Not too long ago I asked a question related to the material conditional that ended up proving just how limited my understanding of the material conditional actually was.

In the meantime, I found a fantastic webpage that hosts the best explanation of the material conditional I have encountered thus far. The page in question: http://philosophy.hku.hk/think/sl/ifthen.php

Now, I have some trouble understanding certain parts. I have to make it clear though; I'd appreciate it if we could leave out any references to the technical nature of the conditional when explaining the conditional. So no "that property of the conditional is true because if you check with the conditional, you'll see that..." Specifically, the author explains the material conditional through a method that tries to preserve some of the logical properties of "if..., then...". As such, try to stick to a similar method without involving things that I'd simply have to assume to be (logically) true (e.g. the one where the teacher says "If you get an A, you can go home early.", but you don't get an A and don't get to go home early, so the teacher didn't lie - even though the teacher might just as well have been lying, but we didn't find out because it didn't happen so we have to assume he was right for the sake of the explanation, thereby making the explanation quite worthless) My first issue arises in this part of the explanation where the author begins motivating two of four truth values of the implication:

First of all, one thing that we accept is that "if φ, then φ" is always true for any statement φ. So to preserve this fact, we need to ensure that the truth-value of "(φ → φ)" is always T whether φ is T or F. Since we are assuming that "→" is truth-functional, this implies that "(φ → ψ)" has the truth-value T whenever φ and ψ have the same truth-value, whatever that is.

(1) Why is "if φ, then φ" always true for any statement φ, regardless of the truth of φ? Is this because φ is simply assumed to be true when one says "if φ...", irrespective of the actual truth of φ? For instance, is "If 1=2, then 1=2" true because, in the case that 1=2, it is necessary that 1=2? Or, 1=2 only if 1=2? Am I overthinking this, or is there some logical structure behind all of this?

(2) According to the author, from (1) it follows that the assumption that "→" is truth-functional implies that "(φ → ψ)" has the truth-value T whenever φ and ψ have the same truth-value, be it F or T. I can sort of intuitively feel (1) is true, but I don't see how he's able to extend this to two different statements φ and ψ. Maybe the reason I don't understand this is because I don't fully grasp (1)? Either way it seems odd that you'd be able to say something specific like that about the nature of the material conditional by simply observing that it's a truth-functional"? I mean, how does (1) lead him to conclude that "Since we are assuming that "→" is truth-functional, this implies that "(φ → ψ)" has the truth-value T whenever φ and ψ have the same truth-value, whatever that is."?

Down the line, he gets to this part:

To fill in the third row of the truth-table, a different kind of argument is needed. This time we consider the properties that we do not want "→" to possess. In particular, consider this sequent:

(P→Q) ⊧ (Q→P)

Surely we do not want this argument to be valid.

(3) I think I have an idea as to why we don't want this argument to be valid; something to the effect of not wanting "If x is an apple, x is a fruit" to be logically equivalent to "If x is a fruit, x is an apple", correct? Of course, some additional background here would be appreciated.



closed as off-topic by Rob Arthan, graydad, Ali Caglayan, pre-kidney, David K Jun 11 '15 at 2:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Rob Arthan, graydad, Ali Caglayan, pre-kidney, David K
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Your question is about philosophy and not about mathematics; if you have problems with $\phi \Rightarrow \phi$, then you need philosophical rather than mathematical assistance. $\endgroup$ – Rob Arthan Jun 10 '15 at 22:27

I will say it again even though you state you don't want it: You're entirely too hung up on the English words "if" and "then". You will not find any explanation that convinces you that the phrase "if this, then that" necessarily has to mean what "$\text{this}\to\text{that}$" means -- for the simple reason that $\text{this}\to\text{that}$ is not the only thing "if this, then that" can mean in plain English.

What is really going on is that mathematical logic needs a concept that behaves like $p\to q$, and once we have that concept we need a way to pronounce it -- and for the pronunciation we simply commandeer the words "if" and "then". This is not necessitated by the plain English meaning of the words; we do it purely with the justification that we can. It may not be right and fair to usurp the meaning of the words in that way, but it's how the world is.

As far as your concrete question goes, "$\to$ is truth-functional" means that whenever $\varphi$ has the same truth value as $\varphi'$, and $\psi$ has the same truth value as $\psi'$, it must be the case that $\varphi\to\varphi$ has the same truth value as $\varphi'\to\psi'$. Or in other words, we can determine the truth of $\varphi\to\psi$ solely by knowing the truth (or falsity) of $\varphi$ and $\psi$ and nothing else about how they relate to each other.

Or, in yet other words, "$\to$ is truth-functional" means neither more or less than it has a truth table, and the truth table tells everything about how it works.

So if we have decided that we want $1=1\to 1=1$ to be true, and also want $1\ne 1\to 1\ne 1$ to be true, that fixes two lines of the truth table.

And since we don't want $p\to q$ and $q\to p$ to mean the same thing, the two remaining lines in the truth table must have different values. We certainly want $T\to F$ to be false, which means that $F\to T$ has to be true.

As for why $p\to p$ is always true: Rather than saying "if $p$ then $q$" as an "explanation" of $p\to q$ it may be more useful to think of $p\to q$ as a promise that "for each proof of $p$ you give me, I will give you a proof of $q$". This promise is easy to keep if $p$ and $q$ are both the same claim, so it ought to be possible to prove $p\to p$ with no need to inquire deeper into what $p$ is -- but if so it has better be true in all circumstances that $p\to p$.

  • $\begingroup$ Of course, the reason we don't want $p \to q$ and $q \to p$ to mean the same thing is based to some extent on the concept of "if/then", or at least based on the "concept that behaves like $p \to q$". +1 $\endgroup$ – Carl Mummert Jun 10 '15 at 22:00
  • $\begingroup$ @Henning It's all clear now, thanks. One thing though: the author of the webpage I sent you says: "The longer answer is that if we use "→" to translate "if ... then ...", we want to make sure that certain logical properties are preserved.". This makes it seem like relied on the linguistic concept of "if..., then..." to define it in mathematical terms. You, on the other hand, make it seem like this is not the case by saying "mathematical logic needs a concept that behaves like p→q, and once we have that concept we need a way to pronounce it" Which is the correct way of thinking about this? $\endgroup$ – Ius Klesar Jun 11 '15 at 0:26
  • $\begingroup$ @Luke: Well, I of course think my way is the more honest one :-) but it's not the one most authors take. Since it's you who needs to end up at peace with how mathematics uses the words and symbols it's ultimately up to you which point of view you find most convincing. It's not even important that you agree with anyone in particular about the metaphysics of it all, as long as you know what the technical properties of the notation are. $\endgroup$ – Henning Makholm Jun 11 '15 at 1:16

For (1), surely we want $\varphi\rightarrow \varphi$ - that is, "if $\varphi$ then $\varphi$" - to be true: "if $\varphi$ . . ." means "assuming $\varphi$ to be true . . .". I think you're over-thinking this one.

For (2), remember that truth-functional means "depending only on the truth values of the propositions involved." So suppose we know $\varphi$ and $\psi$ have the same truth values. So, since $\rightarrow$ is truth-functional, $$\varphi\rightarrow \psi$$ must have the same truth value as $$\varphi\rightarrow\varphi$$ (since $\varphi$ and $\psi$ have the same truth values, and so are interchangeable as long as $\rightarrow$ is in fact truth-functional).

For (3), yes, that's exactly it. I'm not sure what "extra background" you're expecting; I think, as with (1), you're over-thinking things here.

Let me take a stab, seperately, at motivating the material conditional. Consider the sentence "Every even integer $>2$ can be expressed as the sum of two primes" (Goldbach's conjecture) - that is, $$\text{"For all $x$, ($x$ an integer $>2$ $\implies$ $x=p+q$ for some primes $p, q$)."}$$ Note that this has the form "For all $x$, $\varphi(x)\rightarrow \psi(x)$." What would a counterexample to this look like?

  • $-3$ can't be written as the sum of two primes, but that's not a counterexample; $-3$ isn't an integer $>2$. (That is, $\varphi(-3)$ doesn't hold.)

  • $5$ is a non-even integer which can be written as the sum of two primes, but surely that's not a counterexample: Goldbach just says that every even number is the sum of two primes, not that the only numbers which are the sums of two primes are even. (In conjunction with the previous bullet, this means that if $\varphi(x)$ doesn't hold, we really don't care about $\psi(x)$.)

  • Really, the only way Goldbach would be wrong is if there were an even number which was not the sum of two primes - that is, if there were some $a$ such that $\varphi(a)$ but not $\psi(a)$.

And this is exactly the truth table of the material conditional.

  • $\begingroup$ Thanks for the response, (1) and (3) were to be expected, really. (2) is a tad bit harder for me to wrap my head around. The logical property of "φ → φ" always being true - no matter the truth value of φ - in some way preserves the causal properties of the linguistic "if..., then..." If it is the case that φ, then it is the case that φ. It's straight-forward and clear. However, when dealing with "if φ, then ψ", you lose that sense of one-to-one correspondence that "φ → φ" has; now it's two unrelated statements φ and ψ that could both be true or false without making any sense. $\endgroup$ – Ius Klesar Jun 10 '15 at 21:42
  • $\begingroup$ Well, this is where the truth-functionality assumption comes into play: either we have to accept $\varphi\rightarrow \psi$ whenever we know that $\varphi$ and $\psi$ have the same truth value, or "$\rightarrow$" is not a truth functional. Non-truth-functional connectives are much more complicated, but they can be worked with - for instance, in modal logic the expression "$\Box(\varphi\rightarrow\psi)$", read as "$\varphi$ necessarily implies $\psi$," is not truth functional, and can be used to for instance talk about causality. (cont'd) $\endgroup$ – Noah Schweber Jun 10 '15 at 21:46
  • $\begingroup$ But in modal logic, the semantics is correspondingly more complicated. Semantics for classical propositional logic just consists of an assignment of truth values to each propositional variable; for modal logic, we have multiple such assignments, each attached to a world in a Kripke frame (see e.g. en.wikipedia.org/wiki/Kripke_semantics). $\endgroup$ – Noah Schweber Jun 10 '15 at 21:48

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