Logic Exericse:

(1) There are three people in the story: The sister, her nephew and Doc Brady

(2) Someone in the story sent the letter to the sister

(3) Anyone who sent the letter to the sister had a motive

(4) But the nephew didn’t have a motive—and neither did Doc Brady

:. The sister send the letter to herself

Let:      s = the sister n = the nephew d = Doc Brady

            Sx = x is in the story

            Exy = x sent the letter to y

            Mx = x had a motive

Represent the argument and prove the conclusion.

(1)       (Ss & Sn & Sd) & (∀x)(Sx → (x=s v x=n v x=d)

(2)       (∃x)(Sx & Exs)

(3)       (∀x)[(Exs → Mx]

(4)       ~Mn & ~Md

:. Ess

(5)      Sa* & Ea*s                                         2, ∃-Elimination

(6)       Sa*                                                      5, &-Elimination

(7)       (∀x)(Sx → (x=s v x=n v x=d)          1, &-Elimination

(8)      Sa* → (a*=s v a*=n v a*=d)           7, ∀-Elimination

(9)      (a*=s v a*=n v a*=d)                      6, 8 Modus Ponens

(10)                a*=n                                       Assumption for RAA

(11)                Ea*s                                        5, &-Elimination

(12)                Ens                                          10, 11 =Elimination

(13)                Ens → Mn                               3, ∀-Elimination

(14)                Mn                                           12, 13 Modus Ponens

(15)                ~Mn                                        4, &-Elimination

(16)                Mn & ~Mn                              14, 15 &-Introduction

(17)    ~(a*=n)                                             10-16 RAA

(18)    (a*=s v a*=d)                                    9, 17 Disjunctive Syllogism

(19)                a* = d                                     Assumption for RAA

(20)                Ea*s                                        5, &-Elimination

(21)                Eds                                          19, 20 =Elimination

(22)                Eds → Md                               3, ∀-Elimination

(23)                Md                                           21, 22 Modus Ponens

(24)                ~Md                                        4, &-Elimination

(25)                Md & ~Md                              23, 24 &-Introduction

(26)    ~(a*=d)                                             19-25 RAA

(27)    a*=s                                                    18, 26 Disjunctive Syllogism

(28)    Ess                                                      11, 27 =Elimination

Logic Exercise: Inspired by Murder, She Wrote, “Capitol Offense” (1985)

[1] If the suspect killed the victim, then there would be blood and makeup on his hands. [2] But if the suspect killed the victim, then he also moved the victim. [3] The suspect moved the victim’s corpse only if he touched the victim’s shirt. [4] But if the suspect touched the victim’s shirt, then anything on the suspect’s hands would be on the victim’s shirt. [5] But there’s no makeup on the victim’s shirt. So the suspect didn’t kill the victim

Let:      s = the suspect, h = the suspect’s hands, i = the victim’s shirt

Bx = x is blood

Kx = x killed the victim

            Mx = x is makeup

            Vx  = x moved the victim

            Oxy = x is on y

            Txy = x touched y

Represent the argument and prove the conclusion.

Solution:

[1] If the suspect killed the victim, then there would be blood and makeup on his hands. [2] But if the suspect killed the victim, then he also moved the victim. [3] The suspect moved the victim’s corpse only if he touched the victim’s shirt. [4] But if the suspect touched the victim’s shirt, then anything on the suspect’s hands would be on the victim’s shirt. [5] But there’s no makeup on the victim’s shirt. So the suspect didn’t kill the victim

(1)       Ks → (∃x)(∃y)(Bx & My & Oxh & Oyh)       Premise

(2)      Ks → Vs                                                           Premise

(3)      Vs → Tsi                                                          Premise

(4)      Tsi → (∀x)(Oxh → Oxi)                                 Premise

(5)      ~(∃x)(Mx & Oxi)                                           Premise

:. ~Ks

(6)                   Ks                                                        Assumption for RAA

(7)                  (∃x)(∃y)(Bx & My & Oxh & Oyh)     1, 6 Modus Ponens

(8)                  (∃y)(Ba* & My & Oa*h & Oyh)         7, ∃-Elimination

(9)                  Ba* & Mb* & Oa*h & Ob*h                8, ∃-Elimination

(10)                Vs                                                        2, 6 Modus Ponens

(11)                Tsi                                                       3, 10 Modus Ponens

(12)                (∀x)(Oxh → Oxi)                                4, 11 Modus Ponens

(13)                Ob*h                                                   9, &-Elimination

(14)                Ob*h → Ob*i                                       12, ∀-Elimination

(15)                Ob*i                                                     13, 14 Modus Ponens

(16)                Mb*                                                     9, &-Elimination

(17)                Mb* & Ob*i                                         15, 16 &-Introduction

(18)                (∃x)(Mx & Oxi)                                  17, ∃-Introduction

(19)                (∃x)(Mx & Oxi) & ~(∃x)(Mx & Oxi)            5. 18 &-Introduction

(20)    ~Ks                                                                 6-19 RAA

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.