Wednesday 22 December 2010

Understanding Symbolic Logic, Virginia Klenk, Prentice Hall, 5th edition, 2008, Unit 20, Ex. 1(g)

The argument is phrased as below. 'Exactly one' is simply another way of saying that our variable 'x' is 'the' current president of the US, hence a definite description.
There is exactly one current president of the United States. One is a commander-in-chief of the U.S. armed forces if and only if one is the current president of the United States. Therefore, there is exactly one commander-in-chief of the U.S. armed forces.
  1. (∃x)[Px • (y)(Py ⊃y = x)]
  2. (x)(Cx ≡ Px)
  3. ∴(∃x)[Cx • (y)(Cy ⊃y = x)]
  4. Pa • (y)(Py ⊃y = a) ......... 1EI x/a
  5. Ca ≡ Pa ......... 2UI x/a
  6. (Ca ⊃Pa) • (Pa ⊃ Ca) ......... 5BE
  7. Pa ......... 4Simp.
  8. Pa ⊃Ca ......... 6Simp.
  9. Ca ......... 7,8MP
  10. * ¬ (∃x)[Cx • (y)(Cy ⊃y = x)] ......... AIP
  11. * (x)[ ¬ Cx ∨(∃y)(Cy • ¬ (y = x)] ......... 10CQ
  12. * ¬ Ca ∨(∃y)(Cy • ¬ (y = a) ......... 11UI x/a
  13. * (∃y)(Cy • ¬ (y = a) ......... 9,12DS
  14. * Cm • ¬ (m = a) ......... 13EI y/m
  15. * Cm ≡ Pm ......... 2UI x/m
  16. * (Cm ⊃Pm) • (Pm ⊃ Cm) ......... 15BE
  17. * Cm ......... 14Simp.
  18. * Cm ⊃ Pm ......... 16Simp.
  19. * Pm .........17,18MP
  20. * (y)(Py ⊃y = a) ......... 4Simp.
  21. * Pm ⊃m = a ......... 20UI y/m
  22. * m = a ......... 19,21MP
  23. * ¬ (m = a) ......... 14Simp.
  24. * (m = a) • ¬ (m = a) ......... 22,23Conj.
  25. ¬ ¬ (∃x)[Cx • (y)(Cy ⊃y = x)] ......... 10-24IP
  26. (∃x)[Cx • (y)(Cy ⊃y = x)] ......... 25DN

Friday 17 December 2010

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 5th edition, 2008, Unit 20, ex. 1(c), p. 380

The argument is stated thus:

The person who owns a Ferrari is illegally parked. The person who is illegally parked will be towed. So, someone who owns a Ferrari will be towed.
  1. (∃x){Px • Fx • (y)[(Py • Fy) ⊃ y = x] • Ix}
  2. (∃x){Px • Ix • (y)[(Py • Iy) ⊃ y = x] • Tx}
  3. ∴(∃x){Px • Fx • Tx)
  4. Pa • Fa • (y)[(Py • Fy) ⊃ y = a] • Ia ......... 1EI x/a
  5. Pm • Im • (y)[(Py • Iy) ⊃ y = m] • Tm ......... 2EI x/m
  6. Pa • Ia ......... 4Simp.
  7. (y)(Py • Iy) ⊃ y = m ......... 5Simp.
  8. (Pa • Ia) ⊃ a = m ......... 7UI y/a
  9. a = m ......... 6,8MP
  10. Fa ......... 4Simp.
  11. Fm ......... 9,10Id
  12. Pm • Tm......... 5Simp.
  13. Pm • Tm • Fm ......... 11,12Conj.
  14. Pm • Fm • Tm ......... 13Comm.
  15. (∃x){Px • Fx • Tx) ......... 14EG

Friday 10 December 2010

The Mandy Andy argument

The argument:

Everyone likes Mandy. Mandy likes nobody but Andy. Therefore, Mandy and Andy are the same person

Some notes on deductive reasoning:

When we say: ‘Everyone likes Mandy,’ then ‘everyone’ is ‘everyone’. I can pick any representative of the ‘Everyone’ set at random, including Mandy, without violating the truth of the sentence, and this is what I’ll do.

Since we do not use reflexive pronouns in logic (i.e. herself, etc), the sentence: ‘Mandy likes herself,’ is simply ‘Mandy likes Mandy.’ We will use both versions below in spoken explanations.

All arguments are set in a universe of discourse, which is the set of things being talked about on a given occasion. In our case, the universe of discourse is simply ‘persons’.

Analysis:

The first premise says: ‘Everyone likes Mandy.’ Let F be our first premise:

F = {… Betty likes Mandy, Sarah likes Mandy, Clive likes Mandy, Mandy likes Mandy, Sandra likes Mandy, Eddie likes Mandy, …}

The second premise says: ‘Mandy likes nobody but Andy.’ This can be paraphrased as ‘Mandy likes only Andy.’ The adverb ‘only’ requires that we show that not only does Mandy like Andy, but that she does not like anybody else. Let S be our second premise:

S = {… Mandy does not like Betty, Mandy does not like Sarah, Mandy does not like Clive, Mandy likes Andy, Mandy does not like Sandra, Mandy does not like Eddie, …}

Another way of saying ‘Mandy likes only Andy’ is, ‘If a person is not Andy, then Mandy does not like him / her.’ Mandy herself, however, is a person. We can legitimately then substitute Mandy for ‘person’ in our conditional sentence.

Our argument boils down to this:

Mandy likes herself.
If Mandy is not Andy, then Mandy does not like herself.

We can use contraposition in our ‘If …, then …’ sentence and simply flip around the sides while remembering to change the signs. Thus, we get:

Mandy likes herself.
If Mandy likes herself, then Mandy is Andy.

Many people find it easier to change the order of the premises. Let’s do that:

If Mandy likes herself, then Mandy is Andy.
Mandy likes herself.

Using Modus Ponens (affirming the antecedent), we infer:

Mandy is Andy.

In real life, there are of course many Mandies, whereas we take Mandy to be a unique individual. However, this is an irrelevance as we can easily assign a number or some other unique designation to each individual in our set. Further, in Natural language (English, Polish, etc) when we say ‘Everyone likes Mandy,’ we probably think of Mandy as standing apart from everyone else. This can be easily expressed in logic as ‘For every person, such that that person is different from Mandy, every person likes Mandy,’ but this is not what the original premise says. Finally, natural language is imprecise. Conversation overheard on a train last week:

She: Shopping is no fun if you don’t have money.
He: You can’t shop at all if you don’t have money, I would have thought.
She: I mean, you know, when you have little money.

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 5th edition, 2008, Unit 20, Ex. 1(a)

Once again we have to prove the argument. Unlike in the Mandy Andy argument in an earlier post, the conclusion here accords with our intuition.

The sports car buff who owns a Maserati is unemployed. Mary is a sports car buff and owns a Maserati. So, Mary is unemployed.
  1. (∃x){Bx • Mx • (y)[(By • My)⊃y = x] • ¬ Ex}
  2. Bm • Mm
  3. ∴¬ Em
  4. Ba • Ma • (y)[(By • My)⊃y = a] • ¬ Ea ......... 1EI x/a
  5. (y)[(By • My)⊃y = a] ......... 4Simp.
  6. (Bm • Mm)⊃m = a ......... 5UI y/m
  7. m = a ......... 2,6MP
  8. ¬ Ea ......... 4Simp.
  9. ¬ Em ......... 7,8Id

Thursday 2 December 2010

Destructive Dilemma

One of the inference methods involves a chain of reasoning which goes like this:

If my calculations are correct, then my bank charges me a mystery fee on top of all the other disclosed fees and charges. If my suspicions are correct, then I bankroll the bank manager's lunches. Either the bank doesn't charge me a mystery fee or the bank manager buys his own lunches. Therefore, either my calculations are incorrect or my suspicions are unfounded.

I think I like the name of this type of reasoning more than I like the reasoning itself, as it does not present itself readily to the mind when we speak. But it does what it says on the tin: it knocks down the assumptions we have set off with.

The mechanism is quite simple: the disjunctive second premise denies the consequents of the two conditionals. In doing so we are simply using the Modus Tollens argument form twice, the result being that the final conclusion denies the antecedents. The negated antecedents are a disjunction.

We can do without the destructive dilemma in our deduction proofs – it is a derivation of other inference methods: Modus Tollens, Modus Pollens, simplification of conjunctive expressions, and addition by means of disjunction. It is therefore redundant. In fact, the destructive dilemma is like the constructive dilemma in reverse, where we postulate two conditional sentences and a disjunction of their antecedents in order to obtain disjunction of their consequents.

I find the application of the destructive dilemma in teaching counterfactual conditionals to those students who profess to have heard it all before and need to be teased or else put in their place. The first premise, the one containing two conditionals, sketches out a more or less hypothetical situation:

If you hadn’t failed Latin at school, you would be a judge now; and if you had spelt your name correctly on the exam, you wouldn’t be a miner now.

Set the second promise like this:

I am guessing that either you are not a judge or you are a miner.

Set the question like this: what can you infer from this set of sentences? Answer:

Either you had failed Latin at school or you had misspelt your name on the exam paper.

Symbolic Logic, Dale Jacquette, Wadsworth, 2001, Chapter 8, Exercise IV, problem 5

We are asked here to symbolize the argument, verify the translation by the truth tree method, and give a natural deduction proof. I don't know how to do a truth tree using blogger tools - I can just about do it in Word, with a little patience, so only the translation and the deduction follow.

Everything is either an unconscious or immaterial entity. All sentient beings, if they are physically embodied, are material entities. Hence, no sentient, physically embodied beings are conscious.

Using the glossary provided (C, M, E, S, B, P), we get:
  1. (x)[Ex ⊃(¬ Cx ∨ ¬ Mx)]
  2. (x){(Sx • Bx) ⊃[Px ⊃ (Mx • Ex)]
  3. ∴(x)[Sx • Px • Bx) ⊃¬ Cx]
  4. * Sx • Px • Bx ......... ACP
  5. * (Sx • Bx) ⊃[Px ⊃ (Mx • Ex)] ......... 2UI x/x
  6. * Sx • Bx ......... 4Simp.
  7. * Px ⊃ (Mx • Ex) ......... 6,5MP
  8. * Px ......... 4Simp.
  9. * Mx • Ex ......... 8,7MP
  10. * Mx ......... 9Simp.
  11. * Ex ......... 9Simp.
  12. * Ex ⊃(¬ Cx ∨ ¬ Mx) ......... 1UI x/x
  13. * ¬ Cx ∨ ¬ Mx ......... 11,12MP
  14. * ¬ Cx ......... 10,13DS
  15. Sx • Px • Bx) ⊃¬ Cx ......... 4-14CP
  16. (x)[Sx • Px • Bx) ⊃¬ Cx] ......... 15UG

Saturday 27 November 2010

Nouns and sets

Nouns are categories into which we put other nouns. Sets are collections of objects. Sets can also be elements of bigger sets. There is nothing inventive about me saying so other than that thinking about one can help us feel more warmly about the other.

In logic, when we say: ‘The horse is an animal,’ we are simply saying that for any individual that is a horse it belongs to a class of animals. Complex sentences and action verbs can be reduced to this talk: ‘The apples have dropped onto the ground,’ is ‘Something that belongs to the class of apples is such that it belongs to things that have dropped onto the ground.’ The efficiency of such language cannot compete with the efficiency of natural language, but this is quite irrelevant for the purposes for which sentences are analysed in this manner. A lot of our speech is simply set talk.

Nouns are countable and uncountable. So are sets. The parallels are more intriguing than they are exact, but they are there nevertheless. A set is countable when it is a finite set or a countably infinite set, that is, it has the same size as the set N of natural numbers. Nouns are countable when we can count the objects they refer to.

Some people get confused about this business of counting. As in grammar so in mathematics, ‘to count’ does not necessarily mean to say how many, although this is indeed what it means in the case of finite sets. We say we can count the elements of an infinite set A if we can find a function from N (set of natural numbers) to A that is both one-to-one and onto. In other words the sets are the same size, A = N, or have the same cardinality. A very close equivalent of a countably infinite set in language is the plural form of a noun, for example:

apples

When we use it like that, without the definite article or any other restricting phrase, we are talking about an infinite collection of apples, yet one for which we can easily find a one-to-one and onto function of the type f: NA. Suppose we want to name the finite set. In that case, we say:

the apples

There is then a natural number n such that f: nA. The expression:

apple sauce

is an example of an uncountable noun, matching closely the definition of an infinite set which is not countable. That is, we can’t even begin to think of how we would go about finding a function from the set of natural numbers N to the set S (set representing apple sauce). However, since we can also say:

the apple sauce

as in ‘I spilled the apple sauce,’ it would follow that there must be some kind of finite dimension of this otherwise uncountable infinite set. Well, that’s where the parallel is not exact in my view, because as far as I know there are no such things in set theory, but we can, since we have defined S as a set, using Cantor’s axiom of existence and axiom of equality, construct the set {S}, whereby X = {S}, which is equivalent to a container with one element in it. Hence, we can find a function from N to X. And that is not a far cry from what we mean when we say ‘I spilled the apple sauce,’ – the container we spilled it from defines the apple sauce.

Enough about apple sauce. This rambling has focused (if rambling can be focused) on the property of nouns and sets involving countability, but the operations of union and intersection of sets, subsets, and power sets also have parallels in the realm of nouns.

Friday 26 November 2010

Deduction, Daniel Bonevac, Blackwell Publishing 2003, 2nd edition, 8.3 problem 3, p. 238

The argument is:

Everyone likes Mandy. Mandy likes nobody but Andy. Therefore, Mandy and Andy are the same person.

We translate it and prove it:
  1. (x)Lxm
  2. (x)[¬ (x=a) ⊃¬ Lmx] • Lma
  3. ∴m = a
  4. Lmm ......... 1UI x/m
  5. (x)[¬ (x=a) ⊃¬ Lmx] ......... 2Simp.
  6. ¬ (m=a) ⊃¬ Lmm ......... 5UI x/m
  7. m = a ......... 4,6MT

Friday 19 November 2010

Deduction, Daniel Bonevac, Blackwell Publishing 2003, 2nd edition, 8.3, problem 2, p. 238

The argument goes:

Everything other than God had a creator. So, given any two things, at least one had a creator.

We can choose the level of detail we want our predicates to capture in symbolizing this argument. I've opted for the more complete translation here. Glossary: Cxy - x created y, g - God. The proof follows.
  1. (x)[¬(x=g) ⊃(∃y)(Cyx • ¬ Cyg)]
  2. ∴(x)(y)[¬ (x=y) ⊃(∃z)(Czx ∨Czy)]
  3. * ¬ (x)(y)[¬ (x=y) ⊃(∃z)(Czx ∨Czy)] ......... AIP
  4. * (∃x)(∃y)[ ¬ (x=y) • (z)(¬Czx • ¬ Czy)] ......... 3CQ
  5. * (∃y)[ ¬ (a=y) • (z)(¬Cza • ¬ Czy)] ......... 4EI x/a
  6. * ¬ (a=m) • (z)(¬Cza • ¬ Czm) ......... 5EI y/m
  7. * ¬ (a=m) ......... 6Simp.
  8. * (z)(¬Cza • ¬ Czm) ......... 6Simp.
  9. * ¬(a=g) ⊃(∃y)(Cya • ¬ Cyg) ......... 1UI x/a
  10. * ¬Cya • ¬ Cym ......... 8UI z/y
  11. * ¬Cya ......... 10Simp.
  12. * ¬Cya ∨Cyg ......... 11Add.
  13. * (y)(¬Cya ∨Cyg) ......... 12UG
  14. * (y)¬ (Cya • ¬ Cyg) ......... 13DeM
  15. * ¬ (∃y)(Cya • ¬ Cyg) ......... 14CQ
  16. * a = g ......... 15,9MT
  17. * ¬ (g = m) ......... 16,7Id
  18. * ¬ (m = g) ......... 17Comm.
  19. * ¬(m=g) ⊃(∃y)(Cym • ¬ Cyg) ......... 1UI x/m
  20. * (∃y)(Cym • ¬ Cyg) ......... 18,19MP
  21. * Crm • ¬ Crg ......... 20EI y/r
  22. * Crm ......... 21Simp.
  23. * ¬Cra • ¬ Crm ......... 8UI z/r
  24. * ¬ Crm ......... 23Simp.
  25. * Crm • ¬ Crm ......... 22,24Conj.
  26. ¬ ¬ (x)(y)[¬ (x=y) ⊃(∃z)(Czx ∨Czy)] ......... 3-25IP
  27. (x)(y)[¬ (x=y) ⊃(∃z)(Czx ∨Czy)] ......... 26DN

Tuesday 9 November 2010

Deduction, D. Bonevac, Blackwell Publishing, 2003, 8.3 problem 1

A simple argument and a two-step proof:

Jones's killer weighed over 200 pounds. Smith weighs less than 200 pounds. So, Smith isn't Jones's killer.

Regarding the translation, if we take 'Wx' for 'x weighed over 200 pounds', then 'weighs less than 200 pounds' can be translated as a negation of the former, that is 'It is not the case that Smith weighs over 200 pounds'.
  1. (x)(Kxj ⊃Wx)
  2. ¬ Ws
  3. ∴¬ Ksj
  4. Ksj ⊃Ws ......... 1UI x/s
  5. ¬ Ksj ......... 2,4MT

Friday 5 November 2010

Swiss beauty or Swiss joke?

Someone somewhere, in Geneva perhaps or on Madison Avenue, knows why the time on advertised luxury watches is 10:10 and is keeping schtum. It’s when I suffer from a serious lack of things to worry about that I worry about things like that, and I would very much welcome that person telling me why. Otherwise I’m left to my own devices, and these could take me anywhere.

There is the smiley face explanation (which the hands of the clock form at this hour), the various arguments from numerology (time of death of some famous people), the watchmaker’s name set off to good effect by that particular configuration of the hands. Aah! There is the clue. The angle!

So what angle is that? Well, I thought 120 degrees. It is pretty symmetrical and divides the face 3 ways. There are 20 minutes between the number 10, which is where the hour hand is pointing, and the number 2, which is where the minute hand is pointing. 20 minutes corresponds to 120 degrees. Except if the angle was 120 degrees, neither the hour hand would be in the number 10 position nor the minute hand in the number 2 position.

Say, we start counting at precisely 10 o’clock, because that way we can be sure the hour hand is exactly on the number 10 mark and the minute hand exactly on the number 12 mark. The hour hand moves at 5 min per hour, the minute hand at 60 min per hour. The distance between the hands must be 20 min (for the 120 degrees angle), but the minute hand has a 10 min headstart over the hour hand. This gives us the formula: d/5 = [(d + 20) – 10] / 60, where ‘d’ is the distance travelled by the hour hand. Accordingly, d = 0.9, so the time at which the hour hand and the minute hand are at 120 degrees is roughly 10:10:54. The 54 seconds (following rounding) is not a big difference, but it means that the second hand would have to be between the hour hand and the minute hand, which would ruin the symmetry.

The catch is that the time is not 10:10:54 at all. Closer inspection reveals that the majority of watches are set to something like 10:09:36. Working backwards (the formula would now be d/5 = 9.6/60) we learn that d = 0.8, or about 48 seconds (the distance the hour hand has nudged above the number 10 mark). The distance between the hour hand and the minute hand now is 18 minutes and 48 seconds, which is equivalent to about 112.8 degrees. What is so special about 112.8 degrees? Is this a Swiss sense of beauty or a Swiss sense of humour?

Harmony, symmetry and proportion - the constituents of beauty – go back to Pythagoras. We learn from Palladio, for example, that for a column to be pleasant to look at the height of the column should be nine times its diameter. Modern commerce has developed the idea of ‘buy one get one free’ or 9.99, and my guess is not because they imply a bargain but because they sound or look pleasing (compare: ‘buy three get one free’ and 9.98). So, if someone knows what it is about 112.8 degrees on watch faces, come on, out with it!

Thursday 4 November 2010

Symbolic Logic, Irving M. Copi, Prentice Hall, 1979, 5th edition, p. 150, problem 10

We are asked to prove the validity of the following argument, where Ax - x was an accompanist, Bx - x was a bagpiper, and Cx - x was in the cabin. The argument goes:
All accompanists were bagpipers. All bagpipers were in the cabin. At most two individuals were in the cabin. There were at least two accompanists. Therefore, there were exactly two bagpipers.
  1. (x)(Ax ⊃Bx)
  2. (x)(Bx ⊃ Cx)
  3. (x)(y)(z)[(Cx • Cy • Cz) ⊃(x = y ∨x = z ∨ y = z)]
  4. (∃x)(∃y)[Ax • Ay • ¬ (x = y)]
  5. ∴(∃x)(∃y){Bx • By • ¬ (x = y) • (z)[Bz ⊃(z = x ∨z = y)]}
  6. * ¬ (∃x)(∃y){Bx • By • ¬ (x = y) • (z)[Bz ⊃(z = x ∨z = y)]} ......... AIP
  7. * (x)(y){[Bx • By • ¬ (x = y)] ⊃(∃z)[Bz • ¬ (z = x) • ¬ (z = y)]} ......... 6CQ
  8. * (∃y)[Aa • Ay • ¬ (a = y)] ......... 4EI x/a
  9. * Aa • Am • ¬ (a = m) ......... 8EI y/m
  10. * Aa ⊃ Ba ......... 1UI x/a
  11. * Aa ......... 9Simp.
  12. * Ba ......... 11,10MP
  13. * Am ⊃ Bm ......... 1UI x/m
  14. * Am ......... 9Simp.
  15. * Bm ......... 14,13MP
  16. * ¬ (a = m) ......... 9Simp.
  17. * Ba • Bm • ¬ (a = m) ......... 12,15,16Conj.
  18. * (y){[Ba • By • ¬ (a = y)] ⊃(∃z)[Bz • ¬ (z = a) • ¬ (z = y)]} ......... 7UI x/a
  19. * [Ba • Bm • ¬ (a = m)] ⊃(∃z)[Bz • ¬ (z = a) • ¬ (z = m)] ......... 18UI y/m
  20. * (∃z)[Bz • ¬ (z = a) • ¬ (z = m)] ......... 17,19MP
  21. * Bh • ¬ (h = a) • ¬ (h = m) ......... 20EI z/h
  22. * Bh ⊃Ch ......... 2UI x/h
  23. * Bh ......... 21Simp.
  24. * Ch ......... 22,23MP
  25. * Ba ⊃Ca ......... 2UI x/a
  26. * Ca ......... 12,25MP
  27. * Bm ⊃Cm ......... 2UI x/m
  28. * Cm ......... 14,27MP
  29. * Ca • Cm • Ch ......... 26,24,28Conj.
  30. * (y)(z)[(Ca • Cy • Cz) ⊃(a = y ∨a = z ∨ y = z)] ......... 3UI x/a
  31. * (z)[(Ca • Cm • Cz) ⊃(a = m ∨a = z ∨ m = z)] ......... 30UI y/m
  32. * (Ca • Cm • Ch) ⊃(a = m ∨a = h ∨ m = h) ......... 31UI z/h
  33. * a = m ∨a = h ∨ m = h ......... 29,32MP
  34. * a = h ∨ m = h ......... 16,33DS
  35. * ¬ (h = a) ......... 21Simp.
  36. * m = h ......... 35,34DS
  37. * ¬ (h = m) ......... 21Simp.
  38. * h = m ......... 36Comm.
  39. * h = m • ¬ (h = m) ......... 37,38Conj.
  40. ¬ ¬ (∃x)(∃y){Bx • By • ¬ (x = y) • (z)[Bz ⊃(z = x ∨z = y)]} ......... 6-39IP
  41. (∃x)(∃y){Bx • By • ¬ (x = y) • (z)[Bz ⊃(z = x ∨z = y)]} ......... 40DN

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.

Thursday 30 September 2010

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill 2004, 10.4E, 12(b)

We are required to symbolise the following argument and show that it is valid. Despite the name used in the agrument, here abbreviated to lower case 'g', there is nothing that is cock-eyed about the argument.
Groucho Marx doesn't stay in any hotel that is willing to have him as a guest. Any hotel not willing to have Groucho Marx as a guest doesn't. Therefore, Groucho Marx doesn't stay in any hotel.
  1. (x)[(Hx • Wxg) ⊃¬ Sgx]
  2. (x)[(Hx • ¬ Wxg) ] ⊃¬ Sgx]
  3. ∴(x)(Hx ⊃¬ Sgx]
  4. * Hx ......... ACP
  5. * (Hx • Wxg) ⊃¬ Sgx ......... 1UI x/x
  6. * Hx ⊃(Wxg ⊃¬ Sgx) ......... 5Exp.
  7. * Wxg ⊃¬ Sgx ......... 4,6MP
  8. * (Hx • ¬ Wxg) ] ⊃¬ Sgx ......... 2UI x/x
  9. * Hx ⊃ (¬ Wxg ⊃¬ Sgx) ......... 8Exp.
  10. * ¬ Wxg ⊃¬ Sgx ......... 4,9MP
  11. * Sgx ⊃Wxg ......... 10Contrap.
  12. * Sgx ⊃¬ Sgx ......... 7,11HS
  13. * ¬ Sgx ∨¬ Sgx ......... 12MI
  14. * ¬ Sgx ......... 13Taut.
  15. Hx ⊃¬ Sgx ......... 4-14CP
  16. (x)(Hx ⊃¬ Sgx) ......... 15UG

Thursday 23 September 2010

Fog in many areas

I don’t like M.C. Escher’s art that much, but I can’t resist comparisons between his spirals and staircases to the English in some business documents I often have to read. It is amazing that business is done, money changes hands, buildings and bridges do not collapse (usually) despite the fact that what people say on paper is either vague, ambiguous or simply just drivel. Does this makes sense to you:

“The location of the investment shall be determined pursuant to the local zoning plan. If the local zoning plan has not been adopted for a specific area, the location of the public investment shall be determined by a decision on determining the location of a public investment, whereas other investments by a decision on development conditions.”

It always cracks me up when the recipient of a letter, a lawyer or consultant with whom I work closely, says they understand what the author meant to say even though they agree that what he actually says is impenetrable, and then draft a letter in reply to match or go one better in the impenetrability stakes. One wonders what the other side thinks of it!

And yet, they all understand one another! This must be where Grice’s conversational implicature comes in, and specifically the principle of cooperation. We do not, willingly, seek to misunderstand.

But if language of the stated kind has me tearing my hair out and thinking unprintable thoughts, I actually purposefully seek out ambiguity and test the limits of understanding for pleasure. And, interestingly, many of such verbal ambiguities resemble Escher’s art. The ones relying on lexical ambiguity are merely fun; the ones exploiting syntactic ambiguity are one level up on the former; while the ones hinging on both are simply brilliant. Here are a few:

A: Where are you going?
B: I am going down to the bank to get some money.
A: Who do you bank with?
B: I’m sorry, I don’t understand?
A: You said you were going down to the bank to get some money.
B: And so I am; I keep my money buried in a chest down by the river.
(Descriptions, Stephen Neale)

Humphrey Lyttelton used to ask on I’m Sorry I Haven’t a Clue: - How many legs have donkeys? The panelists wrestled with the question for a while whereupon Humph would deadpan: - Legs don’t have donkeys.

P.G. Wodehouse, Mark Twain, B.J. Priestly, Ambrose Bierce and Groucho Marx have delivered in the last category, to quote but one example: ‘Time flies like an arrow. Fruit flies like a banana.’

The quote resembles a cube in the oblique plane (hence the Escher connection), with the front and rear faces flipping back and forth.

I have been playing around with my nemesis ‘the mole’ and have come up with this: There are different ways in which you can remove moles? A/ with laser, B/ with explosives, C/ by blowing their cover, but coming up with a piece of dialogue where the ambiguities are cancelled one by one, but not until they’ve had their run, has proved elusive so far. [mole: a/ a growth on the human skin; b/ a breakwater running out into the sea, c/ a spy, d/ animal that burrows underground (not intended by the question yet temptingly close)].

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 2008, Unit 18, Ex.1, problem (r), p. 354

We have to derive the conclusion from the premises given. Hint: It is tempting to use indirect proof because the conclusion is a simple existential statement, but this would not work very effectively here. It is easier to instantiate the premises and work through them until we isolate, by Modus Ponens, Wz on the first line.
  1. (x){[Ax • ¬ (y)(Dxy ⊃Rxy)] ⊃(z)(Tzx ⊃Wz)}
  2. ¬ (∃x)(∃y)(Tyx • ¬ Dxy)
  3. ¬ (x)[Ax ⊃(y)(Tyx ⊃Rxy)
  4. ∴(∃x)Wx
  5. (∃x)[Ax • (∃y)(Tyx • ¬ Rxy) ......... 3CQ
  6. (x)(y)(Tyx ⊃Dxy) ......... 2CQ
  7. Aa • (∃y)(Tya • ¬ Ray) ......... 5EI x/a
  8. Aa ......... 7Simp.
  9. (∃y)(Tya • ¬ Ray) ......... 7Simp.
  10. Tma • ¬ Ram ......... 9EI y/m
  11. Tma ......... 10Simp.
  12. ¬ Ram ......... 10Simp.
  13. (y)(Tya ⊃Day) ......... 6UI x/a
  14. Tma ⊃Dam ......... 13UI y/m
  15. Dam ......... 11,14MP
  16. Dam • ¬ Ram ......... 15,12Conj.
  17. (∃y)(Day • ¬ Ray) ......... 16EG
  18. (∃y)¬ ( ¬ Day ∨ Ray) ......... 17DeM
  19. (∃y)¬ (Day ⊃ Ray) ......... 18MI
  20. ¬ (y)(Day ⊃Ray) ......... 19CQ
  21. [Aa • ¬ (y)(Day ⊃Ray)] ⊃(z)(Tza ⊃Wz)} ......... 1UI x/a
  22. Aa • ¬ (y)(Day ⊃Ray) ......... 8,20Conj.
  23. (z)(Tza ⊃Wz) ......... 22,21MP
  24. Tma ⊃Wm ......... 23UI z/m
  25. Wm ......... 11,24MP
  26. (∃x)Wx ......... 25UG

Saturday 18 September 2010

Deductive reasoning and common sense

The idea that it is impossible for an argument to be valid if the premises are true and the conclusion is false is central to all logic and mathematics. So if a sequence takes us from truth to falsehood, we can suspect an error in reasoning. All other combinations are perfectly acceptable, including from falsehood to truth, from falsehood to falsehood, and from truth to truth, of course.

The rules of deduction place three restrictions on us. The first is easy to state and quick to dispatch: an assumption must not be extended to all items of a certain kind until it is discharged. Suppose we assume that a car picked at random is of the colour brown. At this stage we can’t say that every car is brown – this would be making an unjustified leap in our reasoning. However, once we get to a stage, through valid reasoning, where we can say: ‘If a car picked at random is brown, then it is made in the US,’ by which we have discharged our assumption, we can say that ‘All cars which are brown are made in the US.’

The second restriction says that if, in the course of our reasoning, we introduce an individual specified by name next to an individual which we have earlier picked at random (strictly in this order), then, likewise, we can’t extend the relation between the two to a relation between all individuals and the one we have specified by name. Thus, by proceeding from ‘Every car in the car park belongs to someone,’ (true) through ‘A car in the car park picked at random belongs to Jamie,’ (possibly true), we come to ‘All cars in the car park belong to Jamie,’ (possibly false) and eventually ‘Someone is the owner of all cars in the car park,’ (possibly false), we have made an error in reasoning. We’ve gone from a true premise to a false conclusion.

The last restriction is hardest to state in plain English, but it is quite straightforward. Suppose that, figuratively speaking, we put into the same box in our head the things that we pick at random and that bear a relationship to one another. We can’t then, in the course of our reasoning, break the bond that connects them and put one thing into one box and the other into another box. Thus, from ‘Every car is identical to itself,’ (true) we can go to ‘A car picked at random is identical to a car picked at random’ (possibly true), and back to ‘Every car is identical to itself,’ but not to ‘Every car is identical to every car,’ as the latter is definitely false.

We can of course rehearse the last argument in a language approaching real life situations, as done earlier, for example, ‘Every driver admires himself,’ where, again, it would be incorrect to conclude that ‘Every driver admires every driver (that is, every other driver and himself too), but with such sentences, of course, we can’t easily demonstrate the truthfulness of the premise(s) or the falsity of the conclusion.

I will run through this again using mathematical examples another time.

Thursday 16 September 2010

Symbolic Logic, Irving M.Copi, Prentice Hall, 5th edition, 1973, p. 150, problem 9

The argument is set as follows:

There is exactly one penny in my right hand. There is exactly one penny in my left hand. Nothing is in both my hands. Therefore, there are exactly two pennies in my hands.

To prove it is not a challenge, to translate it into logical notation is.
  1. (∃x){Px • Rx • (y)[(Py • Ry) ⊃y = x]}
  2. (∃x){Px • Lx • (y)[(Py • Ly) ⊃y = x]}
  3. ¬ (∃x)(Lx • Rx)
  4. ∴(∃x)(∃y){Px • Py • Rx • Ly • ¬ (x = y) • (z){[Pz • (Rz ∨Lz)] ⊃z = x ∨z = y}}
  5. * ¬ (∃x)(∃y){Px • Py • Rx • Ly • ¬ (x = y) • (z){[Pz • (Rz ∨Lz)] ⊃z = x ∨z = y}} ......... AIP
  6. * (x)(y){[Px • Py • Rx • Ly • ¬ (x = y)] ⊃(∃z)[Pz • (Rz ∨Lz) • ¬ (z = x) • ¬ (z = y)]} ......... 5CQ
  7. * Pa • Ra • (y)[(Py • Ry) ⊃y = a] ......... 1EI x/a
  8. * Pm • Lm • (y)[(Py • Ly) ⊃y = m] ......... 2EI x/m
  9. * (x)(Lx ⊃¬ Rx) ......... 3CQ
  10. * Lm ⊃¬ Rm ......... 9UI x/m
  11. * Lm ......... 8Simp.
  12. * ¬ Rm ......... 11,10MP
  13. * Ra ......... 7Simp.
  14. * ¬ (a = m) ......... 12,13Id.
  15. * Pa ......... 7Simp.
  16. * Pm ......... 8Simp.
  17. * Pa • Pm • Ra • Lm • ¬ (a = m) ......... 15,16,13,11Conj.
  18. * (y){[Pa • Py • Ra • Ly • ¬ (a = y)] ⊃(∃z)[Pz • (Rz ∨Lz) • ¬ (z = a) • ¬ (z = y)] ......... 6UI x/a
  19. * [Pa • Pm • Ra • Lm • ¬ (a = m)] ⊃(∃z)[Pz • (Rz ∨Lz) • ¬ (z = a) • ¬ (z = m)] ......... 18UI y/m
  20. * (∃z)[Pz • (Rz ∨Lz) • ¬ (z = a) • ¬ (z = m)] ......... 17,19MP
  21. * Pr • (Rr ∨Lr) • ¬ (r = a) • ¬ (r = m) ......... 20EI z/r
  22. * (y)[(Py • Ry) ⊃y = a] ......... 7Simp.
  23. * (Pr • Rr) ⊃r = a ......... 22UI y/r
  24. * ¬ (r = a) ......... 21Simp.
  25. * ¬ (Pr • Rr) ......... 23,24MT
  26. * ¬ Pr ∨¬ Rr ......... 25DeM
  27. * Pr ......... 21Simp.
  28. * ¬ Rr ......... 27,26DS
  29. * Rr ∨Lr ......... 21Simp.
  30. * Lr ......... 28,29DS
  31. * (y)[(Py • Ly) ⊃y = m] ......... 8Simp.
  32. * (Pr • Lr) ⊃r = m ......... 31UI y/r
  33. * Pr • Lr ......... 27,30Conj.
  34. * r = m ......... 33,32MP
  35. * ¬ (r = m) ......... 21Simp.
  36. * r = m • ¬ (r = m) ......... 34,35Conj.
  37. ¬ ¬ (∃x)(∃y){Px • Py • Rx • Ly • ¬ (x = y) • (z){[Pz • (Rz ∨Lz)] ⊃z = x ∨z = y}} ......... 5-36IP
  38. (∃x)(∃y){Px • Py • Rx • Ly • ¬ (x = y) • (z){[Pz • (Rz ∨Lz)] ⊃z = x ∨z = y}} ......... 37DN

Thursday 9 September 2010

Simple is difficult

Would that people asked only for complex things to be explained. Making a living would be very easy if one was in the business of providing the answers. It is the simple things that are a problem and it’s the simplicity that often brings tuts of derision.

The laws of reasoning are supposed to be very simple. Three, in particular, sound almost trivial. Suppose I look out the window and see a fresh molehill in the garden. My sentence is:

There is a fresh molehill in the garden.

It seems like nothing, but it’s a start. What is the meaning of this sentence? The meaning is: ‘true’, because it is a fact, and ‘true’, because first order deductive logic does not recognize other meanings than ‘true’ or ‘false’.

What can I infer from it? I could repeat it, for example, but it would be no use. I would end up with a tautology. I can conjoin to it any sentence I like using the disjunctive word ‘or’, including its own negation: ‘It is not the case that there is a fresh molehill in the garden’, because the meaning of ‘or’ guarantees that, in the first case, at least one of the sentences will be true, and, in the second case, precisely one sentence will be true. Either way, the whole new sentence will be true. The latter represents the Law of the Excluded Middle: either something is of a certain kind or it isn’t. There is no third option.

Suppose I don’t look out of the window, but hold the thought in my head. What is this thought? It is an assumption, or hypothesis. What is its meaning? I do not know. I can repeat my hypothesis at will because, it being a hypothesis, I am not committing myself to anything, only probing and exploring. However, assumptions are not made to be just random thoughts floating freely in my head – that is a stream of consciousness. We make an assumption in order to prove something.

I can, of course, attach to my assumption another sentence via ‘or’:

There is a fresh molehill in the garden or the gate is open.

This in itself suffices as my conclusion – something I derived from the original sentence ‘There is a molehill in the garden’, but I’m not sure I know what to do with it. Far more convenient, if less illuminating, is to repeat the hypothesis and stop at that:

Assumption: There is a molehill in the garden.
Repetition: There is a molehill in the garden.
Conclusion: If there is a molehill in the garden, then there is a molehill in the garden.

This may not be much of a conclusion, but it is in fact an illustration of the Law of Identity, which, like the Law of the Excluded Middle, holds regardless of fact. When I look out and see a molehill, the conditional sentence is true. When I look out and see no molehill, the conditional sentence is also true. If I now choose to paraphrase my two sentences:

There is a fresh molehill in the garden or it is not the case that there is a fresh molehill in the garden.

If there is a fresh molehill in the garden, then there is a fresh molehill in the garden.

then it turns out that I can paraphrase each with:

It is not the case that both there is a fresh molehill in the garden and there isn’t.

The key word in this sentence is ‘and’. The sentence illustrates the Law of Non-contradiction. We can’t say that something both is and isn’t. A closer look reveals that all these laws are in fact one law. At any rate, all three hold regardless of the facts on the ground, so to speak.

These three laws were known to the ancient Greeks, and we use them so automatically that we never give them a second thought. When the Skeptics searched for answers as to what is and each avenue led them to deeper skepticism, they turned to these fundamental laws as something unchangeable throughout time, something that could be relied on. I rely on them alright but the more I try to explain them, including to myself, the more tortuous the explanations get.

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.5E, 3(d), p. 563

Is the following a theorem of predicate logic: (x)(Ax ⊃Bx) ∨(∃x)Ax ? Yes, it is provable without premises, and the proof is straightforward.
  1. (x)(Ax ⊃Bx) ∨(∃x)Ax
  2. * ¬ (x)(Ax ⊃Bx) ......... ACP
  3. * (∃x)(Ax • ¬ Bx) ......... 2CQ
  4. * An • ¬ Bn ......... 3EI x/n
  5. * An ......... 4Simpl.
  6. * (∃x)Ax ......... 5EG
  7. ¬ (x)(Ax ⊃Bx) ⊃(∃x)Ax ......... 2-6CP
  8. ¬ ¬ (x)(Ax ⊃Bx) ∨(∃x)Ax ......... 7MI
  9. (x)(Ax ⊃Bx) ∨(∃x)Ax ......... 8DN

Thursday 2 September 2010

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.5E, problem 3f, p. 563

We are asked to show that the following statement is a theorem in predicate logic: (x)(∃y)(Ax ∨By) ≡ (∃y)(x)(Ax ∨By). Under Claim 1 we cannot just drop the quantifiers and restore them in reverse order because of the restriction saying that if every x is in a relation to some y, it doesn't follow that there is a y such that every x is in this relation to it. We are not bound by such a restriction under Claim 2.
  1. (x)(∃y)(Ax ∨By) ≡ (∃y)(x)(Ax ∨By)
  2. Claim 1: (x)(∃y)(Ax ∨By) ⊃ (∃y)(x)(Ax ∨By)
  3. * (x)(∃y)(Ax ∨By) ......... ACP
  4. * * ¬ (∃y)(x)(Ax ∨By) ......... AIP
  5. * * (y)(∃x)(¬ Ax • ¬ By) ......... 4CQ
  6. * * (∃y)(Ax ∨By) ......... 2UI x/x
  7. * * Ax ∨Bm ......... 6EI y/m
  8. * * (∃x)(¬ Ax • ¬ Bm) ......... 5UI y/m
  9. * * ¬ Aa • ¬ Bm ......... 8EI x/a
  10. * * ¬ Bm ......... 8Simp.
  11. * * Ax ......... 10,7DS
  12. * * ¬ Aa ......... 8Simp.
  13. * * (y)Ay ......... 11UG
  14. * * (∃y) ¬ Ay ......... 12EG
  15. * * ¬ Ar ......... 14EI y/r
  16. * * Ar ......... 13UI y/r
  17. * * Ar • ¬ Ar ......... 16,15Conj.
  18. * ¬ ¬ (∃y)(x)(Ax ∨By) ......... 4-17IP
  19. * (∃y)(x)(Ax ∨By) ......... 18DN
  20. (x)(∃y)(Ax ∨By) ⊃ (∃y)(x)(Ax ∨By) ......... 3-19CP
  21. Claim 2: (∃y)(x)(Ax ∨By) ⊃(x)(∃y)(Ax ∨By)
  22. * (∃y)(x)(Ax ∨By) ......... ACP
  23. * (x)(Ax ∨Bm) ......... 22EI y/m
  24. * Ax ∨Bm ......... 23UI x/x
  25. * (∃y)(Ax ∨By) ......... 24EG
  26. * (x)(∃y)(Ax ∨By) ......... 25UG
  27. (∃y)(x)(Ax ∨By) ⊃(x)(∃y)(Ax ∨By) ......... 22-26CP
  28. {(x)(∃y)(Ax ∨By) ⊃ (∃y)(x)(Ax ∨By)} • {(∃y)(x)(Ax ∨By) ⊃(x)(∃y)(Ax ∨By)} ......... 20,27Conj.
  29. (x)(∃y)(Ax ∨By) ≡ (∃y)(x)(Ax ∨By) ......... 28BE

Thursday 26 August 2010

'in' and 'at'

The preposition ‘at’ in English is a funny kind of preposition. I can make an equally strong case in my head for not having it in the language as for keeping it firmly where it belongs. If I give too much thought to one side of the argument, that’s the way the pendulum swings at that particular time.

A friend once told me that a local newspaper printed her father’s obituary saying he had died of a heart attack in the Whitstable harbour. The information was only partly true. He died of a heart attack at Whitstable Harbour. He was walking along the quayside when the fatal attack struck and he fell over. Crucially, he didn’t fall into the water – he fell on the ground.

I was reminded of this story the other day while listening to a report about miners trapped in a coalmine in Chile. The news reader said, as one would expect, that the BBC reporter was at the time ‘at the coalmine’ from where he sent the report. Had he been ‘in the coalmine’, he wouldn’t have been able to file the report, of course.

The distinction between ‘in’ and ‘at’ is necessary, but, pointless as it is, it is interesting to speculate by what reasoning our ancestors arrived at it. Were they thinking: ‘in’ for things being inside, and ‘at’ for things being … well, precisely, inside, outside, beside, neither here nor there, kind of generally in the vicinity? Or perhaps: we’ve got most relations covered by now, but where no preposition seems to fit, ‘at’ is the default option. Or maybe, after the fashion of Donald Rumsfeld: there are named and unnamed relations, and of the unnamed ones there are those we know to exist and those we don’t know to exist, and those that we don’t know we don’t know to exist, and them perhaps are best left to ‘at’.

It would seem that ‘on the top rung of the ladder’ was prior to ‘at the top of the ladder’ if we adopt the Darwinian view of language evolution: more complex concepts come after simpler concepts. The concept ‘at the top of the ladder’ is more complex because it is more ambiguous: we can mean the top rung or the second rung from the top, or perhaps even the third rung down from the top one, if the ladder is sufficiently long.

It is easy to see that ‘at’ cannot be a favourite of logicians and mathematicians. It is not precise enough. In a world consisting of objects in a plane, with no value attached to the objects (cones, cubes, spheres, etc), there is no place for relations involving ‘at’. All relations can be handled by: next to, to the left of, to the right of, over, under, etc. However, having said that, what about two non-parallel lines which meet ‘at’ point A? Well, then we come to what a point is in geometry. The point is usually left undefined. But if we try to define it, say, a set of coordinates in the Cartesian system or a circle of radius 0, then we get very precise indeed, and that in turn conflicts with our intuitive understanding of ‘at’, which is used for imprecise descriptions. A vicious circle.

Prepositions have fixed opposites: in / out, on / off, over / under, to / from, or context-dependent opposites: across, through, along, around, etc. What is the opposite of ‘at’? I can’t really say that the opposite of ‘I’ll meet you at the theatre’ is ‘I’ll meet you inside the theatre’.

There is a sense in which ‘in’ meets ‘at’ and a sense in which it doesn’t. Some distinctions seem to me to have very little basis indeed, like ‘arrive at’ and ‘arrive in’. The most satisfying test in choosing between ‘in’ and ‘at’ is to look at the implications: the Whitstable and Chile examples above, or ‘spent a lot of time in the sea’, so suffers from hypothermia; ‘spent a lot of time at sea’, so feels homesick, and so on, but that doesn’t change the fact that I feel kind of rudderless with ‘at’ and at sea without it.

Symbolic Logic, Dale Jacquette, Wadwsworth, 2001, Chpt. 8, Ex. III, problem 17, p. 434

Demonstrate that the following is a tautology:├ ¬ (z)[(∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy)] ⊃¬(x)(Fx ⊃Gx). We proceed by showing that we can derive the implication without the use of any premises other than what we assume.
  1. ├ ¬ (z)[(∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy)] ⊃¬(x)(Fx ⊃Gx)
  2. (x)(Fx ⊃Gx) ⊃(z)[(∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy)] ......... 1Contrapositive
  3. * (x)(Fx ⊃Gx) ......... ACP
  4. * * (∃y)(Fy • Hzy) ......... ACP
  5. * * Fa C Hza ......... 4EI y/a
  6. * * Fa ......... 5Simp.
  7. * * Fa ⊃Ga ......... 3UI x/a
  8. * * Ga ......... 6,7MP
  9. * * Hza ......... 5Simp.
  10. * * Ga • Hza ......... 8,9Conj.
  11. * * (∃y)(Gy • Hzy) ......... 10EG
  12. * (∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy) ......... 4-11CP
  13. (x)(Fx ⊃Gx) ⊃(z)[(∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy)] ......... 2-12CP
  14. ¬ (z)[(∃y)(Fy • Hzy) ⊃(∃y)(Gy • Hzy)] ⊃¬(x)(Fx ⊃Gx) ......... 13Contrap.

Thursday 19 August 2010

Symbolic Logic, Dale Jacquette, Wadsworth, 2001, Chpt. 8, Ex. III, problem 16, p. 434

The task: give a natural deduction proof for the following tautology, ├ (x)(Fx ⊃Gx) ⊃(y)[(∃z)(Fz • Hyz) ⊃(∃z)(Gz • Hyz)].

├ (x)(Fx ⊃Gx) ⊃(y)[(∃z)(Fz • Hyz) ⊃(∃z)(Gz • Hyz)]
  1. * (x)(Fx ⊃Gx) ......... ACP
  2. * * (∃z)(Fz • Hyz) ......... ACP
  3. * * Fa • Hya ......... 2EI z/a
  4. * * Fa ⊃Ga ......... 1UI x/a
  5. * * Fa ......... 3Simp.
  6. * * Ga ......... 5,4MP
  7. * * Hya ......... 3Simp.
  8. * * Ga • Hya ......... 6,7Conj.
  9. * * (∃z)(Gz • Hyz)] ......... 8EG
  10. * (∃z)(Fz • Hyz) ⊃(∃z)(Gz • Hyz) ......... 2-9CP
  11. (x)(Fx ⊃Gx) ⊃(y)[(∃z)(Fz • Hyz) ⊃(∃z)(Gz • Hyz)] ......... 1-10CP

Thursday 12 August 2010

Deduction, Daniel Bonevac, Blackwell Publishing, 2nd Edition, 2003, chpt. 8.3, problem 15

Show that (∃x)[Gx • (Fx ⊃Ha)] ≡ [(∃x)(Gx • Fx) ⊃Ha] is a consequence of (∃x)(y)(y = x ≡ Gy).


  1. (∃x)(y)(y = x ≡ Gy)
  2. ∴(∃x)[Gx • (Fx ⊃Ha)] ≡ [(∃x)(Gx • Fx) ⊃Ha]
  3. * (∃x)[Gx • (Fx ⊃Ha)] ... ACP
  4. * * (∃x)(Gx • Fx) ... ACP
  5. * * Gm • Fm ... 4EI x/m
  6. * * (y)(y = h ≡ Gy) ... 1EI x/h
  7. * * m = h ≡ Gm ... 6UI y/m
  8. * * (m = h ⊃Gm) •(Gm ⊃m = h) ... 7BE
  9. * * Gm ... 5Simp.
  10. * * Gm ⊃m = h ... 8Simp.
  11. * * m = h ... 9,10MP
  12. * * Gr • (Fr ⊃Ha) ... 3EI x/r
  13. * * r = h ≡ Gr ... 6UI y/r
  14. * * (r = h ⊃Gr) •(Gr ⊃r = h) ... 13BE
  15. * * Gr ... 12Simp.
  16. * * Gr ⊃r = h ... 14Simp.
  17. * * r = h ... 15,16MP
  18. * * h = m ... 11Id.
  19. * * r = m ... 17,18Id.
  20. * * Fr ⊃Ha ... 12Simp.
  21. * * Fm ... 5Simp.
  22. * * Fr ... 19,21Id.
  23. * * Ha ... 22,20MP
  24. * (∃x)(Gx • Fx) ⊃Ha ... 4-23CP
  25. [(∃x)[Gx • (Fx ⊃Ha)] ⊃[(∃x)(Gx • Fx) ⊃Ha] ... 3-24CP
  26. * (∃x)(Gx • Fx) ⊃Ha ... ACP
  27. * * ¬ (∃x)[Gx • (Fx ⊃Ha) ... AIP
  28. * * (x)[Gx ⊃(Fx • ¬ Ha) ... 27QC
  29. * * (y)(y = m ≡ Gy) ... 1EI x/m
  30. * * m = m ≡ Gm ... 29UI y/m
  31. * * (m = m ⊃Gm) •(Gm ⊃m = m) ... 30BE
  32. * * m = m ⊃Gm ... 31Simp.
  33. * * m = m ... Id.
  34. * * Gm ... 33,32MP
  35. * * Gm ⊃(Fm • ¬ Ha) ... 28UI x/m
  36. * * Fm • ¬ Ha ... 34,35MP
  37. * * ¬ Ha ... 36Simp.
  38. * * Fm ... 36Simp.
  39. * * Gm • Fm ... 34,38Conj.
  40. * * (∃x)(Gx • Fx) ... 39EG
  41. * * Ha ... 26,40MP
  42. * * Ha • ¬ Ha ... 41,37Conj.
  43. * ¬ ¬ (∃x)[Gx • (Fx ⊃Ha) ... 27-42IP
  44. * (∃x)[Gx • (Fx ⊃Ha) ... 43DN
  45. [(∃x)(Gx • Fx) ⊃Ha] ⊃[(∃x)[Gx • (Fx ⊃Ha)] ... 26-44CP


  46. [(∃x)[Gx • (Fx ⊃Ha)] ⊃[(∃x)(Gx • Fx) ⊃Ha] • [(∃x)(Gx • Fx) ⊃Ha] ⊃[(∃x)[Gx • (Fx ⊃Ha)] ... 25,45Conj.


  47. (∃x)[Gx • (Fx ⊃Ha)] ≡ [(∃x)(Gx • Fx) ⊃Ha] ... 46BE

Thursday 5 August 2010

The Sting: a cerebral movie

The Sting (directed by Roy Hill, 1973), besides its many other endearing qualities, is a two hour documentary of what happens in the mind, in a fraction of a second, when we are trying to solve a mathematical or logical puzzle. I had an opportunity to convince myself of this recently while watching it again 25 years after I first saw it.

The overall effect of the movie on the viewer is like the moment when the solution presents itself to you after you’ve been staring at an equation or a proof. It is neat and elegant, and seemingly effortless, but between the moment you think you don’t know and the moment you do the brain will have performed dozens of operations. Just how? Well, I can only know retrospectively by writing each step out on paper, and even then I can’t be sure my account is really a reflection of what has happened in my head, or only some kind of simplification.

Those steps, which I write on the right hand side of a proof, are like the title cards in the movie. And, uncannily, the title cards actually correspond to the sort of operations required in a logical proof. The Players (step 1) is a set of premises which I must list, together with the conclusion. The Set-Up (step 2) is the planning: the laying out of a strategy which I will follow. This involves thinking many steps ahead, predicting what will happen, say, on line 17 while I’m still on line 7. It’s a process which requires that many items of information be held in the head simultaneously, without letting go of any of them, while the brain is mapping out the route.

The Hook (step 3) is very much like the assumption that I must make to break up an intractable string of symbols. Like in the movie, what I do at this stage will pay off much later, more handsomely than anything I would have achieved without the assumption. But also like in the movie, I have to discharge the assumption, that is, I can’t have a sentence starting with ‘then’, ‘therefore’, ‘thus’ or ‘so’, without being bound by the ‘if’ of the assumption in the same sentence. In the movie, once Gondorff pulls the first stunt on Lonnegan at the poker table, there is no turning back. He and the boys are now committed to working towards closing the scam.

The Tale (step 4) is basically working restlessly through the material now at hand applying the various rules of inference until a workable formula crystallizes. In the movie, it is the putting together of various parts of the elaborate off-track betting den hoax.

The Wire (step 5) and The Shut-Out (step 6) are unconventional or counterintuitive moves, which may surprise the problem solver himself – something such as bringing an entirely new expression into the proof by means of a disjunction (known as addition) or considering two alternatives (proof by cases) whereby you can assure yourself of the correctness of your reasoning by getting the same result from both alternatives.

Finally, there is The Sting (step 7) – the last title card in the movie – when all the strands come together and all the loose ends are tied up. Beats me.

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.5E, problem 3b, p. 562

Show that the sentence: (x)[Ax ⊃(Ax ⊃Bx)] ⊃(x)(Ax ⊃Bx), is a theorem of predicate logic.
  1. (x)[Ax ⊃(Ax ⊃Bx)] ⊃(x)(Ax ⊃Bx)
  2. * (x)[Ax ⊃(Ax ⊃Bx)] ... ACP
  3. * * Ax ... ACP
  4. * * Ax ⊃(Ax ⊃Bx) ... 2 UI x/x
  5. * * Ax ⊃Bx ... 3,4 MP
  6. * * Bx ... 3,5 MP
  7. * Ax ⊃Bx ... 3-6 CP
  8. * (x)(Ax ⊃Bx) ... 7UG
  9. (x)[Ax ⊃(Ax ⊃Bx)] ⊃(x)(Ax ⊃Bx) ... 2-8CP

Sunday 1 August 2010

Predicate logic semantics: conditional v universally quantified statements

It is important in logic and mathematics to be able to distinguish between statements which are implications and statements which are universally quantified, because their truth value analysis can produce different results. I claim no special insight into the problem other than an observation that the otherwise flawless explanations that I often come across miss some key element which could help grasp the idea faster.

First, it is a lot easier to work with mathematical concepts in truth value analysis than with everyday English sentences. The reason: fish are aquatic animals, but suppose you were asked the question: ‘Are all fish aquatic animals?’ during a TV game show. Should you suspect a trap? Some people might. We live in an age of knowledge fragmentation where one never sees the full picture, and so can’t be sure of objective truths. The ancients and the medievals, by contrast, may have dwelled on fiction but they sought to consolidate knowledge and, conceivably, were less prone to relativist dilemmas.

Let our universe of discourse, UD, be integers then. Let Fx stand for ‘x is even’ and Gx for ‘x is evenly divisible by 2’. Our two sentences are:

(1) If a number is an even integer, then every integer is evenly divisible by 2.

(x)Fx ⊃ (x)Gx

(2) If a number is an even integer, then it is evenly divisible by 2.

(x)(Fx ⊃ Gx)

Sentence (1) is an example of material implication. Material implication is an example of a compound statement, with its truth value dependent on the logical operator ‘⊃’. In plain English, the two halves of the sentence are independent, or each is capable of taking on its own truth value. We know that sentences like this are false if and only if the consequent, here (x)Gx, is false while the antecedent, here (x)Fx, is true. Sentence (1) is patently false, because not every integer is evenly divisible by 2.

Sentence (2) can be translated to look like a conditional sentence, and in the ordinary sense of the English grammar it certainly is a conditional: For all x, if x is an even integer, then x is evenly divisible by 2. But the scope of the universal quantifier extends over the whole sentence, which means that the ‘x’ in the second half of the sentence has the same reference as the ‘x’ in the first half of the sentence, thus the sentence says in fact: All even integers are evenly divisible by 2. For this to be false, we would need to find only one counterexample, that is, some even integer that is not evenly divisible by 2. I can’t find such a counterexample. The sentence is true.

Sentences (1) and (2) are not equivalent then. However, where a conditional statement and a universally quantified statement are equivalent for some A and B, as in:

(3) (∃x)Ax ⊃ B

(4) (x)(Ax ⊃ B)

the truth values will of course be the same. The difference is in how we choose to express ourselves: (3) is false when B is false and (∃x) Ax is true, and (4) is false when for at least one A, B does not hold, but the fact remains that when (3) is false (4) will be false, and when (3) is true (4) will be true too.

Thursday 29 July 2010

Symbolic Logic, Dale Jacquette, Wadsworth, 2001, Chpt. 8, Ex. III, problem 15

The task is to prove the tautology:├ (∃x)[(Fx • ¬ Gx) ∨¬ Gx] ⊃(∃y)[(Fy ∨¬ Gy) • ¬ Gy].
  1. ├ (∃x)[(Fx • ¬ Gx) ∨¬ Gx] ⊃(∃y)[(Fy ∨¬ Gy) • ¬ Gy]
  2. * (∃x)[(Fx • ¬ Gx) ∨¬ Gx] / ACP
  3. * * ¬ (∃y)[(Fy ∨¬ Gy) • ¬ Gy] / AIP
  4. * * (y)[(Fy ∨¬ Gy) ⊃Gy] / 3QC
  5. * * (Fa • ¬ Ga) ∨¬ Ga / 2EI x/a
  6. * * (Fa ∨¬ Ga) • (¬ Ga ∨¬ Ga) / 5DeM
  7. * * ¬ Ga / 6Simp.
  8. * * Fa ∨¬ Ga / 6Simp.
  9. * * (Fa ∨¬ Ga) ⊃Ga / 4UI y/a
  10. * * Ga / 8,9MP
  11. * * Ga • ¬ Ga / 10,7Conj.
  12. * ¬ ¬ (∃y)[(Fy ∨¬ Gy) • ¬ Gy] /IP
  13. * (∃y)[(Fy ∨¬ Gy) • ¬ Gy] / 12DN
  14. (∃x)[(Fx • ¬ Gx) ∨¬ Gx] ⊃(∃y)[(Fy ∨¬ Gy) • ¬ Gy]