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?
Best Answer
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 $Q$.