Thursday 29 September 2011

Donkey sentences

Time was I would pore over donkey sentences not to work out what quantification told me about the number of donkeys involved but to understand what puzzled logicians so much about these sentences. Frankly I couldn’t see what the fuss was about. Donkey sentences, by the way, are sentences of the type:

If a farmer owns a donkey, he beats it.

A farmer who owns a donkey beats it.

Time has dimmed my memory of all the logical nuances working themselves out in this type of sentences, but basically – I think – the problem presented itself thus: we imagine a farmer and his donkey – that is “any” farmer and one donkey per farmer. Next, we imagine any or every donkey-owning farmer, except that this time we add another donkey, or more donkeys, to his stable. Does the singular reference of ‘it’ still capture the plurality of donkeys?

I don’t know. One moment I think it does, the next it doesn’t. My problem was of a different kind: how to quantify over “farmer” and “donkey” if I think of them as definite entities? I just couldn’t get my head around the overlapping scopes. In other words, how to symbolize a sentence when I think of one definite farmer and one definite donkey, the donkey that he owns and beats:

If the farmer owns the donkey, he beats it.

The farmer who owns the donkey beats it.

I put pen to paper the other day and came up with this:

(∃x)(∃y)[Fx • Dy • (z)(Dz ⊃z=y) • (w)(Fw ⊃w = x) • (Oxy ⊃Bxy)]

(∃x)(∃y){Fx • Dy • Oxy (z)(Dz ⊃z=y) • (w)[(Fw • Owy) ⊃w = x) • Bxy]}

At the moment, I can’t see what is wrong with it. The symbolization can even accommodate a non-defining ‘who owns the donkey’ clause: "The farmer, who owns the donkey, beats it."

(∃x)(∃y)[Fx • Dy • (z)(Dz ⊃z=y) • (w)(Fw ⊃w = x) • Oxy Bxy]

So, will someone remind me what my problem was about? Too much cheese before bedtime? Translating natural language into the language of symbolic logic is a skill that must be practiced daily. It goes rusty really fast.

Wednesday 28 September 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 1(a), p. 571

Show that the following: a = b ⊃ b = a is a theorem:
  1. * a = b ......... ACP
  2. * b = a ......... Id
  3. a = b ⊃ b = a ......... 1-3 CP

Wednesday 21 September 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 3(a), p. 572

We have to show that: (x)[x = x ¬ (x = x)] is a theorem. There are a number of ways we could go. I choose this:
  1. * ¬ (x = x) ......... ACP
  2. * ¬ (x = x) ∨ x = x ......... Add.
  3. * ¬ (x = x) ......... 1,2 DS
  4. ¬ (x = x) ¬ (x = x) ......... 1-3 CP
  5. ¬ ¬ (x = x) ∨¬ (x = x) ......... 4 MI
  6. x = x ∨¬ (x = x) ......... 5 DN
  7. (x)[x = x ¬ (x = x)] ......... 6 UG

Thursday 15 September 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 2(a), p. 572

As usual, we prove the validity of the argument.
  1. a = b ¬ Bab
  2. ¬ (x)Bxx
  3. a = b ......... 1 Simp.
  4. ¬ Bab ......... 1 Simp.
  5. ¬ Baa ......... 3,4 Id
  6. (∃x)¬ Bxx ......... 5 EG
  7. ¬ (x)Bxx ......... 6 QC

Thursday 8 September 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 2(c), p. 572

The task - to show that the argument is valid:
  1. (z)[Gz ⊃(y)(Ky ⊃ Hzy)]
  2. (Ki • Gj) • i = j
  3. ∴ Hii
  4. Gj ......... 2 Simp.
  5. Gj ⊃(y)(Ky ⊃ Hjy) ......... 1 UI z/j
  6. (y)(Ky ⊃ Hjy) ......... 4,5 MP
  7. Ki ......... 2 Simp.
  8. Ki ⊃Hji ......... 6 UI y/i
  9. Hji ......... 7,8 MP
  10. i = j ......... 2 Simp.
  11. Hii ......... 9,10 Id

Thursday 1 September 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 2(e), p. 572

Show that the argument is valid:
  1. a = b
  2. ∴Ka ¬ Kb
  3. * Kb ......... ACP
  4. * Ka .......... 1,3 Id
  5. Kb ⊃Ka ......... 3-4 CP
  6. ¬ Kb ∨ Ka ......... 5 MI
  7. Ka ¬ Kb ......... 6 Comm.