# Pinocchio logic test

The other day, I watched the YouTube video "Can you pass this logic test from Brazil?" by MindYourDecisions.

We are told that Pinocchio always lies, and that Pinocchio says 'All my hats are green'.
We are asked to find which of the following 5 sentences we can 'conclude' from Pinocchio's sentence, considered as a mathematically false statement.

A) Pinocchio has at least one hat.
B) Pinocchio has only one green hat.
C) Pinocchio has no hats.
D) Pinocchio has at least one green hat.
E) Pinocchio has no green hats.

Not a logician or mathematician myself, I sort of figured that the negation of Pinocchio's sentence P = 'All my hats are green' is S = 'I have at least one hat that is not green'.

As S must be true, Pinocchio must have at least one hat, and one is for sure not green.

Watching the video, I was very puzzled by how the sentences were eliminated.
I would have looked for direct contradictions between A-E and S.
This would have ruled out B and C.
That's where I would have been stuck, though. I wouldn't know how to eliminate D and E.
For D my attempt would have been '1 green hat', which is compatible with D but contradicts S, so it is not safe to conclude D from S.
But for E, I don't see the contradiction. '1 blue hat', or '2 blue hats' are all compatible with both E and S.

Instead, in the video, B, D, and E were eliminated by making an example that contradicted them, and showing that this was compatible with P also not being true.
E.g. to eliminate B, the argument was: if Pinocchio had 2 green hats and 1 blue hat (which contradicts B), he could say P and he would be lying, therefore B cannot be concluded from not(P).

I am clearly missing something, as I do not get how this works.

Can anyone please explain how the elimination method in the video works, logically/mathematically?

As extra food for thought, I imagined the case if Pinocchio had said 'None of my hats are green'.

Wouldn't that lead to the same conclusion, A?

EDIT further attempt to elaborate/solve

Assuming that 'what can we conclude' means 'in which case $$S \to X, X \in \{A,B,C,D,E\}$$ is always true, i.e. can never be false.

$$S \to X = \bar S \lor X$$ is false only when $$S$$ is true and $$X$$ is false. In all other cases it's true.

So $$S \to X$$ is always true when $$\bar S \lor X$$ is always true, i.e. when there is no instance of its opposite $$S \land \bar X$$ being true.

Given that $$S = \bar P$$, perhaps the method in the video is based on showing that $$\bar P \land \bar X$$ is true in at least one case (i.e. that $$P$$ and $$X$$ can both be false at the same time), which proves that $$S \to X$$ is not always true.

Am I on the right track or completely off?

• Although this question may appear basic, it is thoroughly explained. I don’t think this deserves downvotes Jun 30, 2022 at 18:59
• @insipidintegrator Not so. If I have no hats, the statement "All my hats are green" is a vacuous truth. As a liar, Pinocchio is not allowed to state any truths, including vacuous ones. Jun 30, 2022 at 19:10
• Thank you @FShrike . I don't mind downvotes. I have a genuine scientific curiosity about this topic, so if anyone is generous enough to try and help me understand, I will appreciate that very much; those who take some bizarre pleasure in humiliating others for no reason whatsoever, can safely knock themselves off with the downvoting button; I for sure am not going to lose any sleep over it :) Jun 30, 2022 at 19:12
• I also dislike anonymous downvoting, where no pre-existing comment communicates the downvoter's thinking. Unfortunately, this is a very common practice at MathSE. Jun 30, 2022 at 19:27
• A general suggestion: don't take YouTube videos too seriously ... Jun 30, 2022 at 22:19

The elimination approach can be thought of as considering a universe of situations $$S$$. (Or worlds.) For any situation $$x \in S$$, we say $$P(x)$$ iff all of Pinocchio's hats are green in $$x$$.

In order for us to conclude some other predicate $$Q$$ from $$\neg P$$ (that is, that Pinocchio is lying), it must not be possible for us to find an $$x$$ such that $$\neg P(x)$$ but also $$\neg Q(x)$$. The existence of even a single such $$x$$ breaks the implication.

For example, suppose $$Q =$$ "Pinocchio has no hats." Could $$\neg P$$ imply $$Q$$? No: Consider the situation $$x =$$ "Pinocchio has one red hat and no other hat." We see that $$\neg P(x)$$ (it is not the case that all of Pinocchio's hats are green, since there is a red one), but also $$\neg Q(x)$$ (it is not the case that Pinocchio has no hats, since we has exactly one). Thus, it cannot be the case that $$\neg P$$ implies $$Q$$.

• Thanks Brian, I accepted your answer as it's the one I understand best. I believe it's what they call a proof by contrapositive, as mentioned by others, too(?). math.colgate.edu/faculty/valente/math250/lecturenotes/… . The thing that still unsettles me is that one makes up a case for $Q$ almost at random, with the only condition that it falsifies $P$. I cannot immediately think of a real-life case where this would make intuitive sense. BTW I like the Brian Griffin icon, with the dry Martini :D Jul 1, 2022 at 16:43
• Wait... I get it now... If $P =$ 'all my hats are green', $\bar P = S =$ 'I have at least one hat, at least one is not green' and $E =$ 'I have no green hats', to prove that $S \to E$ is not always true, I only need to find one case where $S$ is true and $E$ is false (e.g. 1 green hat and 1 blue hat). Instead I was looking for cases where $E$ was always false when $S$ was true. The latter works for $B,C$ because they are directly contradicted by $S$, but not for $D,E$. What confused me was the fact that the case made to falsify $E$ seemed disconnected from $S$. Sorry... Jul 1, 2022 at 18:15
• @user6376297: Yes, that's right, a single instance in which $S$ is true and $E$ is false is sufficient to disprove $S \implies E$. Jul 2, 2022 at 2:42
• @user6376297: Oh, and yes, I like my avatar too. He can be a bit of an obnoxious blowhard, and I'm afraid to say that that sometimes is true of me too. :-P Jul 24, 2022 at 18:12

A nice instructive question indeed —with the proviso that one should not appeal to formal devices.

Our task is to draw a decidedly true conclusion from Pinocchio's false statement. Hence, we want to obtain this position: Pinocchio's statement is false while our statement is true.

It must be relatively trivial to eliminate (B) and (D): There are situations that make Pinocchio's statement be false, but our statement be false as well. Examples: 2 green hats, 1 blue hat for (B) and 2 blue hats for (D).

(E) certainly falsifies Pinocchio. But there are situations that make our statement be false at the same time. Example: 1 green hat, 1 blue hat.

(C) is tricky: In daily speech, we use such statements with existential import, that is, the listener understands also that Pinocchio actually has got hats (since the quantifier 'all' usually connotes plurality). However, we have been warned that we should not regard the question in daily speech conventions. So, we shall seek a falsifying instance. If there is no falsifying instance, then the statement is accepted as true. Hence, Pinocchio may have no hats and that makes whatever he says about his hats be true. In order that what Pinocchio has said to be false, he must have at least one non-green hat as a counter-instance. So, if (C) were our statement, Pinocchio's could be true, too.

We can take (A) as a continuation of (C): If not (C), then (A).

Let us check it formally: Pinocchio's statement can be expressed as $$\forall x(Px\rightarrow Gx)$$. Its negation is $$\exists x(Px\wedge\neg Gx)$$. That is, at least, Pinocchio has one hat and it is not green —by formal analysis, we secure more information in a much swifter fashion.

the negation of Pinocchio's sentence P = 'All my hats are green' is S = 'I have at least one hat that is not green'.

Not necessarily. From sentence $$P$$'s wording alone
All my hats are green
All my hats are green
All my hats are green
we know only that Pinocchio is lying about at least one of the following: the colour of his existent hats (note that if they don't exist, then whatever claim he makes about them is vacuously true), the portion of his existent collection of hats being green, or the type of possession of which his entire collection is green (in this case, the collection of his existent hats is not entirely green).

In other words, we know that Pinocchio has at least one hat (let's symbolise this as $$L),$$ and that $$¬X(g) \lor S \lor ¬Y(h).$$

The task is to determine which of the given options is/are (must be) implied by this compound premise.

in the video, B, D, and E were eliminated by making an example that contradicted them, and showing that this was compatible with P also not being true.

Yes, this is a natural approach. For each of options $$\textbf A$$-$$\textbf E,$$ let's assume for the sake of contradiction that it is false; if a contradiction is derived from the above premise, then our assumption must have been false, that is, that option must be true (note that that option is not necessarily false if we aren't able to derive some contradiction).

A more direct approach (using option $$\textbf C$$ as an example) is this: we want to check whether $$L\land\big(¬X(g) \lor S \lor ¬Y(h)\big)\implies \textbf C;\tag1$$ equivalently, by contrapositive, whether $$\lnot \textbf C\implies \lnot L\lor\big(X(g) \land ¬S \land Y(h)\big).$$

Implication $$(1)$$ is true only for option $$\textbf A.$$ On the other hand, for each of options $$\textbf A$$-$$\textbf E$$ (still using option $$\textbf C$$ as an example), $$\bigg(L\land\big(¬X(g) \lor S \lor ¬Y(h)\big)\bigg) \land\lnot \textbf C$$ is satisfiable, meaning that implication $$(1)$$ is false.

imagine the case if Pinocchio had said 'None of my hats are green'.

Wouldn't that lead to the same conclusion, A?

Yes: if $$\textbf A$$ is false, that is, Pinocchio has no hat, then "None of my hats are green" is true, so Pinocchio wasn't lying, which is a contradiction, which means that $$\textbf A$$ must actually be true, contrary to our supposition.

EDIT Assuming that 'what can we conclude' means 'in which case $$S \to X, X \in \{A,B,C,D,E\}$$ is always true

The premise is actually $$L\land\big(¬X(g) \lor S \lor ¬Y(h)\big)$$ instead of merely $$S.$$