Tag: Heyting

some skew Smullyan stumpers

Published September 21, 2023 by lieven

Raymond Smullyan‘s logic puzzles are legendary. Among his best known are his Knights (who always tell the truth) and Knaves (who always lie) puzzles. Here’s a classic example.

“On the day of his arrival, the anthropologist Edgar Abercrombie came across three inhabitants, whom we will call $A$, $B$ and $C$. He asked $A$: “Are you a Knight or a Knave?” $A$ answered, but so indistinctly that Abercrombie could not understand what he said.

He then asked $B$: “What did he say?” $B$ replies: “He said that he is a knave.” At this point, $C$ piped up and said: “Don’t believe that; it’s a lie!”

Was $C$ a Knight or a Knave?”

If you are stumped by this, try to figure out what kind of inhabitant can say “I am a Knave”.

Some years ago, my friend and co-author Karin Cvetko-Vah wrote about a much stranger island, the island of two truths.

“The island was ruled by a queen and a king. It is important to stress that the queen was neither inferior nor superior to the king. Rather than as a married couple one should think of the queen and the king as two parallel powers, somewhat like the Queen of the Night and the King Sarastro in Mozart’s famous opera The Magic Flute. The queen and the king had their own castle each, each of them had their own court, their own advisers and servants, and most importantly each of them even had their own truth value.

On the island, a proposition p is either FALSE, Q-TRUE or K-TRUE; in each of the cases we say that p has value 0, Q or K, respectively. The queen finds the truth value Q to be superior, while the king values the most the value K. The queen and the king have their opinions on all issues, while other residents typically have their opinions on some issues but not all.”

The logic of the island of two truths is the easiest example of what Karin and I called a non-commutative frame or skew Heyting algebra (see here), a notion we then used, jointly with Jens Hemelaer, to define the notion of a non-commutative topos.

If you take our general definitions, and take Q as the distinguished top-element, then the truth tables for the island of two truths are these ones (value of first term on the left, that of the second on top):

\[
\begin{array}{c|ccc}
\wedge & 0 & Q & K \\
\hline
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\begin{array}{c|ccc}
\vee & 0 & Q & K \\
\hline
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \quad
\begin{array}{c|ccc}
\rightarrow & 0 & Q & K \\
\hline
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\begin{array}{c|c}
& \neg \\
\hline
0 & Q \\
Q & 0 \\
K & 0
\end{array}
\]

Note that on this island the order of statements is important! That is, the truth value of $p \wedge q$ may differ from that of $q \wedge p$ (and similarly for $\vee$).

Let’s reconsider Smullyan’s puzzle at the beginning of this post, but now on an island of two truths, where every inhabitant is either of Knave, or a Q-Knight (uttering only Q-valued statements), or a K-Knight (saying only K-valued statements).

Again, can you determine what type $C$ is?

Well, if you forget about the distinction between Q- and K-valued sentences, then we’re back to classical logic (or more generally, if you divide out Green’s equivalence relation from any skew Heyting algebra you obtain an ordinary Heyting algebra), and we have seen that then $B$ must be a Knave and $C$ a Knight, so in our new setting we know that $C$ is either a Q-Knight or a K-Knight, but which of the two?

Now, $C$ claims the negation of what $B$ said, so the truth value is $\neg 0 = Q$, and therefore $C$ must be a Q-Knight.

Recall that in Karin Cvetko-Vah‘s island of two truths all sentences have a unique value which can be either $0$ (false) or one of the non-false values Q or K, and the value of combined statements is given by the truth tables above. The Queen and King both have an opinion on all statements, which may or may not coincide with the actual value of that statement. However, if the Queen assigns value $0$ to a statement, then so does the King, and conversely.

Other inhabitants of the island have only their opinion about a subset of all statements (which may be empty). Two inhabitants agree on a statement if they both have an opinion on it and assign the same value to it.

Now, each inhabitant is either loyal to the Queen or to the King (or both), meaning that they agree with the Queen (resp. King) on all statements they have an opinion of. An inhabitant loyal to the Queen is said to believe a sentence when she assigns value $Q$ to it (and symmetric for those loyal to the King), and knows the statement if she believes it and that value coincides with the actual value of that statement.

Further, if A is loyal to the Queen, then the value of the statement ‘A is loyal to the Queen’ is Q, and if A is not loyal to the Queen, then the value of the sentence ‘A is loyal to the Queen’ is $0$ (and similarly for statements about loyalty to the King).

These notions are enough for the first batch of ten puzzles in Karin’s posts

Just one example:

Show that if anybody on the island knows that A is not loyal to the Queen, then everybody that has an opinion about the sentence ‘A is loyal to the Queen’ knows that.

After these two posts, Karin decided that it was more fun to blog about the use of non-commutative frames in data analysis.

But, she once gave me a text containing many more puzzles (as well as all the answers), so perhaps I’ll share these in a follow-up post.

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Boolean and Heyting islands

Published March 26, 2021 by lieven

Raymond Smullyan‘s logic puzzles frequently involve Knights (who always tell the truth) and Knaves (who always lie).

In his book Logical Labyrinths (really a first course in propositional logic) he introduced islands where the lying or truth-telling habits can vary from day to day—that is, an inhabitant might lie on some days and tell the truth on other days, but on any given day, he or she lies the entire day or tells the truth the entire day.

An island is said to be Boolean if is satisfies the following conditions:

$\mathbf{N}$ : For any inhabitant $A$ there is an inhabitant who tells the truth on all and only those days on which $A$ lies.
$\mathbf{M}$ : For any inhabitants $A$ and $B$ there is an inhabitant $C$ who tells the truth on all and only those days on which $A$ and $B$ both tell the truth.
$\mathbf{J}$ : For any inhabitants $A$ and $B$ there is an inhabitant $C$ who tells the truth on all and only those days on which either $A$ tells the truth or $B$ tells the truth (or both). (In other words, $C$ lies on those and only those days on which $A$ and $B$ both lie.)

On any given day there are only Knights and Knaves on the island, but these two populations may vary from one day to the other. The subsets (of all days) for which there is an inhabitant who is a Knight then and a Knave on all other days form a Boolean algebra with operations $\wedge = \cap$ ($\mathbf{M}$eet), $\vee= \cup$ ($\mathbf{J}$oin) and $\neg=$ set-complement ($\mathbf{N}$egation).

Here’s a nice puzzle from Smullyan’s book:

Solomon’s Island also turned out to be quite interesting. When Craig arrived on it, he had the following conversation with the resident sociologist:

Craig : Is this island a Boolean island?
Sociologist : No.
Craig : Can you tell me something about the lying and truth-telling habits of the residents here?
Sociologist : For any inhabitants $A$ and $B$, there is an inhabitant $C$ who tells the truth on all and only those days on which either $A$ lies or $B$ lies (or both).

Show that the sociologist didn’t go native, and that his research is lousy.
(My wording, not Smullyan’s)

Smullyan’s version: This interview puzzled inspector Craig; he felt that something was wrong. After a while he realized for sure that something was wrong, the sociologist was either lying or mistaken!

Extending Smullyan’s idea, we can say that an island is Heyting if, in addition to $\mathbf{M}$ and $\mathbf{J}$ is satisfies the following rules

$\mathbf{T}$ : at least one inhabitant tells the truth on all days.
$\mathbf{F}$ : at least one inhabitant lies on all days.
$\mathbf{I}$ : For any inhabitants $A$ and $B$ there is an inhabitant $C$ sharing Knight-days with $A$ only when $B$ tells the truth, and there are no inhabitants doing this while telling the truth on more days than $C$.

Let’s give an example of an Heyting island which is not Boolean.

On Three-island there are only three kinds of people: Knights, Knaves and Alternates, who can neither lie nor tell the truth two days in a row. All Alternates tell the truth on the same days.

Here’s a riddle:

You meet John, who is a Knight, James, an Alternate, and William, a Knave. You don’t know who is who. You can only ask one question containing at most four words, giving you a Yes or No answer, to just one of the three. The answer must tell you whether that person is James or not.

You may like to watch Smullyan on the Carson show for a hint.

Or, you might just watch it reminiscing long forgotten times, when talkshow-hosts still listened to their guests, and could think for themselves…

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The strange logic of subways

Published March 5, 2021 by lieven

“A subway is just a hole in the ground, and that hole is a maze.”

“The map is the last vestige of the old system. If you can’t read the map, you can’t use the subway.”

Eddie Jabbour in Can he get there from here? (NYT)

Sometimes, lines between adjacent stations can be uni-directional (as in the Paris Metro map below in the right upper corner, 7bis). So, it is best to view a subway map as a directed graph, with vertices the different stations, and directed arrows when there’s a service connecting two adjacent stations.

Aha! But, directed graphs form a presheaf topos. So, each and every every subway in the world comes with its own logic, its own bi-Heyting algebra!

Come again…?

Let’s say Wally (or Waldo, or Charlie) is somewhere in the Paris metro, and we want to find him. One can make statements like:

$P$ = “Wally is on line 3bis from Gambetta to Porte des Lilas.”, or

$Q$ = “Wally is traveling along line 11.”

Each sentence pinpoints Wally’s location to some directed subgraph of the full Paris metro digraph, let’s call this subgraph the ‘scope’ of the sentence.

We can connect such sentences with logical connectives $\vee$ or $\wedge$ and the scope will then be the union or intersection of the respective scopes.

The scope of $P \vee Q$ is the directed subgraph of line 11 (in both directions) together with the directed subgraph of line 3bis from Gambetta to Porte des Lilas.

The scope of $P \wedge Q$ is just the vertex corresponding to Porte des Lilas.

The scope of the negation $\neg R$ of a sentence $R$ is the subgraph complement of the scope of $R$, so it is the full metro graph minus all vertices and directed edges in $R$-scope, together with all directed edges starting or ending in one of the deleted vertices.

For example, the scope of $\neg P$ does not contain directed edges along 3bis in the reverse direction, nor the edges connecting Gambetta to Pere Lachaise, and so on.

In the Paris metro logic the law of double negation does not hold.

$\neg \neg P \not= P$ as both statements have different scopes. For example, the reverse direction of line 3bis is part of the scope of $\neg \neg P$, but not of $P$.

So, although the scope of $P \wedge \neg P$ is empty, that of $P \vee \neg P$ is not the full digraph.

The logical operations $\vee$, $\wedge$ and $\neg$ do not turn the partially ordered set of all directed subgraphs of the Paris metro into a Boolean algebra structure, but rather a Heyting algebra.

Perhaps we were too drastic in removing all “problematic edges” from the scope of $\neg R$ (those with a source or target station belonging to the scope of $R$)?

We might have kept all problematic edges, and added the missing source and/or target stations to get the scope of another negation of $R$: $\sim R$.

Whereas the scope of $\neg \neg R$ always contains that of $R$, the scope of $\sim \sim R$ is contained in $R$’s scope.

The scope of $R \vee \sim R$ will indeed be the whole graph, but now $R \wedge \sim R$ does no longer have to be empty. For example, $P \wedge \sim P$ has as its scope all stations on line 3bis.

In general $R \wedge \sim R$ will be called the ‘boundary’ $\partial(R)$ of $R$. It consists of all stations within $R$’s scope that are connected to the outside of $R$’s scope.

The logical operations $\vee$, $\wedge$, $\neg$ and $\sim$ make the partially ordered set of all directed subgraphs of the Paris metro into a bi-Heyting algebra.

There’s plenty more to say about all of this (and I may come back to it later). For the impatient, there’s the paper by Reyes and Zolfaghari: Bi-Heyting Algebras, Toposes and Modalities.

Right now, I’m more into exploring whether this setting can be used to revive an old project of mine: Heyting Smullyanesque problems (btw. the algebra in that post is not Heyting, oops!).

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