Logical Conditional Truth Table Rationale 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?
 A: 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.
