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?

Answer 1

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.

Answer 2

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.