on the conncetive $\to$, you can reject this assumption and try with other types of conditionals. A new world is open... but of course, once you have left the "safe harbour" of truth-functional math logic, you have to face the troubles of philosophy. — Mauro ALLEGRANZA, Mar 29 '17 at 14:38
See e.g. Jonathan Bennett, A Philosophical Guide to Conditionals, Oxford (2003) and David Sanford, If P Then Q: Conditionals and the Foundations of Reasoning, Routledge (1989). — Mauro ALLEGRANZA, Mar 29 '17 at 14:40
An intuition for the "annoying case": suppose p -> q, and r -> q. Suppose p. Then we have q. What about r? Can it be false? Consider the statements p := "it is raining", r := "the sprinkler is on", and q := "the ground is wet". It is easy to have a world where ~r, q, and r -> q are all true. — James Lea, Mar 29 '17 at 16:11

score 4 · Answer 1 · edited Sep 24 '17 at 19:07

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:

Modus Ponens is valid
$P \rightarrow P$ is a tautology
$\to$ 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.

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. — amWhy, Sep 24 '17 at 19:10

score 2 · Answer 2 · answered Aug 08 '23 at 08:16

First, you should know that it is better (and more standard) to draw the truth table as follows:

$$\begin{array}{c|c|c} P & Q & P\implies Q \\ \hline \text{T} & \text{T} & \text{T} \\ \hline \text{T} & \text{F} & \text{F} \\ \hline \text{F} & \text{T} & \text{T} \\ \hline \text{F} & \text{F} & \text{T} \end{array}$$

This is just a definition, but let's see what the motivation behind it is. We can interpret $P\implies Q$ as "if $P$ is true, then $Q$ must also be true". If you think a little, you will realize that "if $P$ is true, then $Q$ must also be true" is logically equivalent to "There is no case where $P$ is true and $Q$ is false". This is the only assertion of the statement and the statement makes no claim about what should happen when $P$ is false. So if there is a case where $P$ is true and $Q$ is false, then the statement becomes false, and if there is no case where $P$ is true and $Q$ is false, then the statement becomes true; Therefore, the only case that $P\implies Q$ is false, is when $P$ is true and $Q$ is false and this is exactly the motivation behind the above truth table.

How does one obtain $P\implies Q$ truth table?

2 Answers2

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