Asyllogistic Inference by c2e7V2

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									Asyllogistic Inference




     Kareem Khalifa
 Department of Philosophy
   Middlebury College
                 Overview
•   What are asyllogistic inferences?
•   Why do such inferences matter?
•   Translation tricks
•   Proofs
•   Exercises
 What are Asyllogistic Inferences?
• A syllogistic inference is any inference consisting
  only of singular propositions and/or quantified
  statements that can be turned into A, E, I, or O.
   – Recall A = All F’s are G’s; E = No F’s are G’s; I =
     Some F’s are G’s; O = Some F’s are not G’s.
• An asyllogistic inference is any inference
  containing at least one statement that is neither
  a singular proposition nor a quantified statement
  that can be turned A, E, I, or O.
        Why do they matter?
• Quick answer: because we use asyllogistic
  inferences all the time, e.g.,
  – Northern New England states are rural and
    have lax gun control laws.
  – Some northern New England states are
    Democratic strongholds.
  – So some rural states with lax gun control laws
    are Democratic strongholds.
      A teaser about proofs…
• There’s nothing new here!
  – You apply your same four rules (UI, UG, EI,
    EG) along with all the rules for propositional.
• But…

        BEWARE OF
       TRANSLATION!!!
          First Translation Trap
•  All F’s are either G or H. (All foxes are either
   gentle or they’re hungry).
1. Temptation: (x)(Fx  Gx) v (x)(Fx  Hx)
    –   Wrong!
2. Proper Translation: (x)(Fx  (Gx v Hx))
• Clearly, the English statement allows some
   foxes to be gentle but not hungry, so long as
   the rest are hungry.
• However, Translation 1 doesn’t allow this.
      Second Translation Trap
•  F’s and G’s are H. (Fools and goons are
   horrible people)
1. Temptation: (x)((Fx & Gx)  Hx)
• Wrong!
2. Correct: (x)((Fx v Gx)  Hx)
• The English statement clearly requires
   someone who is a fool but not also a goon to
   be horrible, but Translation 1 prohibits this.
• Ditto for goons who are not fools.
      Third Translation Trap
•  All except F’s are G’s. (Except for the
   French people, everyone was genuine.)
1. Temptation: (x)(Fx  ~Gx)
• Half Wrong!
2. Correct: (x)(Fx  ~Gx)
• Translation 1 permits non-French people
   to be not genuine. However, the English
   statement denies this.
        Returning to proofs…
• If you translate right, there’s no
  difference between asyllogistic and
  syllogistic inferences…
• Use EI and UI to turn the predicate logic
  proof into a propositional logic proof, prove
  it accordingly, and then use EG and UG to
  turn it back into predicate logic.
                    Example
•   Except for the French speakers, everyone was
    genuine. None of the Swiss students were genuine. So
    the Swiss students speak French.
1. (x)(Fx  ~Gx)                             A
2. (x)(Sx  ~Gx)                             A
(x)(Sx  Fx)
3. Fa  ~Ga                                  1 UI
4. Sa  ~Ga                                  2 UI
5. ~Ga  Fa                                  3 E
6. Sa  Fa                                   4, 5 HS
7. (x)(Sx  Fx)                              6 UG
     Sample Exercises, p. 473
• A3: No car is safe unless it has good
  brakes.
  – (x)((Cx  (~Sx v Bx))
• A5: A gladiator wins if and only if he is
  lucky.
  – (x)(Gx -> (Wx  Lx))
• A6: A boxer who wins if and only if he is
  lucky is not skillful.
  – (x)((Bx & (Wx  Lx))  ~Sx)
           More sample exercises
• A8: Not all tools that are cheap are either soft or
  breakable.
     – (x)((Tx & Cx) & (~Sx & ~Bx))
• A11: In America, everything is permitted that is not
  forbidden. In Germany, everything is forbidden that is not
  permitted. In France, everything is permitted even if it’s
  forbidden. In Russia, everything is forbidden even if it’s
  permitted.
     – This says everything you would need:
•   (x) ((((Ax & ~Nx)  Px) & ((Gx & ~Px)  Nx)) & ((Fx  Px) & (Rx  Nx)))
     – But if you want to get a more literal translation of the “even if”
       statements, you could conjoin the following:
• (x) (((Fx & Nx)  Px) & ((Rx & Px)  Nx))

								
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