8
$\begingroup$

How do you explain that the implication $p \implies q$ has the value TRUE when $p$ is FALSE and $q$ is TRUE? It is difficult for students to accept that FALSE $\implies$ TRUE should have the value TRUE.

I can't say I've developed a compelling argument. When $p$ is FALSE, there is no implication, so $q$ is unaffected and can be either TRUE or FALSE---both compatible with the expression value of TRUE.

I'd be interested to hear more convincing approaches.

$\endgroup$
5
  • $\begingroup$ Who is the target audience here? Mathematics university students without bachelor's degree? $\endgroup$
    – Tommi
    Dec 16, 2023 at 10:30
  • 4
    $\begingroup$ In my country we're used to people saying "If (something blatantly false) then I'm the queen of England" or "If (something blatantly false) then chicken have teeth" or "With ifs, you can put Paris in a bottle" so people are not too shocked when they discover that mathematical "if" behaves just like everyday "if". $\endgroup$
    – Stef
    Dec 16, 2023 at 12:18
  • $\begingroup$ @Isaiah: Good find! But quite useful to hear of other approaches and examples. $\endgroup$ Dec 16, 2023 at 14:28
  • 1
    $\begingroup$ I have an informal explanation of vacuous truth that might be useful to you here: math.stackexchange.com/a/1257728/132640 $\endgroup$
    – msouth
    Dec 16, 2023 at 22:49
  • 1
    $\begingroup$ And see my explanation here: math.stackexchange.com/questions/1551320/… $\endgroup$
    – mweiss
    Dec 17, 2023 at 0:31

10 Answers 10

15
$\begingroup$

It makes more intuitive sense if you view an implication as a "promise". The truth value of the expression represents whether the promise can be broken (the promise is considered true unless broken).

From that point of view, it makes intuitive sense that in a situation when the antecedent is false, the promise is irrelevant to the situation, so the promise remains unbroken (i.e. true) regardless of the consequent.

For instance, suppose the promise is that "if you eat your vegetables then you can have dessert." Today you didn't eat your vegetables, but I still let you have dessert. Have I broken my promise? No. I have been true to my word. My promise has nothing to do with the situation at hand. I never said anything about what would happen if you didn't eat your vegetables.

$\endgroup$
2
  • $\begingroup$ I think one should add that the "promise" is considered "true" if it's unbroken (as a convention). Otherwise consider this reasonable scenario: I say "if you eat your vegetables then you can have dessert" but I lie - I have no intention to keep that promise. Then you don't eat your veggies and don't get dessert. Was my promise actually true? $\endgroup$ Dec 16, 2023 at 21:53
  • $\begingroup$ Along these lines, "there is nothing you can do to guarantee desert at this point" fits the OP's query pretty well. $\endgroup$
    – Cort Ammon
    Dec 16, 2023 at 22:48
9
$\begingroup$

I use the example

If your flight is on time, then I will pick you up from the airport.

If your friend tells you this, the only time they have lied is if the premise is true and the conclusion false. If your flight is late, no matter what happens afterwards, they won't have lied.

It is the same with formal statements: the only time $P \to Q$ is false is when the premise is true and the conclusion false. If you don't want them to be false, logic demands that the other situations must be true.

$\endgroup$
5
$\begingroup$

I convert $p \Rightarrow q$ to conjunctions and disjunctions as follows.

Consider a simple, familiar theorem of the form $p(x) \Rightarrow q(x)$. For my students, who have had at least a year of calculus, the following has seemed perfect:

If $f$ is differentiable at $c$, then $f$ is continuuous at $c$.

I then ask them to come up with example for $f$ and $c$ such $p$ and $q$ exhibit all four combinations of TRUE and FALSE, or explain why a particular combination is impossible. This in effect fills in the truth table with possible/impossible instead of TRUE/FALSE; but let's not dwell on that.

Instead, we agree (the class and I) that $p \Rightarrow q$ is equivalent to $$(p \wedge q) \mathrel{\vee} (\lnot p \wedge q) \mathrel{\vee} (\lnot p \wedge \lnot q) \,.$$ And we agree that if $p \Rightarrow q$ is a theorem then $p \wedge \lnot q$ cannot be TRUE. And that fills in the truth table.

I used to use more natural language example, but biological examples, including humans, seem full of exceptions. It seems safer to build on mathematics they know. And it seems good for the student to reinforce mathematics they should know, in case they don't.

$\endgroup$
5
$\begingroup$

One informal interpretation is that $p \implies q$ means "$q$ is at least as true as $p$".

Since TRUE is "at least as true as" FALSE, the implication FALSE $\implies$ TRUE is true.

$\endgroup$
1
  • $\begingroup$ This works even better when p and q are events or predicates. With events "$P \implies Q$ is equivalent to $P \subseteq Q$. With predicates, $\forall x, p(x) \implies q(x)$ is equivalent to $\operatorname{supp}(p) \subseteq \operatorname{supp}(q)$ (where $\operatorname{supp}$ means "support", i.e. $\operatorname{supp}(p) = p^{-1}(\{\text{True}\})$). $\endgroup$
    – Stef
    Dec 16, 2023 at 12:22
2
$\begingroup$

A nice thing about the convention we use for the Booleans of $p\rightarrow q$ is that when $q$ is true, then $p\rightarrow q$ is "true" independent of the truthiness of the premises $p$.

If we know that $q$ is true, for reasons independent of the premises $p$, it would be awkward for the Boolean of $p\rightarrow q$ to shift from true to false if we reconsidered the premises $p$ and decided they were flawed.

If we wanted "false implies true" to be false, then we might as well just use the logical operation of "and".

$\endgroup$
2
  • $\begingroup$ I doubt the objections are about "let's create new binary operators for all possible truth tables" and more about labeling this particular binary operator as "if p, then q" or "p implies q". -- "There's another operator called OXD that has this truth table" would likely be accepted with minimal complaint. $\endgroup$
    – R.M.
    Dec 16, 2023 at 12:51
  • 1
    $\begingroup$ Regarding your last sentence, I don't think the misconception is confusing $\Rightarrow$ with AND, but rather with $\Leftrightarrow$. $\endgroup$ Dec 16, 2023 at 12:55
2
$\begingroup$

Logical implication is a binary operator $\implies$ defined for booleans $p,\ q$ such that $p \implies q$ if and only if $\neg\,(\,p \land \neg\,q)$ is true. But by the time I was introduced to this definition, I already had a strong intuitive sense of the meaning of "implication" which did not align with this mathematical definition and confused me.

Suppose that $\mathrm{rain} \implies \mathrm{clouds}$ is true. If it is not raining, then the truth of the implication tells us nothing about whether there are clouds. $\neg\,\mathrm{rain}$ does not "imply" anything. Yet the mathematical definition assigns truth value to both expressions, $\neg\,\mathrm{rain} \implies \mathrm{clouds}$ and $\neg\,\mathrm{rain} \implies \neg\,\mathrm{clouds}$. This way of expressing the lack of information in the proposition $\neg\,\mathrm{rain}$ can feel like a contradiction.

The source of my confusion was the result in classical logic that $p$ entails $q$ if and only if $p \implies q$. In fact, there is a distinction between implication and entailment in philosophy. I was able to accept $\implies$ as a simple mathematical function on booleans which could be meaningfully applied to false $p$, once I realised that the source of my confusion was an significant problem in philosophy which I did not need to resolve in a mathematics class.

$\endgroup$
1
  • 1
    $\begingroup$ I have to disagree. If all we know about weather is that $\text{rain} \implies \text{clouds}$, then there is no way to know whether $(\neg \text{rain}) \implies \text{clouds}$ and $(\neg \text{rain}) \implies \neg \text{clouds}$ are true or false. $\endgroup$
    – Stef
    Dec 16, 2023 at 12:30
2
$\begingroup$

Given a relationship : PREMISE ⟶ CONSEQUENCE

The truthy value for that relationship is: is it possible to get the consequence given the premise?

This relationship is easy to understand for true premises.

PREMISE ⟶ CONSEQUENCE CAN WE GET THAT CONSEQUENCE?
TRUE ⟶ TRUE YES - we get truth from true premises
TRUE ⟶ FALSE NO - we do not get falsehood from true premises

The problem is that we want false premises to work the same way as true ones. But they don’t.

PREMISE ⟶ CONSEQUENCE CAN WE GET THAT CONSEQUENCE?
FALSE ⟶ TRUE YES
FALSE ⟶ FALSE YES

The reason FALSE ⟶ ANYTHING is valid is because FALSE does not really imply anything — the consequence could be either TRUE or FALSE. That is:

A FALSE premise cannot imply any consequence.

Written another way:

A FALSE premise does not have the power to invalidate either TRUE or FALSE consequences.

$\endgroup$
1
  • $\begingroup$ > "A FALSE premise cannot imply any consequence.": YES! This is my preferred explanation. $\endgroup$ Dec 16, 2023 at 22:42
2
$\begingroup$

I think one possible difficulty is the idea itself that $p \Rightarrow q$ might have a value TRUE or FALSE. Reading "$p$ implies $q$" some might interpret it as what programmers would call a "statement" rather than an "expression" that evaluates to something. I think a way to help students make sense of this is to interpret the truth value of $p \Rightarrow q$ as an assessment of whether "the data is consistent with the implication statement". So if $p$=FALSE and $q$=TRUE, the data is consistent with $p \Rightarrow q$ and so its value is TRUE. If instead $p$=TRUE and $q$=FALSE, the data violates $p \Rightarrow q$ which must be FALSE. It is also important that we stop at assessing whether the data is consistent with the implication without trying to assess the truthiness of the implication (in everyday language); as a result the expression "3 is odd $\Rightarrow$ 3 < 12" is true because the data is consistent with the implication albeit the implication is false in the everyday meaning of the term (where implication suggests some form of causality).

$\endgroup$
1
$\begingroup$

Are they familiar with the principle of explosion, or could it be introduced? The idea that deriving a contradiction (or a proof of $\bot$, if you prefer) means you can prove anything is slightly more general, but then explains that $p \implies q$ makes sense whenever $p$ is false (since any value of $q$ could be derived).

$\endgroup$
1
$\begingroup$

How to explain (FALSE => TRUE) is TRUE

You might try to avoid using the literals "TRUE" AND "FALSE" as logical propositions as here.

Try to clarify the meaning in propositional logic of "implies" with an example:

If it is raining ($R$), then it is cloudy ($C$)

$R \implies C$

This does not mean that rain causes cloudiness. It does not even mean that it is always cloudy when it is raining. (That would require some form of predicate logic, e.g. quantifying over time.) It means only that, at the moment, it is not both raining and not cloudy.

$R\implies C ~~\equiv ~~ \neg (R \land \neg C)$

It also means that if it is not raining, then it may or may not be cloudy. Both are still possibilities.

$\endgroup$
4
  • $\begingroup$ I upvoted for the first sentence ("You might try to avoid using the literals "TRUE" AND "FALSE" as logical propositions as here."). However I'm a bit confused by your cloudy argument. Doesn't "If it is raining, then it is cloudy" specifically mean "There is not such thing as a sunshower"? Or is the whole thing just wordplay on the different uses of present tense? $\endgroup$
    – Stef
    Dec 16, 2023 at 12:25
  • $\begingroup$ @Stef In propositional logic, propositions are usually thought of as being about what IS true (present tense). Suppose for example that, at the moment, it is raining implies it is cloudy, then, AT THE MOMENT (but necessarily at other times), a sunshower is ruled out. If you want to talk about things that are always true over all time or over some period of time, you should use quantifiers over time. Suppose, for example, that at any time, if it is then raining, it is cloudy. This would indeed rule out sunshowers over all time. $\endgroup$ Dec 16, 2023 at 17:16
  • $\begingroup$ Well, in my opinion this just confuses the issue further. You're teaching the students "If a triangle ABC has a right angle in C, then AC^2 + BC^2 = AB^2" and then when students get confused about implication, you suddenly change the meaning of present tense propositions to follow an arbitrary unspoken convention "because that's how we do it in propositional logic". $\endgroup$
    – Stef
    Dec 17, 2023 at 11:41
  • $\begingroup$ @Stef In mathematics, we only deal with what IS true (present tense). If you introduce predicates or functions that are time-dependent as in physics, students should know that it is best to introduce variables for time and quantifiers over time which are not found in purely propositional logic. I think students are less likely to get confused by implications in propositional logic if we stick to "real-world" examples in the present tense. $\endgroup$ Dec 17, 2023 at 19:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.