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.
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.
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.
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.
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.
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".
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.
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.
"A FALSE premise cannot imply any consequence.": YES! This is my preferred explanation.
– Joseph O'Rourke Dec 16 '23 at 22:42I 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).
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.
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).
