Lecture Notes, 4 Oct 2010

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§1.5: Rules of Inference (cont'd)

1. Prove that this argument is valid:
   P1:   Lynn works part time or full time.
   P2:   If Lynn doesn't play on the team,
               then she doesn't work part time.
   P3:   If Lynn plays on the team,
               then she's busy.
   P4:   Lynn doesn't work full time.
   C :   ∴ Lynn is busy.

   1. Syntax & semantics of representation of atomic propositions:

      Let pt = Lynn works part time.
      Let ft = Lynn works full time.
      Let plays = Lynn plays on the team.
      Let busy = Lynn is busy.
      

   2. Translation of argument:

      P1:    pt &or ft
      P2:    ¬plays → ¬pt
      P3:    plays → busy
      P4:    ¬ft
      C :    ∴ busy

   3. We want to prove that this is a valid argument
      • without using truth tables

   4. Strategy:

      Want to show:    busy (i.e., C)

      Can derive C from P3 via MP if we can show: plays
      Can derive "plays" from P2 via MT if we can show: pt
      Can derive "pt" from P1 via DS if we can show: ¬ft
      But we have ¬ft = P4

      So, "unwind" this strategy to write the proof:

      start with "¬ft"
      use that & DS on P1 to derive "pt"
      use that & MT on P2 to derive "plays"
      use that & MP on P3 to derive "busy"
      QED!

   5. Here's the proof:

      1.	¬ft	:	P4
      2.	pt ∨ ft	:	P1
      3.	pt	:	1,2; DS
      4.	¬plays → ¬pt	:	P2
      5.	plays	:	3,4; MT
      6.	plays → busy	:	P3
      7.	busy	:	5,6; MP

   ---

2. Digression: Was that really a legitimate use of MT?
   1. After all, MT says:

      From:                    A → B
      &:                       ¬B
      you may infer:    ¬A

   2. But the application above seems be of the form:

      From:               ¬A → ¬B
      &:                       B
      you may infer:    A

   3. Reply: We can handle this in several ways:
      1. We could say (as I implicitly did above) that both of these are MT:
         • After all, at their core, they follow the "same" pattern:
           • From a conditional and the "opposite" of its consequent,
             you may infer the "opposite" of its antecedent.

         • But how would a computer know that?

      2. We could say that there are two versions of MT,
         namely, the two in IIA and IIB, above.
         • That's another way to interpret what I did in the example.

      3. We could prove that version B is derivable from version A; here's how:
         1. ¬A → ¬B : premise
         2. B               : premise
         3. ¬¬B           : b; DN+SL
            • (i.e., line c follows from line b by Double Negation
              (which tells us that they are logically equivalent)
              together with Shakespeare's Law
              (which tells us that we can replace one with the other))

         4. ¬¬A           : a,c; MT (version A)
         5. A               : d; DN+SL

         Now that we have this proof, we can appeal to MT (version B) in the future.

         • It's like a "macro" in a computer program;
           we can "expand" the macro whenever we need it.

   ---

3. Is This Argument Valid?
   P1:   I play golf or tennis.
   P2:   If it's not Sunday, I play golf and tennis.
   P3:   If it's Saturday or Sunday, I don't play golf.
   C :   ∴ I don't play golf.
   1. Syntax & semantics of representation of atomic propositions:
      g = "I play golf"
      t = "I play tennis"
      sun = "It's Sunday"
      sat = "it's Saturday"

   2. Translation of argument:
      P1:  (g ∨ t)
      P2:  (¬sun → (g ∧ t))
      P3:  ((sat ∨ sun) → ¬g)
      C :  ¬g

   3. Strategy:
      1. Try to show ¬g (i.e., C)
         Can derive C from P3 via MP, if we can show (sat ∨ sun)
         Can derive (sat ∨ sun) from "sun" via Addition, if we can show "sun"
         Can derive "sun" from ¬(g ∧ t) via MT, if we can show ¬(g ∧ t)
         Can derive ¬(g ∧ t) from (¬g ∨ ¬t) via DM+SL,
         if we can show (¬g ∨ ¬t)
         Can derive (¬g ∨ ¬t) either from ¬g
         or from ¬t via Addition.
         Can we derive either one of those?

      2. Deriving ¬g won't help; that's what we're trying to do in the first place!
         (This is called "begging the question" or arguing in a circle.)

      3. Can we derive ¬t?
         • P1 contains "t", but there's no rule that would allow us to derive ¬t from P1
         • P2 contains "t", but there's no rule that would allow us to derive ¬t from P2

         • So we probably can't derive ¬t.

      4. So, let's revert to truth tables to see if the argument is valid or not.
         • This exercise is left to the reader :-)
         • Construct a truth table with 2⁴=16 rows
           & find a row with T premises but F conclusion!
           • (The answer is online at the book's website!)

         • Hint: Make
           tval(g) = T
           tval(sat) = tval(sun) = F
           tval(t) = T
           • e.g., it's Monday & I play golf and tennis.

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