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

Sunday 7 March 2010

Predicate Logic, Howard Pospesel, Prentice Hall. 2003, Chpt 11, ex. 20

Symbolize the following argument and show that it is valid.

Logical equivalence is mutual entailment. Every statement entails any logical truth. Therefore, all logical truths are logically equivalent.

Qxy - x is logically equivallent to y
Nxy - x entails y
Tx - x is a logical truth
  1. (x)(y)(Qxy ≡ Nxy)
  2. (x)(y)(Ty ⊃Nxy)
  3. ∴ (x)(y)(Ty ⊃Qxy)
  4. * Ty / ACP
  5. * (y)(Ty ⊃Nxy) / 3UI x/x
  6. * Ty ⊃Nxy / 5UI y/y
  7. * Nxy / 6,7MP
  8. * (y)(Qxy ≡ Nxy) / 1UI x/x
  9. * Qxy ≡ Nxy / 1UI y/y
  10. * (Qxy ⊃Nxy) • (Nxy ⊃Qxy) / 9BE
  11. * Nxy ⊃Qxy / 10Simp.
  12. * Qxy / 7,11MP
  13. Ty ⊃Qxy / 4-12CP
  14. (y)(Ty ⊃Qxy) / 13UG
  15. (x)(y)(Ty ⊃Qxy) / 14UG