How to interpret material conditional and explain it to freshmen?

Question

After studying mathematics for some time, I am still confused.

The material conditional “$\rightarrow$” is a logical connective in classical logic. In mathematical texts one often encounters the symbol “$\Rightarrow$”, which is read as “implies” or “if … then ….” It is customary and reasonable to treat “$\Rightarrow$” as the material conditional, i. e. as equivalent to saying “the antecedent is false or the consequence is true (or both)”.

Now there is some controversy about whether the material conditional really captures conditional statements because it doesn't really say anything about a causal connection between the antecedent and the consequence. This is quite often illustrated by the means of statements in natural languages such as “the moon is made out of cheese $\Rightarrow$ all hamsters are green” – since the moon isn't made out of cheese, is this statement true? This remained problematic to me.

While I came to accept the material conditional as a good way of describing implications and conditionals, I'm having a hard time to explain this usage to freshmen whenever I get asked.

My questions are: How can we best justify the interpretation of “$\Rightarrow$” as a material conditional? Why is it so well-suited for mathematics? How can we interpret or read it to understand it better? Can my confusion about it be led back to some kind of misunderstandig or misinterpretation of something?

I have yet very poor background in mathematical logic (I sometimes browse wikipedia articles about it), but I'd have no problem with a technical answer to this question if it clarifies the situation.

Answer 1

The original question asked "Why is the material conditional so well-suited for mathematics?" Here's a central consideration which others have not touched on.

One thing mathematicians need to be very clear about is the use of statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any $\epsilon$ ... there is a $\delta$ ... And the quantifier-variable notation serves mathematicians brilliantly to regiment statements of multiple generality and make them utterly unambiguous and transparent. (It is when we come to arguments involving generality that borrowing notation from logic to use in our mathematical English becomes really helpful.)

Quantifiers matter to mathematicians, then: that should be entirely uncontentious. OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might write

$\forall n$(if $n$ is even and greater than 2, then $n$ is the sum of two primes)

Note then, we restrict the universal quantifier by using a conditional. So now think about the embedded conditional here.

What if $n$ is odd, so the antecedent of the conditional is false. If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we can't refute Goldbach's Conjecture by looking at odd numbers!! So in these cases, if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent come out true by default. Come out "vacuously" true, if you like. In other words, the embedded conditional will have to be treated as a material conditional which is true when the antecedent is false.

So: to put it a bit tendentiously and over-briefly, if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they have come to know and love, they will have to get used to using material conditionals too.

Answer 2

Perhaps this will help to capture the truth-functional character of material implication:

The truth-value of an inclusion (subset) relation between sets corresponds to the truth-value of an implication relation, where $\subseteq$ corresponds to the $\rightarrow$ relation.

E.g., suppose $A\subseteq B$. Then if it is true that $x\in A$, then it must be true that $x\in B$, since $B$ contains $A$. However, if $x\notin A$ (if it is false that $x \in A$), it does not mean that then $x\notin B$, since if $A\subseteq B$, then $B$ may very well contain elements that $A$ does not contain.

Similarly, suppose we have that $p\rightarrow q$. If $p$ is true, then it must be the case that $q$ is true. But if $p$ is false, that does not necessarily mean then that $q$ is necessarily false. (For all we know, perhaps $q$ is true regardless of whether or not $p$ is true.) So $q$ can be true, while $p$ is false.

I don't know if this analogy helps or not. But it was the above analogy (correspondence) that helped me to firmly grasp the logic of material implication.

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Here's a more down-to-earth example you may have already stumbled upon:

CLAIM: "If (it rains), then (I'll take an umbrella)":

I'd be lying (my assertion would be false) if (it rains = true), and I do not (take an umbrella).

But perhaps it's cloudy out, and I decide I'll take an umbrella , just in case it rains. In this case:

If it doesn't rain (it rains = false), but I took my umbrella (true), my claim above would not be a lie (it would not be false).

Answer 3

The issue seems to be with the behaviour of $p\to q$ when $p$ is false. If $p\to q$ were false when $p$ is false, then you could conclude $p$ from $p\to q$ without any extra premises, and therefore guarantee $q$ as well. That betrays the idea of this being a conditional.

If (you stole the cookie) then (you're a horrible person)

Just from this, which most people with a sweet tooth would accept, I could then conclude that you stole the cookie and are in fact a horrible person. I don't need to assert that you stole the cookie separately, I can just conclude it from the truth of this statement. This behaviour loses the important "if" part of the conditional. To capture

if $P$ (happened), then $Q$ (would happen)

we need to allow for vacuous truths. As a positive example:

If (you won the bet) then (I would have paid you)

Is still true whether you actually win the bet or not.

Answer 4

An intuitive way to understand the material conditional is as a promise. If you hold up your end of the deal (so the antecedent is true), then I must hold up mine (the conclusion is true). But if you break the promise, then I can do what ever I want without me breaking the promise. And if I am going to hold up my end of the bargain no matter what, it doesn't particularly matter if you hold up yours.

And as far as why it works so well in mathematics, it might be because it is truth functional and mimics some part of what implication means in a non-mathematical context. It doesn't have anything to do with causality, which is often how it is used out of math; it only relies on the truth values of the constituent statements.

Answer 5

The material conditional is often simply defined as follows:

$A \implies B ~~\equiv ~~ \neg (A \land \neg B)$

This "definition" turns out to be a theorem of classical propositional logic that can be derived from what might be called "first principles" using a form of natural deduction.

First, we need to prove that $(A\implies B) \implies \neg (A \land \neg B)$

1. $A\implies B~~~~$ (Assume)

2. $A ~\land ~\neg B~~~~$ (Assume, to obtain a contradiction)

3. $A~~~~$ (Elim $\land$, 2a)

4. $\neg B~~~~$ (Elim $\land$, 2b)

5. $B~~~~$ (Elim $\implies$, 1, 3)

6. $B \land \neg B ~~~~$ (Intro $\land$, 5, 4)

7. $\neg (A ~\land ~\neg B)~~~~$ (Intro $\neg$, 2, 6)

8. $(A\implies B) \implies \neg (A \land \neg B)~~~~$ (Intro $\implies$, 1, 7)

Now, we need to prove that $\neg (A \land \neg B)\implies(A\implies B)$

1. $\neg (A \land \neg B)~~~~$ (Assume)

2. $A~~~~ $ (Assume)

3. $\neg B~~~~ $ (Assume, to obtain a contradiction)

4. $A \land \neg B~~~~$ (Intro $\land$, 2, 3)

5. $ (A \land \neg B) \land \neg (A \land \neg B)~~~~$ (Intro $\land$, 4, 1)

6. $\neg \neg B~~~~$ (Intro $\neg$, 3, 5)

7. $B~~~~ $ (Elim $\neg$, 6)

8. $A \implies B~~~~$ (Intro $\implies$, 2,7)

9. $\neg (A \land \neg B)\implies(A\implies B)~~~~ $ (Intro $\implies$, 1, 8)

Combining these two results, we have as required:

$A \implies B ~~\equiv ~~ \neg (A \land \neg B)$

Note that the only properties of the implication operator ($\implies$) that are used here are:

1. Introduction of the $\implies$ operator by means of conditional (or direct) proof resulting in the discharging of an active premise.

2. Elimination of the $\implies$ operator by detachment (modus ponens)