Logic Exercise: Arthur Conan Doyle, "The Sign of the Four"

 Arthur Conan Doyle, "The Sign of the Four" (1890):

"How came he, then?" I reiterated. "The door is locked, the window is inaccessible. Was it through the chimney?"

"The grate is much too small," he answered. "I had already considered that possibility."

"How then?" I persisted.

"You will not apply my precept," he said, shaking his head. "How often have I said to you that when you have eliminated the impossible whatever remains, however improbable, must be the truth? We know that he did not come through the door, the window, or the chimney. We also know that he could not have been concealed in the room, as there is no concealment possible. Whence, then, did he come?"

"He came through the hole in the roof!" I cried.

Let the four hypotheses—the came in through the door, window, chimney and hole in the roof—be represented as (respectively): Hd, Hw, Hc and Hh. Let the a priori probabilities of the first three hypotheses be much more likely than the forth—say,

            P(Hd) = P(Hw) = P(Hc) = 33/100

            P(Hh) = 1/100

Now suppose the first three hypotheses are, as Holmes says, “impossible”—i.e., subsequent evidence is incompatible with the first three hypotheses (but compatible with the last hypothesis):

            P(Hd & E) = P(Hw & E) = P(Hc & E) = 0

            P(Hh & E) ≠ 0

Part (a):

Show that P(E|Hd) = P(E|Hw) = P(E|Hc) = 0

Part (b):

Show that, on the evidence, the last (a priori) “improbable” hypothesis “must be the truth”—i.e., P(Hh|E) = 1

Solution:

Part (a):

By the General Conjunction Rule of probability:

0 = P(Hd & E) = P(Hw & E) = P(Hc & E)

                = P(E|Hd)P(Hd) = P(E|Hw)P(Hw) = P(E|Hc)P(Hc)

                = P(E|Hd)(.33) = P(E|Hw)(.33) = P(E|Hc)(.33)

                = P(E|Hd) = P(E|Hw) = P(E|Hc),

as claimed.

Part (b):

By Bayes’ Rule:

            P(Hh|E) =

                                                P(E|Hh) x P(Hh)                                                       

                         P(E|Hh) x P(Hh) + P(E|Hd) x P(Hd) + P(E|Hw) x P(Hw) + P(E|Hw) x P(Hw)

But by Part (a), the latter three summands are zero, so

P(Hh|E) = P(E|Hh) x P(Hh) / P(E|Hh) x P(Hh) = 1. It "must be the truth," as Holmes claims.

Logic Exercise: Sherlock Holmes vs. Vincent Bugliosi

“Circumstantial evidence is a very tricky thing. It may seem to point very straight to one thing, but if you shift your own point of view a little, you may find it pointing in an equally uncompromising manner to something different” (Sherlock Holmes, “The Boscombe Valley Mystery”)

Let:      Cx =     x is circumstantial evidence

Ax =     x is ambiguous evidence (“if you shift your point of view, [the evidence can point to something different”)

Represent Holmes’ claim as:

(H)       All circumstantial evidence is ambiguous.

“For those who feel a case based on circumstantial evidence is, by definition, not a strong one, let me correct a common misperception,” writes former Los Angeles District Attorney Vincent Bugliosi. “Circumstantial evidence has erroneously come to be associated in the public mind and vernacular with an anemic case….But nothing could be further from the truth. In fact, most first degree murder cases are based on circumstantial evidence. This is so because other than eye-witness testimony (and in some jurisdictions, a confession), which is direct evidence, all other evidence, even fingerprints and DNA, is circumstantial evidence.” (Vincent Bugliosi, The Prosecution of George W. Bush for Murder [Vanguard Press: New York, 2008], page 100)

Let:      Dx =     x is direct evidence

            Ex  =    x is eyewitness testimony

            Ox  =    x is a confession

            d = DNA evidence

Can Bugliosi’s line of thought support a counter-argument to Holmes’s claim? Represent the following argument and prove its validity. Evaluate the argument.

(1) Evidence that’s neither eyewitness testimony nor a confession is not direct evidence.

(2) If something’s not direct evidence, then it’s circumstantial evidence.

(3) DNA evidence is not a confession and it’s not eyewitness testimony.

(4) DNA evidence is unambiguous

Therefore, Not all circumstantial evidence is ambiguous

 Solution:

(3) Sherlock Holmes vs. Vincent Bugliosi

(H)      (∀x)(Cx → Ax)

(1)      (∀x)[~(Ex v Ox) → ~Dx)     Premise

(2)      (∀x)(~Dx → Cx)                    Premise

(3)      ~Od & ~Ed                             Premise

(4)      ~Ad                                         Premise

:. ~(∀x)(Cx → Ax)

(5)                  (∀x)(Cx → Ax)           Assumption for RAA

(6)                  ~(Ed v Od)                 DeMorgan’s Law

(7)                  ~(Ed v Od) → ~Dd    1, ∀-Elimination

(8)                  ~Dd                            6, 7 Modus Ponens

(9)                  ~Dd → Cd                   2, ∀-Elimination

(10)                Cd                                8, 9 Modus Ponens

(11)                Cd → Ad                      5, ∀-Elimination

(12)                Ad                               10, 11 Modus Ponens

(13)                Ad & ~Ad                   4, 12 & Introduction

(14) ~(∀x)(Cx → Ax)                       5-13 RAA