# Where the material conditional gets its truth conditions

Oh, the material conditional. Some love it, some hate it. But can we all agree that *explaining it to the uninitiated* is a perennial headache? If you’ve taught baby-logic, you know how this goes. There you are, giving a just *lucid* shpeel on deductive systems, until you get to this part:

A |
B |
A→B |

T | T | T |

T | F | F |

F | T | T |

F | F | T |

and the tires screech to a halt. *Why* are those bottom two values True? — they demand. The first two rows don’t bother them. But if **A** is *false*, why should it be that **A** → **B** is *true*, regardless of the truth of **B**?

You could just say it’s a convention, get over it. But why is this the convention adopted in classical logic? My colleague Jonathan Livengood and I discussed this, and came up with a better answer:

Suppose we agree on the first two rows of the above truth table. If implication (→) is both

non-trivialandasymmetric, then this itsonly possibletruth table.

Here’s why. Start by writing down all the possibilities for these bottom two rows. There are only four, and **A** → **B** has to be one of them.

A |
B |
(1) | (2) | (3) | (4) |

T | T | T | T | T | T |

T | F | F | F | F | F |

F | T | T | F | F | T |

F | F | F | T | F | T |

**Column*** 1* is trivial, because it has the same values as

**B**. If this were the correct column, then saying

**A → B**would mean the same thing as just saying

**B**. So, assuming → is not trivial, we can throw this column out.

* Column 2* has a symmetry property that implication doesn’t. Namely, it stays the same if we reverse the

**A**and

**B**cells. If this were material implication, then

**A → B**would be true if and only if

**B → A**is true. So, assuming → is asymmetric, we can throw this column out too. (This column is actually the usual truth table for

**A ↔ B**.)

**Column*** 3* has exactly the same problem: it stays the same when we reverse the cells containing

**A**and

**B**. So we can throw it out for the same reason. (This column is actually the usual truth table for

**A&B**. So, plausibly, we can also observe that “implies” should mean something different than “and.”)

And that’s it! If → is non-trivial and asymmetric, then * Column 4* is the only option left: the standard, not-just-conventional truth table for material implication.

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It’s also useful to explain it this way:

A -> B means there are no cases where A is true but B is false. That’s basically what we want implication to mean. And if A is false, then of course there cannot be a case where A is true and B is false!

You’re right, “basically what we want” is that A → B is equivalent to A&~B. But how do you convince a skeptic of that?

Maybe the trick isn’t to focus upon one statement alone. No statements of logic really exist by themselves; each is some sort of little corner of the world that we are trying to represent. As such, we mark unknown states as true, until we have some reason to believe them to be false.

Why should “true” be the default position? One might equally argue that “false” is the default position (making column 3 the correct truth table).

I think people that have problems accepting A->B being equivalent to not -A v B do that on the basis that they do not accept bivalent logic. See for example http://en.wikipedia.org/wiki/Constructive_mathematics

In that case, your analysis would not be complete as there would be more possible values in the table.

I’m with Robert: the problem grasping the truth table for A->B is that the everyday-language meaning of “If A then B” simply doesn’t assign a truth-value in cases where A is false. I don’t think there’s any getting around the fact that the formal-logic meaning of A->B is very different from any natural-language sense.

Never having taught logic, I don’t know the best way to make that happen in the classroom. Maybe the argument presented here would help, but to me it has the feel of an argument that would only convince people who are already used to thinking in this way (and hence who don’t need it).

When I was learning this stuff, I found the A->B truth table very troublesome, and I don’t think that this argument would have helped me. You’d have had a much better chance convincing me that A->B was equivalent to “not(A and not-B)”. (In fact, that’s probably what good old Mr. Jorgensen did back in the day.) But maybe I’m wrong and this argument will help folks in the classroom.

When I was taught basic logic, my teacher got the point across in a different fashion. It’s not rigorous by any means, but it was a nice example.

P = “it rains” and Q = “the party is cancelled.”

P->Q = “If it rains then the party is cancelled.”

It’s raining (P) and the party is cancelled (Q), were you lied to? No, since that’s exactly what was said. That’s the True, True is True case.

It’s raining (P) and the party isn’t cancelled (~Q). Were you lied to? Well, yeah. You expected the party to be cancelled since it was raining. That’s the False, False is False case.

No one had a problem with this, like you said. But he established a basic metric to gauge the truth or falsehood of the statement: whether or not you would call someone a liar for the statement.

Now, it’s not raining (~P) and the party isn’t cancelled (~Q). Were you lied to? No. If it was raining then the party would be cancelled. But since it’s not raining, the party doesn’t need to be cancelled for this to be a true statement. This is more intuitive, since it’s what we expect. When you look outside and see it’s not raining, you wouldn’t expect the party to be cancelled. This is the False, False is True condition.

Then comes the least intuitive condition. It’s not raining (~P) and the party is cancelled (Q). Were you lied to? The party was cancelled because the guest of honor was sick, or maybe because the venue burned down. The rain was a sufficient but not a necessary condition for the cancellation of the party. Again, if it did rain, the party would have been cancelled as well, so there isn’t a lie. The person could easily say “I only mentioned rain! I didn’t say anything else!” and they’d be telling the truth. The statement just didn’t list every possibility. Sure, seeing no rain might give us the expectation that the party will happen, but we also know there could be other extenuating circumstances. That’s the False, True is True condition.

He was clear to stress the fact that the statement never changed with the two conditions. The fact that it’s not raining (~P) does not itself change the statement “if it is raining then the party will be cancelled” (P->Q) to make it “if it is not raining then the party will be cancelled” (~P->Q). It’s not raining (~P) is only used to apply the original statement “if it’s raining then the party will be cancelled” (P->Q) in order to judge the validity of the statement (as we did above). It does not create a new statement to analyze.

This becomes clearer when modus tollens and modus ponens are applied.

P->Q

P

Therefore Q. (Modus Ponens)

We’re given P->Q is true. We know that the person saying “If it rains, then the party is cancelled” is telling the truth. Therefore, if it’s raining the ONLY possibility for P->Q to remain true while P is true is for Q to also be true. Otherwise, the statement lied to us but it can’t lie to us.

P->Q

~Q

Therefore ~P (modus tollens).

Again, we’re given P->Q to be true. We know the statement is telling us the truth. But, the party is not cancelled. So, we know the only way for the statement to still tell us the truth while the party is not cancelled is if it is also not raining. Again, all we care about is the original statement as a whole.

Bryan,

Assuming that A->B is asymmetrical seems problematic: isn’t this something that itself demands proof? Perhaps it can be more easily or more intuitively explained to students then the truth table for A->B (“If you have an apple, then you have a fruit, but if you have a fruit, you don’t necessarily have an apple”)

Here’s a somewhat intuitive way of explaining row three. In a truth table, we’re interested in showing whether certain statements are true or false given certain premises. In row three, we are given the premises ~A and B, and we ask whether “if A then B” is true. The statement “if A then B” can be interpreted as “given the additional premise A, can we show that B is true.” Thus our premises are now ~A, B, and A. Since we are given B as a premise, we already know it’s true. So the answer to the question “given the additional premise A, can we show that B is true” is “yes, of course we can show B is true, because it is given as true.”

An “intuitive” explanation of row four is more difficult, but I find it helpful to think about the proof of row four, which employs the “principle of explosion”:

~A

~B

—

A (ex hypothesi)

~A

┴

B (ex falso quodlibet)

A->B

So row four relies on the notion that “anything follows from a contradiction”–but why should that be? Well, if we are given that both A and ~A, then it seems that any statement can be both asserted and denied. Thus what’s to stop us from both denying B (given) and asserting it? In other words, if we interpret row four as asking “given ~A and ~B, and given the additional premise A, can we show that B”, then the answer is something like “well, if I can assert both A and ~A, then I can also assert both B and ~B; so B.” In other words, if we assume A and ~A, then all bets are off and we can assert or (and!) deny anything.

Reflecting on this, a student may look back at row three and ask “in row three, can’t we use the principle of explosion in a similar manner to prove A->~B?

~A

B

—

A (ex hypothesi)

~A

┴

~B (ex falso quodlibet)

A->~B

Thus from the premises ~A and B we can prove both that A->B and A->~B? The answer is of course “yes,” and for the same reason: if we can both assert and deny a statement, then all bets are off.

To Ed:

thank you thank you thank you!!!

What I teach is that (X&Y) -> X must be a tautology, which means (if you write its t.table out) that A -> B must be T for all assignments of truth values but A=T, B=F (i.e., X&Y=T, X=F is the only assignment that doesn’t appear in the t.table). Then the further assumption that A-> B is not itself a tautology, yields the last line of the table.

That seems better to me, since it delivers all four rows of the t.table, from (what seem to me) weaker assumptions. (The argument comes from Dorothy Edgington.)

As to the question of whether material implication is a good translation of natural language, that is the question of whether it has a truth function meaning at all. (Clearly not in counterfactual cases, but ever? There is a philosophical literature on this question: the subject of Edgington’s article.)

Very creative ideas, thanks everyone!

Robert and Ted:I completely agree: the very idea of truth-functional semantics is a very basic difficulty — what a strange way to capture meaning! I sometimes motivate this idea by explaining logic gates to the students: this binary reduction of meaning is essential for your (standard) desktop computer to work.Ed:I like how your logic teacher explained it. It’s a very intuitive example. I only worry a bit (as I did with Noah) that here, True is being taken as the “default position.” You ask the students: “were you lied to?” — and they say no. But if I would have asked them: “was the truth told?” — I bet they would also say no. As some are suggesting, the problem here may really be that truth-functional semantics constrains us in non-intuitive ways. You can’t tell a binary computer that the sentence A→B is meaningless in the case that A is false. You have to assign 0 or 1 toeverything.Keith:Your apples/fruit example is excellent. In my experience, students find the asymmetry of → to be acceptable, for just this kind of reason. But I’m worried your argument from a contradiction would equally (and incorrectly) indicate that if A is true and B is false, then A→B is true.Anon:I really like the Edgington argument, thanks for sharing! Very simple and clever. The only danger is that trusting assumptions like “X&Y → X is a tautology” might lead students astray. A student might equally try to assume that “X → ~X” is a contradiction, letting A=X, B=~X, and applying the same trick. But this gives the wrong truth table (it’s either column 2 or 3). After all, X→~Xisn’ta contradiction — though students’ intuitions may suggest otherwise…This is a nice question. One reason for the conventional choice is that it lets our logical reconstructions preserve our ordinary reasoning patterns. For example, we want ‘affirming the consequent’ to be fallacious in our proofs, so line three had better be true. Of course, students aren’t always so clear that affirming the consequent is a mistake.

But, Bryan, I’m afraid I don’t see your objection to Keith. Keith’s standard is that the conditional is true iff the consequent can be derived from the antecedent once you’ve been told what the truth values of the two sub-sentences are. In your example, you’re given, A, ~B, and A. This isn’t going to give you B, so there’s no danger of A→B being true.

I propose to think about the “implied conditional” as follows: You can’t logically label a statement false if it can be true. Thus: saying A “implies” B, or, “if A then B,” is not the same thing as saying “A causes B.” Thus, if B is true, the statement can be true. And since A does not cause B, if B is false, the statement still has the potential of being true.

The only problem I have with what you say is that you don’t explain why the material conditional SHOULDN’T BE given a trivial or symmetric interpretation. People have difficulty accepting the material conditional interpretation of conditional statements because there are circumstances where:

I take it to be true that A,

but I take it to be false that if ~A, then B,

but “if ~A, then B” is true, if A is true, according to the material conditional interpretation.

For example, suppose I’m a Democrat, and:

I take it to be true that the Republicans will win the election

I take it to be false that if the Republicans don’t win the election, then the Democrats will increase taxes.

But “if the Republican don’t win the election, then the Democrats will increase taxes” is true, if it is true that the Republicans will win the election, according to the material conditional interpretation.

So, if we accept the material conditional interpretation, it looks like I’m holding a false belief.

http://plato.stanford.edu/entries/conditionals/

I think the answer lies in applying a possible worlds interpretation to all our conditional beliefs (indicative and subjunctive).

What about relevance logic? (I assume that the concept of truth table is misnomer in that case).

In simple modal logic we have a function (ie truth table) for each world

TruthValue(PropositionalVariable, World). The nature of the truth itself is preserved in each world, we just have a lot more worlds to consider. Does not relevance logic also preserve simple truth tables = functions?

Hi, i’m doing a coursera course on mathematical thinking. This thing, the material conditional comes up and i don’t want to memorize it, i want to understand it. Can someone explain how you come to the possibilities for the bottom two rows? And the upper two rows two too.

Hi Nick, I’ve just fixed this page, as the tables seem to have been messed up in the Great Escape from Blogger in 2012. Is it any clearer now?

The idea here is to just assume the first two columns accurately represent what “A implies B” means when A is true and B is true, and when A is true and B is false. We then just write down all the possible combinations for the last two rows. That’s where these rows come from — they are the only four possibilities!

I think you can also derive the vacuous truth conditions from the truth table for inclusive disjunction and conjunction.

Start with the definition of inclusive or on propositions A and B. Then ask what happens to one of the disjuncts if the other is false. If we equate the truth values for A v B that evaluate to true with this notion of implication, we get something like

A v B (~B => A) & (~A => B)

A bit of substitution shows that the implication must be true whenever the antecedent is false.

Shouldn’t the last combination

F F F

Not be

F F T

Or are you trying to make a point and I don’t understand the English introduction?

Yep, typo — thanks!