## Three logicians walk into a bar: a formal explanation

spikedmath.com/445.html

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Let E refer to the propositions expressed by the sentence:

“Everyone wants beer”

“everyone” refers to the three people on the right. Let’s call them “a”, “b”, “c” from left to right. Let “Wx” =df “x wants beer”.

We could try to show this but let’s just take it intuitively that the following holds:

Everyone wants beer ↔ a wants beer and b wants beer and c wants beer

Formalizing we get:

E↔(Wa∧Wb∧Wc)

Now, assume that to begin with, a, b, c does not know whether the two others want beer or not. This is technically ‘left open’ in the comic, but it is not irrelevant.

Now, assume every person knows if he wants beer or not, or rather, he either knows that he wants beer, or he knows that he does not want beer. Without this, it doesn’t work either. Introducing “Kx(P)” to mean “K knows that P” and formalizing:

∀xKx(Wx)∨Kx(¬Wx)

Now, interpreting “logicians” to be a group of people being perfect at making inferences in at least this case. Such people are sometimes called “ideally rational” or similar. They never make a wrong inference and never miss an inference.

Let’s think about a’s position as he is first to answer the question. If he knows that he wants beer, then he does not know the truth of E. But if he knows that he does not want beer, he can know that E is false. Because: he is part of everyone, so if he does not want beer, it is not the case that everyone wants beer. If a is being truthful etc., he has to answer either “I don’t know” or “No”.

Now b is in almost the same position as a. Obviously, if a has already answered “No”, there is no reason to respond or at least he should respond the same. If a’s answer is “I don’t know”, b still lacks information about whether or not c wants beer [Wc or ¬Wc1]. Likewise, b knows his own state, so he knows either that he wants beer or that he doesn’t want beer. If he knows that he doesn’t, he can infer that E is false, similarly to a. If he knows that he does, then he can’t infer anything about E, and has to answer “I don’t know”.

Now, c has all the information he needs to answer either “Yes” or “No”. Obviously, if any of the previous answers are “No”, then he should also answer “No” or simply not answer. If both earlier answers are “I don’t know”, he can infer that both a and b want beer. He also knows his own state. If he does not want beer, he will answer “No”. If he does, he can infer that E is true, and thus answer “Yes”.

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Pretty similar comic which does need set theory to properly formalize: mrburkemath.blogspot.com/2011/05/coffee-logic.html

### Notes

1Which could also be interpret as: To go to the bathroom or not go to the bathroom? :D

## “Do you still beat your wife?” formalized

Using the formalization system I wrote of earlier, let’s take a look at this famous question.

First we should note that this is a yes/no question which is different from the questions that I have earlier formalized. The earlier questions sought to identity a certain individual, but yes/no questions do not. Instead they ask whether something is the case or not. So this time I cannot use the (x=?) question phrase from earlier, since there is no individual to identify similarly to the earlier cases.

One idea is to simply add a question mark at the end. Like this:

F1. (∃x)(∃y)(Wxy∧Bxy∧Cxy∧x=a)?

Wxy ≡ x’s wife is identical with y

Bxy ≡ x beats y

Cxy ≡ x used to beat y1

a ≡ you

But this fails to capture that it is not all of the things that are being asked whether they are the case or not. It is only the Bxy part that is being asked. The rest is stipulated as true. We can change the formalization to capture this, like this:

F1*. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)?

The question mark is now understood as a predicate that works on whatever is before/to the left of it. (In parentheses for clarity.) Not to the right like with the other monadic predicates and propositional connectors (¬, ◊, □, etc.). In this case the question mark only functions on (Bxy) and not the rest of the formula.

Translated into LE:

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and is it the case that x beats y?

When answered in the positive, the answered version simply removes the question mark. Producing:

F2. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)

Answering in the negative removes the question mark and adds a negation sign to the part of the formula that the question predicate is working on. Producing this:

F3. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧¬(Bxy)

If the produced formula is true, then the question has been answered correctly. However since this question is loaded. Both of the produced formulas are false, that is, it is both false that:

F2. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and x beats y.

and that:

F3. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧¬(Bxy)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and it is not the case that x beats y.

Since they both imply the falsehood:

(∃x)(∃y)(Wxy∧Cxy∧x=a)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you.

1Alternatively one could deepen to formalization to formalize the temporal aspect of this predicate. Though it doesn’t seem important here so I will leave it out.

## Language, the modal fallacy and the symbolic representation of a conditional

“[W]hat follows from a true premiss must be true” (The Problems of Philosophy, p. 60, link)

Wrote Russell as an example of a principle of logic that is more self-evident than the inductive principle. If we were to formalize this we would perhaps write it like this:

E1. □[([∀P][Q∧Q⇒P])→P]1

Or perhaps just just in propositional logic:

E2. □(P→Q) where “P” means P is true and P implies Q.

As the reader of my blog should know by now, the modal fallacy consists of trusting language and placing the modal operator of necessity in the consequence instead of before the conditional:

E3. P→□Q

However we could also put the modal operator somewhere else in our formalization:

E5. P□→Q

Operators solely ‘work on’ whatever is to the right of them.2 Thus the modal operator in (E5) works on the material conditional and not the proposition to the left of it. Similarly in (E2) the modal operator works on the parentheses-set which is treated as a single entity.

(E5) is closer to normal english (and danish) than (E2) which we can express in normal english like this:

E4. Necessarily, if P follows from Q and Q is true, then P is true.

Consider the sentence:

E5. If P follows from Q and Q is true, then P must be true.

(E5) is a reformulation of (E2). (E5) is worded like it would be by a normal english speaking person. In (E5) it may seem as though the modal operator is intended to work on the consequent. Indeed some people think this and commit the modal fallacy.

However the modal operator may also be thought of as working on the second part of the “if, then” clause, that is, the “then” part. The only problem with this interpretation that I know of, is that it makes the operator work on something that is to the left of it instead of to the right of it: Because “then” is to the left of “must” in (E5).

### Notes

1Ignoring the complexities of bivalance.

## Using propositions as variables

I have encountered the following problem a couple of times. This problem is this: When formalizing something in predicate logic, the predicate uses propositions as variables.1 We may refer to this as the predicate acting upon the variable. The predicate is a function similar to functions in mathematics like “F(x) = x4”. Predicates were also written like this (with parentheses) to begin with. Variables are written in the english lower-case letters beginning from x and i.e. {x, y, z, etc.}. For variables that have to do with time ‘t’ is often used. Propositions are written in upper-case english letters beginning with P i.e. {P, Q, R, S, etc.}. What ought one to write, then, when a predicate acts upon propositions? Ought one to switch to lower-case but keep the same letters i.e. {p, q, r, s, t}? We ought to use the way which is the least confusing and which is powerful enough to express whatever meaning clearly we might want to express. Here are some possible ways to formalize it:

1. Upper case beginning from P.
2. Lower case beginning from p.
3. Upper case beginning from X.
4. Lower case beginning from x.
5. Upper case beginning from P in parentheses.
6. Upper case beginning from P in square brackets.
7. Upper case beginning from P in curly brackets.
8. Upper case beginning from P in sub-script.
9. Upper case beginning from P in sup-script.

## Some possible causes of confusion

### Upper case beginning from P

Example: s knows that P is formalized as KsP.

Normally only the predicate is written in upper case e.g. “Fx” and all variables are written in lower case e.g. Rxyz.

### Lower case beginning from p

Example: s knows that P is formalized as Ksp.

Normally propositions are written in the upper case. The confusion may be enhanced when the proposition used in some predicate is also used elsewhere in the analysis without being acted upon by a predicate e.g. Kxp⇒P where “Kxp” means x knows p.

### Upper case beginning from X

Example: s knows that P is formalized as KsX.

Normally propositions are not referred to as “X”. “X” is usually reserved for variables in predicates. So this may cause confusion.

### Lower case beginning from x

Example: s knows that P is formalized as Ksx.

For the same reason as above and that propositions are normally written in the upper case.

### Upper case beginning from P in parentheses.

Example: s knows that P is formalized as Ks(P).

This is sometimes used. Parentheses are normally used for other purposes, so that may cause confusion. The good thing about using some sort of surrounding formalization is that it makes it much less confusing to have the propositional variable be complex e.g. an agent might know that someone knows a proposition logically implies that that proposition is true. Formalized as Ka(Kx(P)⇒P).

### Upper case beginning from P in square brackets

Example: s knows that P is formalized as Ks[P].

I have used this from time to time. I have not seen anyone else use it. It suffers from the same problem as above although to a lesser degree since square brackets are not as commonly used as parentheses. When square brackets are used, then they are usually used as alternate parenthesis. When some wff becomes so complicated that multiple sets of parentheses are needed, then one alternates between regular parentheses and square brackets e.g. (P→[Q→(R↔T)]). It makes it easier to keep track of the parenthesis sets. I sometimes use square brackets to mean that some wff is a formalization of something that was written in normal language. One might combine square brackets with parentheses so one can alternate surrounding characters even within some predicate e.g. an agent might know that someone knows a proposition logically implies that that proposition is true. Formalized as Kx[Kx(P)⇒P].

### Upper case beginning from P in curly brackets

Example: s knows that P is formalized as Ks{P}.

I have never seen this used and I don’t know of any other uses of curly brackets that may be confused with this.

### Upper case beginning from P in sub-script

Example: s knows that P is formalized as KsP.

Sub-script is sometimes used to number propositions e.g. (P1, P2, P3 … Pn.) to mean all P propositions. I suppose one could formalize this in second order predicate logic as (∀P)(P). It is usually employed for infinite sets or ‘changing’ sets e.g. a set of premises of an argument.2 It is also sometimes used to symbolize an alternate meaning of a word e.g. atheist2 but words are not used in wff’s, so that should not be a source of confusion. Another thing is that it may be hard to read since the letters are so small. It’s hard to even see that the “P” is in upper case.

### Upper case beginning from P in sup-script

Example: s knows that P is formalized as KsP.

I don’t know of any usage of sup-scripts in formalization at all. So I don’t know of any potential sources of confusion. The problem with readability also applies here.

## Summary and conclusion

Given the 9 suggestions above I think one should go with the method of surrounding the acted upon proposition with alternatingly parentheses and square brackets. The reason for choosing this method is that it introduces no new characters, so it is characteristic parsimonious which is important. The less different characters used the better all other things being equal. This is a general fact of language. The curly bracket solution introduces two new characters, “{“ and “}”, but doesn’t have any large advantages as far as I can see over the chosen method. The other methods are either confusing or too hard to read. To make it very simple: The chosen method is both parsimonious and powerful.

1Predicates with more than one variable are sometimes called relations. I shall just refer to them all as predicates in this essay.

2That set is ‘changing’ in the sense that the number of premises vary from argument to argument.