Friday, 9 April 2010

Deduction, Daniel Bonevac, Blackwell Publishing, 2nd edition, 2003, Ex. 8.3, problem 13

As elsewhere in this set of problems, the task is to show that the formula (∃x)(y)(x = y ≡ Gy) implies the conclusion of the argument reproduced below. Since the conclusion itself is a biconditional, we show first that the formula on the left implies the formula on the right, and then the reverse. We are free to use the same constants from line 18 on as in the first assumption because that assumption has been discharged on line 17.
  1. (∃x)(y)(x = y ≡ Gy)
  2. ∴(∃x)(Gx • Fxx) ≡ (∃x)(∃y)(Gx • Gy • Fxy)
  3. * (∃x)(Gx • Fxx) / ACP
  4. * Ga • Faa / 3EI x/a
  5. * (y)(h = y ≡ Gy) / 1EI x/h
  6. * h = a ≡ Ga / 5UI y/a
  7. * (h = a ⊃ Ga) • (Ga ⊃h = a) / 6BE
  8. * Ga / 4Simp.
  9. * Ga ⊃h = a / 8Simp.
  10. * h = a / 8,9MP
  11. * Faa / 4Simp.
  12. * Fah / 10,11Id
  13. * Gh / 10,8Id
  14. * Ga • Gh •Fah / 9,12,13Conj.
  15. * (∃y)(Ga • Gy • Fay) / 14EG
  16. * (∃x)(∃y)(Gx • Gy • Fxy) / 15EG
  17. (∃x)(Gx • Fxx) ⊃(∃x)(∃y)(Gx • Gy • Fxy) / 3-16CP
  18. * (∃x)(∃y)(Gx • Gy • Fxy) / ACP
  19. * (∃y)(Ga • Gy • Fay) / 18EI x/a
  20. * Ga • Gm • Fam / 19EI y/m
  21. * (y)(h = y ≡ Gy) / 1EI x/h
  22. * h = a ≡ Ga / 21UI y/a
  23. * (h = a ⊃ Ga) • (Ga ⊃h = a) / 22BE
  24. * Ga ⊃h = a / 23Simp.
  25. * Ga / 20Simp.
  26. * h = a / 24,25MP
  27. * h = m ≡ Gm / 21UI y/m
  28. * (h = m ⊃ Gm) • (Gm ⊃h = m) / 27BE
  29. * Gm / 20Simp.
  30. * Gm ⊃h = m / 28Simp.
  31. * h = m / 29,30MP
  32. * m = h / 31Id
  33. * m = a / 32,26Id
  34. * Fam / 20Simp.
  35. * Faa / 33,34Id
  36. * Ga • Faa / 25,35Conj.
  37. * (∃x)(Gx • Fxx) / 36EG
  38. (∃x)(∃y)(Gx • Gy • Fxy) ⊃ (∃x)(Gx • Fxx) / 18-37CP
  39. [(∃x)(Gx • Fxx) ⊃(∃x)(∃y)(Gx • Gy • Fxy)] • [(∃x)(∃y)(Gx • Gy • Fxy) ⊃ (∃x)(Gx • Fxx)] / 17,38Conj.
  40. (∃x)(Gx • Fxx) ≡ (∃x)(∃y)(Gx • Gy • Fxy) / 39BE

Monday, 5 April 2010

Which way is forward?

When the clocks change I always entertain myself for a few days with the idea of consistency as we apply it in our efforts to describe what is actually happening to time. Consistency is at the heart of logic. Strictly speaking, a set of statements is consistent if and only if it is possible for all the statements to be true.

Assigning values of true or false takes place against a set of criteria, and in order to make sense of our comparisons we assign the same criteria each time we compare similar things. If my criterion for ‘bright light’ is ‘sufficient to illuminate my desk to work at’, then it is false that a candle is bright and true that a 100W bulb is bright. Change the criteria by substituting ‘football pitch’ for ‘my desk’, and both sentences are false.

What are we to make of the following English sentences then?

The clocks go forward in the spring by an hour.
We’ve brought the meeting forward by an hour.

Does ‘forward’ suggest a movement from an earlier to a later time or from a later time to an earlier time? How important is our perceived position relative to the change? I can best describe what I do when I advance the hands of my clocks as ‘turning them forward’, so that what was 7am is now 8am. By this logic the first sentence is true. By the same logic the second sentence is false, because in bringing the meeting forward we are moving from a later time to an earlier time.

There are a number of inconsistencies like this in English, many of them only apparent. This problem, for example, is only superficially similar to that of ‘an alarm going off’, where, if one were to choose a sense of termination or completion as a criterion for ‘off’, or the shutting off of operation of electric devices, one would be led to believe that the alarm has just stopped ringing. The facts are that the verb ‘to go off’ is much older than commercial electricity and that it was traditionally used to describe bursts of energy, explosions, and so on.

Coming back to ‘forward’ and ‘back’ (consider that to move a meeting back by an hour is to postpone it by that amount of time), I have aged by an hour by putting my clocks forward. Fortunately this loss will be reversed in the autumn, but is aging itself a movement forward or back? On McTaggart’s time line, whereby an event which is now present, was future, and will be past, aging is a backward movement. Or is it something to do with another one of our dearly held conventions, namely that the forward movement is a movement from left to right?

In an item heard last week scientists have tested our intuitions again by endowing ‘forward’ with the meaning it has in the sentence about rescheduling meetings rather than changing the clocks:

The seasons are moving forward across the northern hemisphere.

Thursday, 1 April 2010

Deduction, Daniel Bonevac, Blackwell Publishing, 2nd edition, 2003, Ex. 8.3, problem 12

Show that the given statement is a consequence of a formula. To show that it is so, we construct an argument with the first statement as a premise and the second as a conclusion. Two things worth noting about this proof. One: 'y' from line 4 is instantiated to 'a' on line 5 by universal instantiation even though, or perhaps because, we have already instantiated into 'a' by existential instantiation on line 4. Two: this allows us to set up a biconditional which can be easily simplified and the remainder dispatched by Modus Ponens after having first introduced the identity a = a.
  1. (∃x)(y)(y = x ≡ Gy)
  2. ∴(x)Fx ⊃(∃x)(Gx • Fx)
  3. * (x)Fx / ACP
  4. * (y)(y = a ≡ Gy) / 1EI x/a
  5. * a = a ≡ Ga / 4UI y/a
  6. * (a = a ⊃Ga) • (Ga ⊃a = a) / 5BE
  7. * a = a ⊃Ga / 6Simp.
  8. * a = a / Id
  9. * Ga / 8,7MP
  10. * Fa / 3UI x/a
  11. * Ga • Fa / 9,10Conj.
  12. * (∃x)(Gx • Fx) / 11EG
  13. (x)Fx ⊃(∃x)(Gx • Fx) / 3-12CP

Friday, 26 March 2010

Symbolic Logic, Dale Jacquette, Wadsworth, 2001, Chapter 8, Ex. III, problem 11

Prove the theorem. There is more than one way of doing it. If we go past the contradiction on line 16, we can bring in just about any statement, including (x)Fxx, because anything follows from a contradiction.
(x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] ⊃(x)Fxx
  1. * (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] / ACP
  2. * * ¬ (x)Fxx / AIP
  3. * * (∃x) ¬ Fxx / 2QC
  4. * * ¬ Faa / 3EI x/a
  5. * * (y)[Fay ≡ (z)(Gaz ≡ Gyz)] / 1UI x/a
  6. * * Faa ≡ (z)(Gaz ≡ Gaz) / 5UI y/a
  7. * * [Faa ⊃(z)(Gaz ≡ Gaz)] • [(z)(Gaz ≡ Gaz) ⊃Faa] / 6BE
  8. * * (z)(Gaz ≡ Gaz) ⊃Faa / 7Simp.
  9. * * ¬ (z)(Gaz ≡ Gaz) / 4,8MT
  10. * * (∃z) ¬ (Gaz ≡ Gaz) / 9QC
  11. * * (∃z) ¬ [(Gaz ⊃ Gaz) • [(Gaz ⊃ Gaz)] / 10BE
  12. * * ¬ [(Gam ⊃ Gam) • [(Gam ⊃ Gam)] / 11EI z/m
  13. * * ¬ (Gam ⊃ Gam) ∨ ¬ [(Gam ⊃ Gam)] / 12DeM
  14. * * ¬ (Gam ⊃ Gam) / 13Tuatology
  15. * * ¬ (¬ Gam ∨ Gam) / 14MI
  16. * * Gam • ¬ Gam / 15DeM
  17. * * Gam / 16Simp.
  18. * * Gam ∨¬ (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] / 17Add.
  19. * * ¬ Gam / 16Simp.
  20. * * ¬ (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] / 19,18DS
  21. * ¬ (x)Fxx ⊃¬ (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] / 2-20IP
  22. * (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] ⊃(x)Fxx / 21Contrap.
  23. * (x)Fxx / 1,22MP
  24. (x)(y)[Fxy ≡ (z)(Gxz ≡ Gyz)] ⊃(x)Fxx / 1-23CP

Sunday, 21 March 2010

Definite Donkey

Supply exceeds demand in research related to donkey sentences, and posts like this only drive the price further down. So, I will say this: I have no answers of my own to the intractable donkey sentence problem, but I also don’t share the dilemmas which others readily admit. Instead, I have my own donkey-sentence-related dilemmas.

In a nutshell, donkey sentences are like (1):

(1) If a farmer owns a donkey, he beats it.

The pronouns ‘he’ and ‘it’ are anaphoric on ‘farmer’ and ‘donkey’, and the latter pair especially has polarized philosophers, logicians and linguists. Is the sentence true of, say, ten farmers, each of whom has one donkey which he beats, or is it true of ten farmers, each of whom has an unspecified number of donkeys of which he beats every single donkey that he owns.

By convention, first-order logic follows the second approach. It is convenient because the use of a universal quantifier allows us to bind the variable ‘y’ in the predicate expression ‘Bxy’.

(x){Fx ⊃ (y)[(Dy • Oxy) ⊃Bxy]}

It suits me. The indefinite article ‘a’ in English is called so for a reason. It is intended to capture both ‘one’ and ‘every’ of the items mentioned. It could be argued though that this strategy distorts the meaning (truth) of some sentences, such as:

(2) If a gentleman fancies a lady, he marries her.

I can live with it. It is not the only place where logic and language come apart. I tend to agree with Quine: the logic is clear, it is the language that is ambiguous. Where the problem becomes interesting for me is the preservation or loss of equivalence upon conversion of conditional sentences into sentences containing relative clauses as we go from indefinite to definite descriptions, as can be seen from the following pairs of sentences:

(3a) If a gentleman compliments a lady, he fancies her.
(3b) A gentleman who compliments a lady fancies her.

(4a) If a politician criticizes the queen, he envies her.
(4b) A politician who criticizes the queen envies her.

(5a) If the lady fancies a gentleman, she humours him.
(5b) The lady who fancies a gentleman humours him. *

(5a) If the dog sees the postman, it chases him.
(5b) The dog that sees the postman chases him. *

The sentences sound plausible at first, but less so as we move towards the last pair. (I have deliberately kept the Simple Present tense throughout to enable a same-basis comparison.) Sentences (3a) and (3b) are equivalent, with ‘a gentleman’ taking wide scope. Sentences (5a) and (5b), on the other hand, present the problem of scope. If, as in the symbolization of (1), we quantify universally over ‘the postman’ to ensure that all the variables are bound, we will shoot ourselves in the foot over ‘the postman’ being a definite description.

(5a) does not convert elegantly into (5b) in English, and I wonder if it is the question of logic or the question of semantics. If we substitute the verb ‘hate’ for the verb ‘see’, the difficulty works in the opposite direction.

(6a) If the dog hates the postman, it chases him. *
(6b) The dog that hates the postman chases him.

(6b) can be paraphrased as:

(6c) The dog chases the postman it hates.

This can be captured straightforwardly as:

(∃x){Px • (y)(Py ⊃ y = x) • (∃z){Dz • Hzx • (w)[(Dw • Hwx) ⊃w = z] • Czx}}

Plug in proper names, and it is (5a) in turn that becomes easiest to handle (‘If Razor sees Jasper, it chases him,’ is Srj ⊃Crj). Double anaphora coupled with definite descriptions is a mind-bender, and I wonder if there is a systematic treatment that ties all the ends.

Symbolic Logic, Dale Jacquette, Wadsworth, 2001, Chapter 8, Ex. III, problem 9

Show that the following disjunction is a tautology. A disjunction is often easiest to prove by assuming the negation of one of the disjuncts. In this case it is also sensible to write the other disjunct at the bottom of the page and work backwards by simplifying, until we can see what it is that we are aiming to get. Then we make further assumptions: (x)Gx on line 8 and (x)Fx on line 17.

(∃x)(Fx ≡ ¬ Gx) ∨[(∃x) ¬ Fx ≡ (∃x) ¬ Gx]
  1. * ¬ (∃x)(Fx ≡ ¬ Gx) / ACP
  2. * (x) ¬ (Fx ≡ ¬ Gx) / 1QC
  3. * (x) ¬ [(Fx ⊃¬ Gx) • (¬ Gx ⊃Fx)] / 2BE
  4. * ¬ [(Fx ⊃¬ Gx) • (¬ Gx ⊃Fx)] / 3UI x/x
  5. * ¬ (Fx ⊃¬ Gx) ∨¬ (¬ Gx ⊃Fx) / 4DeM
  6. * (Fx ⊃¬ Gx) ⊃(¬ Gx • ¬ x) / 5MI
  7. * (Fx ⊃¬ Gx) ⊃¬ (Gx ∨Fx) / 6DeM
  8. * * (x)Gx / ACP
  9. * * Gx / 8UI x/x
  10. * * Gx ∨Fx / 9Add.
  11. * * ¬ (Fx ⊃¬ Gx) / 10,7MT
  12. * * ¬ (¬ Fx ∨¬ Gx) / 11MI
  13. * * Fx • Gx / 12DeM
  14. * * Fx / 13 Simp.
  15. * * (x)Fx / 14UG
  16. * (x)Gx ⊃(x)Fx / 8-15CP
  17. * * (x)Fx /ACP
  18. * * Fx / 17UI x/x
  19. * * Fx ∨Gx / 18Add.
  20. * * Gx ∨Fx / 19Comm.
  21. * * ¬ (Fx ⊃¬ Gx) / 20,7MT
  22. * * ¬ (¬ Fx ∨¬ Gx) / 21MI
  23. * * Fx • Gx / 22DeM
  24. * * Gx / 23Simp.
  25. * * (x)Gx / 24UG
  26. * (x)Fx ⊃(x)Gx / 17-25CP
  27. * ¬ (x)Fx ⊃¬ (x)Gx / 16Contrap.
  28. * (∃) ¬ Fx ⊃(∃x) ¬ Gx / 27QC
  29. * ¬ (x)Gx ⊃¬ (x)Fx / 26 Contrap.
  30. * (∃) ¬ Gx ⊃(∃x) ¬ Fx / 29QC
  31. * [(∃) ¬ Fx ⊃(∃x) ¬ Gx] • [ (∃) ¬ Gx ⊃(∃x) ¬ Fx] / 28,30Contrap.
  32. * (∃x) ¬ Fx ≡ (∃x) ¬ Gx / 31BE
  33. ¬ (∃x)(Fx ≡ ¬ Gx) ⊃[(∃x) ¬ Fx ≡ (∃x) ¬ Gx] / 1-32CP
  34. ¬ ¬ (∃x)(Fx ≡ ¬ Gx) ∨[(∃x) ¬ Fx ≡ (∃x) ¬ Gx] / 33MI
  35. (∃x)(Fx ≡ ¬ Gx) ∨[(∃x) ¬ Fx ≡ (∃x) ¬ Gx] / 34DN

Saturday, 13 March 2010

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 2008, 5th edition, Unit 20, Ex. 3.e

We are asked to prove a theorem. Since the theorem is an equivalence we have to prove each implication separately and then conjoin them. Proving it left to right is not especially challenging, but proving it in the opposite direction requires that we instantiate (y)(Fy ⊃m = y) twice, first into 'y', then into 'x', so that we could set up Modus Ponens arguments inside the scope of the innermost assumption and derive an identity.
{(∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y]} ≡ (∃x)[Fx • (y)(Fy ⊃x = y)]
  1. * (∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y] / ACP
  2. * (∃x)Fx / 1Simp.
  3. * Fa / 2EI x/a
  4. * (x)(y)[(Fx • Fy) ⊃x = y] / 1Simp.
  5. * (y)[(Fa • Fy) ⊃a = y] / 4UI x/a
  6. * (Fa • Fy) ⊃a = y / 5UI y/y
  7. * ¬ (Fa • Fy) ∨a = y / 6MI
  8. * (¬ Fa ∨¬ Fy) ∨a = y / 7DeM
  9. * ¬ Fa ∨(¬ Fy ∨a = y) / 8Assoc.
  10. * ¬ Fy ∨a = y / 3,9DS
  11. * Fy ⊃a = y / 10MI
  12. * (y)(Fy ⊃a = y) / 11UG
  13. * Fa • (y)(Fy ⊃a = y) / 3,12Conj.
  14. * (∃x)[Fx • (y)(Fy ⊃x = y)] / 13EG
  15. {(∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y]} ⊃ (∃x)[Fx • (y)(Fy ⊃x = y)] / 1-14CP
  16. * (∃x)[Fx • (y)(Fy ⊃x = y)] / ACP
  17. * Fm • (y)(Fy ⊃m = y) / 16EI x/m
  18. * Fm / 17Simp.
  19. * (y)(Fy ⊃m = y] /17Simp.
  20. * * Fx • Fy / ACP
  21. * * Fy / 20Simp.
  22. * * Fy ⊃m = y / 19UI y/y
  23. * * m = y / 21,22MP
  24. * * Fx / 20Simp.
  25. * * Fx ⊃m = x / 19UI y/x
  26. * * m = x / 24,25MP
  27. * * x = m / 26Id
  28. * * x = y / 23,27Id
  29. * (Fx • Fy) ⊃x = y / 20-28CP
  30. * (y)[(Fx • Fy) ⊃x = y] / 29UG
  31. * (x)(y)[(Fx • Fy) ⊃x = y] / 30UG
  32. * (∃x)Fx / 18EG
  33. * (∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y] / 32,31Conj.
  34. (∃x)[Fx • (y)(Fy ⊃x = y)] ⊃ (∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y] / 16-33CP
  35. {(∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y]} ⊃ (∃x)[Fx • (y)(Fy ⊃x = y)] • (∃x)[Fx • (y)(Fy ⊃x = y)] ⊃ (∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y] / 15,34Conj.
  36. {(∃x)Fx • (x)(y)[(Fx • Fy) ⊃x = y]} ≡ (∃x)[Fx • (y)(Fy ⊃x = y)] / 35BE