Thursday, 28 October 2010

In Our Time: Logic; BBC Radio 4, broadcast 21 October 2010

In Our Time is a programme about the history of ideas chaired by Melvyn Bragg. Last week’s topic was Logic. Melvyn’s guests included: A. C. Grayling, Professor of Philosophy at Birkbeck, University of London; Peter Millican, Gilbert Ryle Fellow at Philosophy at Hertford College at the University of Oxford; Rosanna Keefe, Senior Lecturer in Philosophy at the University of Sheffield.

The programme is usually a leisurely-paced tour down the ages tracing the rise of an idea, philosophical, literary or scientific development, and their impact on our times. To this extent, last week’s discussion was no different. The speakers attempt to boil down a vast body of knowledge to a few bitesize pieces which the general public can comprehend, while Melvyn Bragg’s job is to moderate the discussion (read: ‘herd the speakers’), and steer it towards a conclusion (on rare occasions).

In Our Time rounds off my week nicely, but it can be difficult to concentrate for extended periods of time during the programme, especially if one is unfamiliar with the subject matter.

Logic charted the route from Aristotle, through the swamp of Scholastic philosophy, to modern times, and in particular Gottlob Frege. From the comfort of my armchair, I would have liked to add just a few thoughts of my own where, perhaps, the answers were not entirely to my satisfaction.

Thus, to the question of why the Aristotelian categorical propositions had driven the development of logic and attracted so much attention, I’d say it is because practically all declarative sentences that we use in language can be reduced to one of the four proposition types: A, E, I, O. A proposition of the type A, for example, has the structure: All P are Q, or: All sheep are docile. We can plug into this pattern a sentence like: ‘Politicians promised us the moon,’ simply by paraphrasing it as: ‘All politicians are persons who promised us the moon,’ where ‘politicians’ is ‘All P’, and ‘promised us the moon’ is ‘are Q’, that is ‘are persons who promised us the moon.’ Why should it be useful? Well, it is cool to know that all sentences have been written in a key that fits one of the four patterns.

To the questions of why the medievals were so keen on the development of deductive logic, I’d say it’s because they were looking for an argument to clinch all arguments – an argument which would show incontestably that God exists, without having to refer to any empirical evidence. Chief of those was St Anselm of Canterbury’s ontological argument for the existence of God (roughly: God is perfect. Anything that exists is more perfect than anything that doesn't. Therefore, God exists). The enlightened view today is that such exercises amounted to no more than a load of tosh.

Logic today, the kind I’m interested in anyway, brings together quantification theory, set theory, mathematics and linguistics. It is an open chapter, which is constantly being written, and which in itself is quite annoying, but I’d rather that than the certainties of death and taxes.

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 4th Edition, 2004, 10.5E, problem 2(d), p. 562

The task: to show that the argument is valid. It is, indeed. In 7 easy steps we deduce the conclusion.
  1. (z)(Lz ≡ Hz)
  2. (x)¬(Hx ∨¬ Bx)
  3. ∴¬ Lb
  4. (z)[(Lz ⊃Hz) • (Hz ⊃Lz)] ......... 1BE
  5. ¬(Hb ∨¬ Bb) ......... 2UI x/b
  6. ¬ Hb • Bb ......... 5DeM
  7. ¬ Hb ......... 6Simp.
  8. (Lb ⊃Hb) • (Hb ⊃Lb) ......... 4UI z/b
  9. Lb ⊃Hb ......... 8Simp.
  10. ¬ Lb ......... 7,9MT

Friday, 22 October 2010

Deduction, Daniel Bonevac, Blackwell Publishing, 2003, 8.3 problem 16, p. 238

We can say in symbolic terms - according to the instructions to this problem - that there is one and only one God: (∃x)(y)(y = x ≡ Gy). We can further show that the following is a consequence of this formula: (∃x)(Gx • ¬ Fx) ≡ (x)(Gx ⊃¬ Fx). Since the conclusion is an equivalence, we need to prove it for two cases. A conditional proof is required in each case (twice in the first case). The hardest part, as usual, is being able to spot that we have to instantiate the same proposition more than once.


  1. (∃x)(y)(y = x ≡ Gy)
  2. ∴(∃x)(Gx • ¬ Fx) ≡ (x)(Gx ⊃¬ Fx)
  3. CASE 1: (∃x)(Gx • ¬ Fx) ⊃ (x)(Gx ⊃¬ Fx)
  4. * (∃x)(Gx • ¬ Fx) ......... ACP
  5. * Ga • ¬ Fa ......... 4EI x/a
  6. * (y)(y = m ≡ Gy) ......... 1EI x/m
  7. * * Gx ......... ACP
  8. * * x = m ≡ Gx ......... 6UI y/x
  9. * * (x = m ⊃ Gx) • (Gx ⊃x = m) ......... 8BE
  10. * * a = m ≡ Ga ......... 6UI y/a
  11. * * (a = m ⊃ Ga) • (Ga ⊃a = m) ......... 10BE
  12. * * Gx ⊃x = m ......... 9Simp.
  13. * * x = m ......... 7,12MP
  14. * * Ga ⊃a = m ......... 11Simp.
  15. * * Ga ......... 5Simp.
  16. * * a = m ......... 15,14MP
  17. * * m = a ......... 16Comm.
  18. * * x = a ......... 13,17Id
  19. * * ¬ Fa ......... 5Simp.
  20. * * ¬ Fx ......... 18,19Id
  21. * Gx ⊃¬ Fx ......... 7-20CP
  22. * (x)(Gx ⊃¬ Fx) ......... 21UG
  23. (∃x)(Gx • ¬ Fx) ⊃ (x)(Gx ⊃¬ Fx) ......... 4-22CP
  24. CASE 2: (x)(Gx ⊃¬ Fx) ⊃(∃x)(Gx • ¬ Fx)
  25. * (x)(Gx ⊃¬ Fx) ......... ACP
  26. * (y)(y = a ≡ Gy) ......... 1EI x/a
  27. * a = a ≡ Ga ......... 26UI y/a
  28. * (a = a ⊃ Ga) • (Ga ⊃a = a) ......... 27BE
  29. * a = a ......... Id
  30. * a = a ⊃ Ga ......... 28Simp.
  31. * Ga ......... 29,30MP
  32. * Ga ⊃¬ Fa ......... 25UI x/a
  33. * ¬ Fa ......... 31,32MP
  34. * Ga • ¬ Fa ......... 31,33Conj.
  35. * (∃x)(Gx • ¬ Fx) ......... 34EG
  36. (x)(Gx ⊃¬ Fx) ⊃(∃x)(Gx • ¬ Fx) ......... 25-35CP
  37. [(∃x)(Gx • ¬ Fx) ⊃ (x)(Gx ⊃¬ Fx)] • [(x)(Gx ⊃¬ Fx) ⊃(∃x)(Gx • ¬ Fx)] ......... 23,36Conj.
  38. (∃x)(Gx • ¬ Fx) ≡ (x)(Gx ⊃¬ Fx) ......... 37BE

Thursday, 14 October 2010

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill 2004, 10.5E, 4(d), p. 563

Show that the statement ¬ (x)(∃y)[(Ax • Bx) ∨Cy] is equivalent to (∃x)(y)[ ¬ (Cy ∨Ax) ∨¬ (Cy ∨ Bx)]. The task involves primarily the use of DeMorgan's law and distribution. I show the equivalence one way. Of course, we need to show it both ways in order to be sure the statements are indeed equivalent. However, the reverse procedure is quite straightforward and involves practically working backwards through the steps 1-10.
  1. ¬ (x)(∃y)[(Ax • Bx) ∨Cy]
  2. (∃x)(y)[ ¬ (Ax • Bx) • ¬ Cy] ......... 1CQ
  3. (y)[ ¬ (Am • Bm) • ¬ Cy] ......... 2EI x/m
  4. ¬ (Am • Bm) • ¬ Cy ......... 3UI y/y
  5. (¬ Am ∨¬ Bm) • ¬ Cy ......... 4DeM
  6. (¬ Am • ¬ Cy ) ∨ (¬ Bm • ¬ Cy) ......... 5Dist.
  7. (¬ Cy • ¬ Am ) ∨ (¬ Cy • ¬ Bm) ......... 6Comm.
  8. ¬ (Cy ∨Am) ∨¬ (Cy ∨ Bm) ......... 7DeM
  9. (y)[ ¬ (Cy ∨Am) ∨¬ (Cy ∨ Bm)] ......... 8UG
  10. (∃x)(y)[ ¬ (Cy ∨Ax) ∨¬ (Cy ∨ Bx)]

Thursday, 7 October 2010

'Into' and 'onto' functions in translation

The names ‘into’ and ‘onto’ functions are just about as imaginative as the names for an escalator that goes up and an escalator that goes down: an ‘up’ and ‘down’ escalator. Never mind the names though. The ideas is what counts.

The point here is so trivial that it hardly merits a mention: some people just can’t accept that there is no one-to-one correspondence between words in different languages. But when the same idea is drilled into heads in a maths class, the lesson is taken for what it stands, that is, except in all those cases when it just washes over the punters.

An ‘into’ function from X to Y is a function where for every element in X there is an element in Y to which it is mapped, but where it is not the case that for every element in Y there is an element in X that is mapped to it.

An ‘onto’ function from X to Y is a function where for every element in Y there is an element in X that is mapped to it.

A situation where one element in X is mapped to more than one element in Y is not a function, as it goes against the definition of a function whereby for every element in the domain there is one and only one element in the range. However, we say that X has the same size as Y if and only if there exists a one-to-one function that maps X onto Y.

Thus, it will be easy to find many ‘into’ functions between the word-sets in any two languages. I’d hazard a guess that mapping the closest equivalent of the English verb ‘to get’ in any European language to the set containing ‘get’ in English will leave many English ‘get’s’ unpaired. On the other hand, as far as I can ascertain, mapping the English verb ‘to get’ to the semantically closest set in Polish exhausts the list of ‘dostac’. It is an ‘onto’ function.

Are there any same-size word sets in any two languages? For that we would need a one-to-one and an ‘onto’ function. If there are any such sets, they are likely to lie at the peripheries of our communication: either very general or very technical and thus very narrow. It is worth noting, for example, that a lot of units of measurement, which we would expect to be prime candidates for one-to-one and onto, cannot be mapped whole (with all their semantic accretions) to the nearest set in a foreign language. Assuming we have found a match for the word ‘mile’, we may still have to look elsewhere for the equivalents of ‘to be miles ahead’ or ‘to be miles away’.

Are there any word-sets in any two languages where an element in one is mapped to two or more elements in the other (i.e. is not a function)? At first glance, there seem to be many (the example often quoted is that of the English ‘morning star’ and ‘evening star’, being two names for what in another language might well be one), but we may not realize that we are hemmed in by usage in such circumstances, which undermines the one-to-one pairing. It would be unusual to say ‘I saw the morning star in the evening and the evening star in the morning’, but not so unusual to say so in a language where we only have one word for Venus, say ‘X’: ‘I saw X both in the evening and in the morning’.

However, consider flipping it around – going from English into a language with just one word ‘X’ for the morning and the evening star. It is a function because one element in E(nglish) is assigned just one element in X, and it is ‘onto’ because for every element in X (and there is just one) there exists an element in E that is mapped onto it. So, this time the translation ‘I saw an evening star in the evening,’ and ‘I saw a morning start in the morning’ as ‘I saw X in the evening’ and ‘I saw X in the morning’ respectively is perfectly legitimate.

Wednesday, 6 October 2010

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill 2004, 10.5E, 4(b), p. 563

We have to show that the following pair of sentences are equivalent. I may well have done this example before but I haven't been keeping a very scrupulous record of the answers I have already posted. At any rate, two sentences are equivalent if they can each be derived from the other. Thus:


  1. (∃x)(∃y)Axy ⊃Aab
  2. (∃x)(∃y)Axy ≡ Aab
  3. * (∃x)(∃y)Axy ......... ACP
  4. * Aab ......... 3,1MP
  5. (∃x)(∃y)Axy ⊃Aab ......... 3-4CP
  6. * Aab ......... ACP
  7. * (∃y)Aay ......... 6EG
  8. * (∃x)(∃y)Axy ......... 7EG
  9. Aab ⊃(∃x)(∃y)Axy ......... 6-8CP
  10. [(∃x)(∃y)Axy ⊃Aab] • [Aab ⊃(∃x)(∃y)Axy] ......... 5,9Conj.
  11. (∃x)(∃y)Axy ≡ Aab ......... 10BE

This way we have derived (∃x)(∃y)Axy ≡ Aab from (∃x)(∃y)Axy ⊃Aab. To go in the opposite direction it is enough to break down (∃x)(∃y)Axy ≡ Aab into a conjunction of two conditional sentences and simplify by dropping Aab ⊃(∃x)(∃y)Axy. We'll be left with (∃x)(∃y)Axy ⊃Aab.

Saturday, 2 October 2010

More on restrictions in deductive reasoning

By way of reminding ourselves of the restrictions on deductive reasoning discussed informally the other week, we are told that once we’ve made an assumption (sort of like turning away from our opponent for a bit of shadowboxing just to find the punch that works best before we resume the fight), that assumption must not be generalized until it is discharged. To see why, we observe what happens in the following argument:

Let Ex - x is even, Dx - x is evenly divisible by 4, Universe of Discourse: all integers

1. (x)Ex ⊃(x)Dx
2. * Ex ......... Assumption
3. *(x)Ex ......... ERROR!
4. *(x)Dx
5. *Dx
6. Ex ⊃ Dx ......... Assumption discharged
7. (x)(Ex ⊃ Dx)

Our premise on line (1) says that ‘If every number is even, then every number is evenly divisible by 4.’ The sentence is true because the antecedent is false – not every number is even, of course. We need not worry about the consequent ‘every number is evenly divisible by 4,’ because whatever the consequent, if the antecedent is false, the sentence is always true. The conclusion, however, is false, because it says ‘Every even number is evenly divisible by 4.’ This is plainly not the case with 2; the number 2 is not evenly divisible by 4.

The error was made on line (3) where we generalized our assumption from line (2) ‘a number picked at random is even.’ There is no problem with making that kind of assumption, it is just that we are prohibited from generalizing it to ‘every number is even.’

Our second restriction says that if we introduce into our reasoning an individual specified by name after we have picked another individual at random, then the relation between the latter and the former cannot be reversed.

Let Lxy - x is larger than y, a - the integer 2; UD: all integers

1. (x)(∃y)Lxy
2. (∃y)Lxy
3. Lxa
4. (x)Lxa ......... ERROR!
5. (∃y)(x)Lxy

Our premise on line (1) says ‘Every number is larger than some number.’ This sentence is obviously true. We drop the universal quantifier to get something like ‘A number picked at random is larger than some number,’ – still true. Then, we drop the existential quantifier to get ‘A number picked at random is larger than 2.’ This sentence can very well be true. However, when we now restore the quantifiers (we restore them in reverse order or else we wouldn’t be getting very far with our reasoning), on line (4) we get ‘Every number is larger than 2.’ This is patently false.

Finally, when we generalize a variable of a particular kind, we must generalize all of them at once.

Let Dxy - x is evenly divisible by y; UD: all integers

1. (x)Dxx
2. Dxx
3. (x)Dxy ......... ERROR!
4. (y)(x)Dxy

Line (1) says ‘Every number is evenly divisible by itself.’ This is true. Line (2) says ‘A number picked at random is evenly divisible by itself.’ This is also true. Then, on line (3) we generalize only one variable to get ‘Every number is evenly divisible by a number picked at random,’ and we run into trouble.

The idea, as usual, is to avoid being led from true premises to a false conclusion.