Garbage Logic

Question

One sunny afternoon John asks Sally how she's doing, and Sally responds that she isn't happy. Sally then quotes the song by the band Garbage - "I'm only happy when it rains." John then claims Sally has committed a fallacy. Is John correct?

Answer

To get a feel for the question, consider whether you think the following is a good inference:

(i) S is only happy if it rains
(ii) It does not rain
(iii) Hence, S is not happy

Let "It rains" be denoted by Q and "S is happy" by P. If you think (i) has the form (Q->P) then this will be invalid. You might think this because "if" on its own seems to be introducing "It rains" as the antecedent of a material conditional. Then (i) is equivalent to: "It rains, only if S is happy." That is:

(1) It rains only if S is happy (Q->P)
(2) It does not rain (~Q)
(3) Hence, S is not happy (~P)

On the other hand, if you think (i) has the form (P->Q), then the argument is valid. You might think this because "only if" is typically taken to introduce the consequent of a material conditional, and "S is only happy if..." is plausibly read as "S is happy only if...". Then (i) is equivalent to "S is happy only if it rains." That is:

(4) S is happy only if it rains (P->Q)
(5) It does not rain (~Q)
(6) Hence, S is not happy (~P)

There are two further options as the setup permits another reading:

(iv) S is only happy if it rains
(v) S is not happy
(vi) Hence, it does not rain

So that taking "if" to introduce the antecedent, we have the valid:

(7) It rains only if S is happy (Q->P)
(8) S is not happy (~P)
(9) Hence, it does not rain (~Q)

And if you take "only happy if" to introduce the consequent, we have the invalid:

(10) S is happy only if it rains (P->Q)
(11) S is not happy (~P)
(12) Hence, it does not rain (~Q)

So, whether a fallacy has been committed depends on whether Sally intended to infer "it does not rain" or "Sally is not happy, as well as whether "Sally is only happy if it rains" should be read as "Sally is happy only if it rains" or "It rains only if Sally is happy."

Concerning the first, I take Sally to be supporting the claim "I'm not happy" by appealing to the fact that it is not raining, which is common knowledge. If this is correct, then the second premise of the argument should be "It does not rain." So we can rule out setup (vi)-(vi).

Concerning the second, I take Sally to be claiming that if she's happy, then it rains. That is, Sally is making a valid argument and John is incorrect. This reading is also consistent with Sally not being happy, but it raining nevertheless. In other words, all we know is that if Sally is happy then it's raining, and if it's not raining then Sally isn't happy. This is also a bit sadder. Sally isn't even happy all and only those times it rains. Rather, Sally is only happy some of the times it rains.

Trust Logic, Not Tortoises

Carroll's note What Achilles Said to the Tortoise holds many lessons, many of which related to putative justification - or lack of justification - for basic logical inferences. Recently, Romina Padro, pulling from coursework and discussions with Kripke, has argued one more lesson should be added to the list, namely, that under certain conditions adopting basic logical inferences is impossible. I've a few thoughts on this new lesson, in particular how it might play with the old lessons. Check it out a recent draft here!

Now That's a Hat of a Different Color...

Three Hats Puzzle
Three individuals, call them Alex, Barbara, and Cherise, enter a pitch black room, where they are led to a table on which rests five hats, 3 red hats and 2 blue hats. The hats are arranged in no obvious order, and no individual can discern the colors in the dark, but Alex, Barbara, and Charles know how many hats of each color there are. They each select a hat from the table, and wear that hat outside the room into a well-lit area. Alex looks at Barbara and Cherise, and says, “I don’t know what color my hat is.” Barbara looks at Alex and Cherise and says, “I don’t know what color my hat is.” Cherise does not look at anyone else, since Cherise is blind. Nevertheless, Cherise says “I know what color my hat is.” This is all that is said, and they each speak truly.

Challenge
Explain how Cherise knows her hat color.

Note: Like many puzzles, this has numerous lateral solutions, e.g. the red and blue hats are differently shaped, Cherise is colorblind but can see blue, etc. Lateral solutions are easily dismissed without affecting the details of the scenario, e.g. the hats share all properties save color, Cherise is not just colorblind, etc. The challenge is to find a logical solution. A logical solution will follow directly from the details of the scenario, and will not be easily dismissed since doing so will require changing the scenario.

Solution and Discussion
I pose this puzzle to students who are then encouraged to work in small groups (1-3 students) to find a solution. Once students have understood the puzzle and the distinction between lateral and logical solutions, groups are quick to reason in the following manner:

Since Alex speaks truly, she must not see two blue hats. If Alex did see two blue hats, then she would know her hat was red. Then Alex must see either two red hats or one red and one blue hat. Similarly for Barbara, who must see either two red hats or one red and one blue hat

This seems unsurprising; the reasoning involved is direct, it is an immediate consequence of understanding the details of the case. We may show this formally. First, we fix on our notation.

Our language is first order with identity. Our domain consists of eight objects, which we sort into individuals with the predicate “I” and hats with the predicate “H”. Let “a” denote Alex, “b” Barbara, “c”. Let “B” be the predicate applying to blue hats, and “R” the predicate applying to red hats. Sample axioms characterizing the domain include (see here for the full set):

1.      ∀x(Hx v Ix)
Everything in the domain is either a hat or an individual
2.      ~∃x(Hx & Ix)
Nothing in the domain is both a hat and an individual
3.      ∃x∃y∃z(x≠y v x≠z v y≠z & Ix & Iy & Iz & ∀w(Iw -> (x=w v y=w v z=w)))
There are exactly three individuals
4.      ∀x(Bx -> Hx)
All blue hats are hats
5.      ∀x(Rx -> Hx)
All red hats are hats
6.      ∀x(Hx -> (Bx v Rx))
Every hat is either red or blue
7.      ∃x∃y(x≠y & Bx & By & ∀z(Bz -> (x=z v y=z)))
There are exactly two blue hats

…And so on. We also introduce relations.  Let “S” stand for an irreflexive binary relation holding between an individual and a hat with the intended reading being that the individual sees the hat. Let binary “K” hold between individuals with the intended reading that the first individual knows the hat color of the second. Sample characterizing axioms include:

8.      ~∃x(Sxx)
The ‘sees’ relation is irreflexive
9.      ∀x∀y(Sxy -> (Ix & Hy))
Individuals see hats
10.     ∀x∀y(Kxy -> (Ix & Iy))
Only individuals know things
11.      ∀x∀y∀z((Sxy & Sxz & y≠z) -> ∀w(S(x,w) -> (w=y v w=z))
Individuals see at most two hats

…And so on. We also characterize the following facts concerning the case:

12.  ∃x∃y(x≠y & Sax & Say)
Alex sees two things
13.  ∃x∃y(x≠y & Sbx & Sby)
Barbara sees two things
14.  ~∃x(Scx)
Cherise sees nothing
15.  ~Kaa
Alex does not know what color hat she is wearing
16.  ~Kbb
Barbara does not know what color hat she is wearing

With these axioms in hand, we may infer the following additional facts as theorems simply based on the domain and relation constraints:

17.  ∃x∃y(Sax & Say & (Bx & By) v (Rx & Ry) v (Bx & Ry))
Alex sees either a blue/blue, red/red, or blue/red distribution
18.  ∃x∃y(Sbx & Sby & (Bx & By) v (Rx & Ry) v (Bx & Ry))
Barbara sees either a blue/blue, red/red, or blue/red distribution

We add two more plausible facts. Observe, there is a relationship between knowing one’s hat color and the possible hat distribution. Consider Alex. If Alex sees two blue hats she knows what color hat she is wearing, which we may formalize as (watch the scope; avoid the Drinker Paradox!):

19.  ∃x∃y(Sax & Say & x≠y & Bx & By) -> Kaa
If there are two blue hats Alex sees, then Alex knows her hat color

A similar fact pertains to Barbara, but we will leave that aside here. Importantly, given (15) the consequent of (19) is false. Hence, the following theorems can be inferred:

20.  ∀x∀y(Sax & Say & x≠y) -> ~(Bx & By))
If Alex sees two hats they are not both blue
21.  ∃x∃y∃z(Sxy & Sxz & y≠z & ((Ry & Rz) v (By & Rz)))
Someone sees either a red/red or blue/red distribution of hats

In other words, for Alex, Barbara, or Cherise, the only available distributions of colors are red and red, or blue and red. We have then matched the direct reasoning above with our axioms.

Of course, this is not the answer to the puzzle. To solve the puzzle we must infer Cherise knows what color hat she is wearing, i.e. Kcc (Exercise: Why won’t simply adding this fact to the axioms suffice?).

This step seems the trickiest for students. I suspect it is because moving forward in the solution requires indirect reasoning, i.e. assuming something for the sake of a contradiction. The stumbling block, however, often leaves them ready to abandon the puzzle. Don’t let the difficulty of the puzzle stand in the way…we’ll infer the solution next post. In the meantime, play around with the axiom set here. The syntax is readable by Prover9. All theorems were checked with this application. Models were checked with the bundled Mace4 finite model checker.

Square of Individuals

Suppose there are at least two distinct individuals, Alex and Bob, and that Alex is part of Bob.

Ground mereology has the two-place parthood relation as reflexive, i.e. everything is part of itself: the maximal part. The relation is not, however, symmetric as it is intuitively false that if x is part of y, then y is thereby part of x. There are at least two ways to reject symmetry: asymmetry or antisymmetry. On the former, any x part of y entails y is not part of x. On the latter, if x and y are parts of each other, they are identical. The first is too strong, since inconsistent in the presence of reflexivity:

1.      ∀x P(x,x)                                 Premise
2.      ∀x∀y P(x,y) -> ~P(y,x)          Premise
3.            SHOW   !                           DD
4.                  P(a,a)->~P(a,a)         2, ∀ Instantiation
5.                  P(a,a)                          1, ∀ Instantiation
6.                  ~P(a,a)                       4,5 MP
7.                    !                                 5,6 !

Ground mereology accepts instead the weaker antisymmetry. Additionally, parthood is taken to be transitive as it is plausible any part x of y which is part of z entails x is also part of z.

Useful definitions can be constructed from this characterization of parthood. Two individuals are said to overlap if they share a part in common, are discrete if they do not overlap, and an individual is said to overlap the complement of another, if the first shares a part with the complement of the second.

Remarks in hand, return to Alex and Bob, denoting the first with “a” and the second with “b”, and the parthood between them as “P(a,b)”. Observe, parthood entails overlap for these individuals:

1.      P(a,b)                                    Premise
2.      SHOW ∃x P(x,a) & P(x,b)     DD
3.             P(a,a)                             Reflexivity
4.             P(a,a) & P(a,b)               1,3, CI
5.             ∃x P(x,a) & P(x,b)          4, ∃ Introduction

Line 5 reflects that “a” and “b” share a part in common. Hence, if Alex is part of Bob, then Alex overlaps Bob. The converse does not hold (Exercise: Find a countermodel).

Observe next, to say Alex and Bob are discrete, is to deny they overlap. Equivalently, it is to claim they have no common parts. Moreover, we would be saying of, say, Alex, that Alex overlaps some part of the complement of Bob, and vice versa. For Alex to overlap Bob’s complement is for there to be a part of Alex that is not a part of Bob.

1.      ~∃x P(x,a) & P(x,b)                           Premise
2.           SHOW∃x P(x,a) & ~P(x,b)           DD
3.                 ∀x P(x,a) -> ~P(x,b)              1, Substitution
4.                 P(a,a)                                     Reflexivity
5.                 P(a,a) -> ~P(a,b)                   3, ∀ Instantiation
6.                 ~P(a,b)                                   4,5 MP
7.                 P(a,a) & ~P(a,b)                     4,6 CI
8.                 ∃x P(x,a) & ~P(x,b)                7, ∃ Introduction

Hence, if Alex is discrete from Bob, then Alex overlaps the complement of Bob. Since both overlap and complement overlap are symmetric, we can say the same for Bob (Exercise: Prove overlap and complement overlap are symmetric).

Our brief foray into the mereology wilderness permits, given the assumptions with which we began, a square of individuals (cp. Square of Opposition). As is well-known, blindly translating categorical sentences into first-order notation undermines logical relations of the traditional square, as classical logic permits conditionals which are vacuously true. We avoid the problem of existential import by sticking with individuals. Hence, our square parallels the tradition:

Implication holds whenever the first is true then the second must be, and if the second is false so must the first be. Contraries are sentences which may both be false but which may not both be true. Contradictory sentences require that if one is true the other is false, and if one is false the other is true. Subcontraries are sentences which may both be true, but which may not both be false (Exercise: Verify the remaining corners).

There’s ∃x about Mary

PUZZLE:
Larry is married but Nick isn’t. Larry is looking at Mary, and Mary is looking at Nick.

QUESTION:
Is someone married looking at someone not married?

CHOICES:

A.     Yes
B.     No
C.     Not enough information to answer

This puzzle was posed to me by a student (thanks Richard!) after class one morning. A cursory google suggests 80% choose incorrectly. I am skeptical; I’ve found no empirical evidence supporting this claim (I’d be interested if anyone else has). Be warned, searching for the puzzle will likely turn up solutions, so if you’d like to solve it, best settle here and reflect for a bit. I’ll wait. Once you are finished, check your answer by clicking it above. Afterwards, scroll down for a solution and some discussion.

SOLUTION:
The hallmark of a logical solution to a puzzle is that once presented with the solution, nearly everyone agrees it is correct. The Wason Selection Test is an example. Many choose incorrectly. Bentham reports a psychologist once confessed to him that nearly everyone accepts the standard solution as correct once it is explained [1]. The puzzle under discussion strikes me as logical in this sense. I’m curious if you agree.

I’ve translated the puzzle into a standard classical first-order language, where the predicate and relations symbols are obvious. It is straightforward to prove the solution (I’m using Hardegree’s natural deduction system in Symbolic Logic: A First Course). In symbols:

1.      Ml & ~Mn                                            Premise
2.      Llm & Lmn                                          Premise
3.      SHOW∃x∃y(Mx & ~My & Lxy)          ID
4.                   ~∃x∃y(Mx & ~My & Lxy)      AID
5.                    ∀x∀y(~Mx v My v ~Lxy)      4, Substitution
6.                    ~Ml v Mm v ~Llm                  5, Universal Instantiation
7.                     Mm                                         1,2,6 DE
8.                     ~Mm v Mn v ~Lmn               5, Universal Instantiation
9.                     ~Mm                                      1,2,6, DE
10.                     !                                            7,9 Contradiction

I’ve assumed as premises the information provided in the puzzle. On the SHOW line you’ll find a symbolization of the follow-up question. On line 4, I assume the negation of the SHOW line. On line 5, I substitute negated existential quantifiers for universal quantifiers trailed by negation, which is then distributed via De Morgan application. Since both variables in line 5 are under universal scope, I instantiate without restriction; in particular, to the constants denoting Larry and Mary. The result is line 6, which says Larry is unmarried, Mary is, and Larry is not looking at Mary. However, on line 1 we assumed Larry was married. Similarly, on line 2 we assumed Larry was looking at Mary. Hence, by disjunctive syllogisms, we infer Mary must be married on line 7. Instantiating line 5 once more, this time with Mary unmarried, Nick married, and Mary not looking at Nick, results by similar reasoning in Mary not being married on line 9. Since we already inferred that Mary was married, we now find ourselves in a contradiction. Hence, we infer someone is looking at someone not married.

Another way to think about the solution is to observe (or suppose what is plausible) that Mary is either married or not. If married, then since Mary is looking at Nick who is unmarried, someone is looking at someone who is unmarried. If unmarried, then since married Larry is looking at Mary, someone married is looking at someone unmarried. Either way, the answer is "Yes."

DISCUSSION:
Several of my students, when initially posed with the puzzle, claim there is “Not enough information to answer”. I take this to suggest students have trouble thinking to make certain plausible suppositions. To be clear, I do not think students have trouble making plausible suppositions and reasoning from them in general. They do, after all, deliberate about the future. Rather, it just doesn’t naturally occur to them to make even plausible suppositions in certain contexts, such as the context of puzzles and the contexts of proofs.

Works Cited
[1] Bentham, J. (2008). Logic and Reasoning: Do the Facts Matter? Studia Logica. 88:67-84