Question about the FT, FF cases in the conditional:

Question

The conditional operator, $\phi \implies \psi$, is True for the values $TT, FT, FF$ and false for $TF$. I can easily understand why it's true for $TT$ and false for $TF$, but why is it for $FT$ and $FF$? As far as I can see, we can't come to any conclusion in those cases:

If $\phi$ is false but $\psi$ is true, to me that doesn't imply anything about the truth or falsity of $\phi \implies \psi$. Couldn't it be that when $\phi$ is true then $\psi$ is false? Why does this situation imply that $\psi$ follows from $\phi$? e.g.

$\phi : n < 4$, $\psi : n > 2$. If $\phi$ is false then clearly $\psi$ is true, but why does that imply that the truth of $\psi$ follows from the truth of $\phi$? If $\phi$ is true then either $\psi$ is true or $\psi$ is false, depending on the situation.

Similarly, $\phi: n < 4$, $\psi : n < 2$. If $\phi$ is false then $\psi$ is false but if $\phi$ is true then $\psi$ is not necessarily true.

Where did I go wrong with my reasoning?

Answer 1

The point behind the truth table is that you want it to behave in the following way: It is not the case that $\phi$ is true but $\psi$ is false. So you set it up to reflect that.

The reason it is DEFINED for other truth values as you mention above is this: Suppose you want to verify the truth of the statement If $x$ is even, then $x^{2}$ is even. Now suppose that $x=3$. Then $x^{2}=9$. If you didn't define the truth values in the way above, you have that the statement is false (since you would have defined $FF$ is $F$). But this doesn't reflect what we want (intuitively the statement should only be wrong if we can find $x$ even but $x^{2}$ false). Similar examples can be thought of for the other case: Think about "If $x>0$, then $x^{2}>0$".

This does create some weird situations sometimes: If $x\neq{x}$, then dolphins are sharks is a true statement. However this doesn't affect things much when you do math: You use modus ponens to prove things (i.e. if $p\implies{q}$ and $p$, then $q$). Here while If $x\neq{x}$, then dolphins are sharks is a true statement, you don't have $x\neq{x}$ to conclude that dolphins are sharks, so it doesn't really hurt anything.

Answer 2

I think that this explanation of S.C.Kleene, Mathematical Logic (1967) [pag.10 - footnote 12] is the best "short" elucidation of it :

The ordinary usage certainly requires that "If $A$ , then $B$" to be true when $A$ and $B$ are both true, and to be false when $A$ is false but $B$ is false. So only our choice for $T$ in the third and fourth lines can be questioned. But if we changed $T$ to $F$ in both these lines, we would simply get a synonym for $\land$; in the third line only, for $\lnot$ . If we changed $T$ to $F$ in the fourth line only, we would loose the useful property of our implication that "If $A$ , then $B$" and "If $\lnot B$ , then $\lnot A$" are true under the same circumstances [...].

Answer 3

"Follows from" is often not the best interpretation of the conditional operator. ϕ⟹ψ can probably be better understood to mean that ψ is at least as true as ϕ, or no less true than ϕ. If we know that ϕ = F, then whether ψ = T or ψ = F doesn't matter. Either way, ψ is at least as true as ϕ, and not less. The content or meaning of ϕ and ψ don't matter; we are only comparing truth values.

How we establish that ϕ⟹ψ in the first place is another question entirely. There are at least three ways to do this. For one, we could claim ϕ⟹ψ simply from the knowledge that ϕ is false or ψ is true. But while that is perfectly true, it's also totally useless. We can't use that fact to deduce anything else meaningful. For a second, We could simply assume it is true and examine the consequences. That's a common intermediate step in more involved proofs. For a third, we could assume ϕ and through some chain of reasoning deduce ψ. Then, we can say that ψ follows from ϕ. But ϕ⟹ψ alone says nothing about which of these three ways we used, so "follows from" is not the best way to read it.