Saturday, 27 February 2010

The many ways of being wrong

The key pass which I get at reception at North Gate, an office tower in Warsaw, on my way to teaching an English class says: “Return of the card at the exit.” I can’t make up my mind which part of it grates more: the noun ‘return’ or the double use of ‘the’.

One would expect to see: “Return card at exit,” but that expectation is not out of respect for the English grammar with regard to the use of articles, nor out of a sense of confusion arising from a declarative rather than imperative use of ‘return’. In “Return of the card at the exit” the articles are used correctly but unnecessarily, while the choice of a noun instead of a verb may merely reflect the building manager’s quirky notion of telling me about it rather than asking me to do it.

There is something very compelling about missing off the articles in ephemeral messages of this kind, a practice conveying efficiency or a sense of urgency perhaps, so when they are inserted where they properly belong, the messages don’t grab us with anything like the force they have minus articles. As we can usually point at the things in our immediate surroundings (ostensive function of language) which the message refers to, some liberties are allowed. I am as yet mystified though why my Avast virus protection software plays a recorded message with a reassuring pop-up window saying: ‘Virus database has been updated,’ and the accompanying text runs: ‘A new version of virus database has been installed.’

Neil Armstrong is in yet another category with his memorable line: ‘That’s one small step for [a] man, one giant leap for mankind.’ I don’t know who is right – those who claim he didn’t say the ‘a’ or those who claim he did. The latest explanation is that apparently the indefinite article ‘a’ was lost in transmission static. Said a cynic to a sceptic: that shows you how important articles are! One way or another, from the logical point of view, both versions of this proposition are contradictions.

Consider this: if there is a man for whom the step taken on that momentous day was not ‘big’, then it is not true that it was a big step for every man. Alternatively, if Armstrong did miss off the article, thus equating ‘man’ with ‘mankind’, then he effectively said that it was not a big step for man and it was a big step for man, again committing a contradiction.

Fortunately, natural language does not need to follow logic all the time, otherwise the sentence: ‘Joe got drunk and lost his way home’ would be equivalent to ‘Joe lost his way home and got drunk,’ seeing that conjunction is commutative. We understand Armstrong’s line despite the grammar and the logic, rather than because of it.

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 4th edition, 10.6, ex.4(d)

We are asked to symbolize the argument and prove its validity. The argument is:

Rebecca loves those and only those who love her. The brother of Charlie loves Rebecca. Sam is Charlie's brother. So Sam and Rebecca love each other.
  1. (x)(Lxr ≡ Lrx)
  2. (∃x)[Bxc • (y)(Byc ⊃y = x) • Lxr)
  3. Bsc
  4. ∴ Lsr • Lrs
  5. * ¬ (Lsr • Lrs) / AIP
  6. * ¬ Lsr ∨¬ Lrs / 5DeM
  7. * Lsr ≡ Lrs / 1UI x/s
  8. * (Lsr ⊃Lrs) • (Lrs ⊃Lsr) / 7BE
  9. * Lsr ⊃¬ Lrs / 6MI
  10. * Lrs ⊃Lsr / 8Simp.
  11. * Lrs ⊃¬ Lrs / 9,10HS
  12. * ¬ Lrs ∨¬ Lrs / 11MI
  13. * ¬ Lrs / 12Taut.
  14. * Bac • (y)(Byc ⊃y = a) • Lar / 2EI x/a
  15. * (y)(Byc ⊃y = a) / 14Simp.
  16. * Bsc ⊃s = a / 15UI y/s
  17. * s = a / 3,16MP
  18. * Lar / 14Simp.
  19. * Lsr / 17,18Id
  20. * Lsr ⊃Lrs / 8Simp.
  21. * Lrs / 19,20MP
  22. * Lrs • ¬ Lrs / 21,13Conj.
  23. ¬ ¬ (Lsr • Lrs) / 5-22IP
  24. Lsr • Lrs / 23DN

Saturday, 20 February 2010

Rubbish logic

The city has decided to remove litter bins from the streets and lanes feeding commuters from their flats to my local underground station. The pen pushers and bean counters reckoned it was better that the locals should drop litter on the pavement than that the council should have to empty out litter bins overflowing with household rubbish.

There is a rubbish shed round the back of each house where people are supposed to take out their household rubbish. Rubbish collection rates are included in the monthly rent and rubbish collection is weekly or twice weekly.

It is all well and good when everyone has the time to take their rubbish out and put it in a skip inside the rubbish shed. The system usually works at weekends. On weekdays, though, few people have the time to make a special trip to the rubbish shed, usually requiring going in the opposite direction to one’s destination, when they are in a hurry to catch the underground or get in the car and join the race to the city. Instead of letting rubbish accumulate under the kitchen sink from day to day, people make up last night’s bin contents into a small bundle and take it along with them on the way to the car or the underground to be disposed of in one of the litter bins provided by the city.

The city has caught on to their game. Why should the city be paying to have people’s household rubbish taken away? Better to remove the litter bins altogether. The city can’t lose. Either the people will be forced to take out their rubbish to the rubbish shed, or they will deposit it in front of the entrance to their building or else dump it on the pavement. Of the latter two options, the first will sooner or later make it thoroughly unpleasant for all neighbours, who will take matters into their own hands, perhaps by naming and shaming the offenders, or the offenders will get into the crosshairs of traffic wardens patrolling the streets and be forced to pay fines. One nil to the city.

On an ungenerous interpretation, the city’s reasoning can be summed up like this:

If the city provides litter bins, people will dump their household rubbish in them.
If the city removes the litter bins, dumping will cease. Problem solved.

This is a fallacious argument (denying the antecedent), never mind the arrogance. It is a version of laissez faire economics: stand back and let the situation sort itself out. But since the city cannot defend itself (it is unlikely that it will) in these pages, let us be fair and let us redeem the city by pretending that the reasoning was somewhat like this:

If the city doesn’t provide litter bins, people will either dump their rubbish at the entrance to their building or dump it on the street. If they dump it at the entrance to their building, then their neighbours will intervene. If they dump household rubbish on the street, then the traffic wardens will intervene. If either the neighbours intervene or the traffic wardens do, then people will take household rubbish to the rubbish sheds. Therefore, if the city doesn’t provide litter bins, people will take household rubbish to the rubbish sheds.

This is actually a logically valid argument:
  1. ¬ P ⊃(E ∨S)
  2. E ⊃N
  3. S ⊃T
  4. (N ∨T) ⊃R
  5. ∴ ¬ P ⊃R
  6. * ¬ P / ACP
  7. * E ∨S / 6,1MP
  8. * (E ⊃N) • (S ⊃T) / 2,3Conj.
  9. * N ∨T / 7,8CD
  10. * R / 9,4MP
  11. ¬ P ⊃R / 6-10CP

Quad erat demonstrandum.

Symbolic Logic, Irving M. Copi, Prentice Hall, 5th edition, 1979, ex. 2, p.132

The aim is to construct a formal proof of validity for the following argument. Hint: best not to mess around with a straight assumption for conditional proof, however promising it might look, because we are bound to fall foul of the principle whereby the instantiating variable, or the instantiating constant, must be free in at least all those places in the instantiating statement where the general variable is free in the 'decapitated' (with the quantifier removed) general statement. Thus, indirect proof, which produces an elegant deduction.
  1. (x)[(∃y)Byx ⊃(z)Bxz]
  2. ∴ (y)(z)(Byz ⊃Bzy)
  3. * ¬ (y)(z)(Byz ⊃Bzy) / AIP
  4. * (∃y)(∃z)(Byz • ¬ Bzy) / 3CQ
  5. * (∃z)(Baz • ¬ Bza) / 4EI y/a
  6. * Bam • ¬ Bma / 5EI z/m
  7. * Bam / 6Simp.
  8. * ¬ Bma / 6Simp.
  9. * (∃y)Bym ⊃(z)Bmz / 1UG x/m
  10. * (∃z) ¬ Bmz / 8EG
  11. * ¬ (z)Bmz / 10CQ
  12. * ¬ (∃y)Bym / 9,11MT
  13. * (y) ¬ Bym / 12CQ
  14. * ¬ Bam / 13UI y/a
  15. * Bam • ¬ Bam / 7,14Conj.
  16. ¬ ¬ (y)(z)(Byz ⊃Bzy) / 3-15IP
  17. (y)(z)(Byz ⊃Bzy) / 16DN

Sunday, 14 February 2010

Predicate Logic, Howard Pospesel, Prentice Hall, 2nd edition, 2003, chpt. 11, ex. 21

A line from the poem Desiderata: If you compare yourself with others, you may become vain and bitter; for always there will be greater and lesser persons than yourself, is spun out into an argument, which goes like this:

If one compares oneself with greater people, one will become bitter. And if one compares oneself with lesser people, one will become vain. If people compare themselves with others, they will compare themselves with some who are greater as well as some who are lesser. Consequently, if you compare yourself with others, you will become vain and bitter.

The domain (universe of discourse) is people. For simplicity's sake, I keep the symbolisation to the minimum. The aim is to prove the argument. Our glossary: Gxy - x is greater than y, Lxy - x is lesser than y, Bx - x is bitter, Vx - x is vain, Cxxy - x compares oneself to y.


  1. (x)(y)[(Gyx • Cxxy) ⊃Bx]
  2. (x)(y)[(Lyx • Cxxy) ⊃Vx]
  3. (x)(y){Cxxy ⊃[(∃z)(Gzx • Cxxz) • (∃w)(Lwx • Cxxw)]}
  4. ∴ (x)(y)[Cxxy ⊃(Vx • Bx)]
  5. * Cxxy / ACP
  6. * (y){Cxxy ⊃[(∃z)(Gzx • Cxxz) • (∃w)(Lwx • Cxxw)]} / 3UI x/x
  7. * Cxxy ⊃[(∃z)(Gzx • Cxxz) • (∃w)(Lwx • Cxxw)] / 6UI y/y
  8. * (∃z)(Gzx • Cxxz) • (∃w)(Lwx • Cxxw) / 5,7MP
  9. * (∃z)(Gzx • Cxxz) / 8Simp.
  10. * Gax • Cxxa / 9EI z/a
  11. * (y)[(Gyx • Cxxy) ⊃Bx] / 1UI x/x
  12. * (Gax • Cxxa) ⊃Bx / 11UI y/a
  13. * Bx / 10,12MP
  14. * (∃w)(Lwx • Cxxw) / 8Simp.
  15. * Lmx • Cxxm / 14EI w/m
  16. * (y)[(Lyx • Cxxy) ⊃Vx] / 2UI x/x
  17. * (Lmx • Cxxm) ⊃Vx / 16UI y/m
  18. * Vx / 15,17MP
  19. * Bx • Vx / 13,18Conj.
  20. Cxxy ⊃(Vx • Bx) / 5-19CP
  21. (y)[Cxxy ⊃(Vx • Bx)] / 20UG
  22. (x)(y)[Cxxy ⊃(Vx • Bx)] / 21UG

Thursday, 4 February 2010

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 4th edition, 2004, Ex. 10.5, 3(d)

The task is to show that the statement: (x)(Ax ⊃Bx) ∨(∃x)Ax, is a theorem of predicate logic. Before we show the workings, we can reason like this: let 'Ax' be 'x' is a tiger; 'Bx' - 'x' is a man-eater. Then, we have: Either all tigers are man-eaters or tigers exist. Since the statement is a disjunction, it is false if and only if both disjuncts are false. If there is a tiger that is not a man-eater, our left disjunct will be false, but that still makes the whole sentence true, because tigers do exist. If we let: 'Ax' be 'x' is a unicorn, and 'Bx' - 'x' is white, then the existential claim on the right is of course false, but the universal claim on the left is true. Precisely because no unicorns exist, the statement: All unicorns are white, is true. So again, we have failed to force both disjuncts to be false.


In proving this theorem we can go a number of ways. We can assume the negation of one of the disjuncts (doesn't matter which), then discharge the assumption and simplify, or we can negate the whole theorem and show a contradiction. Both methods follow.

Show: (x)(Ax ⊃Bx) ∨(∃x)Ax
  1. * ¬ (∃x)Ax / ACP
  2. * (x) ¬ Ax / 1CQ
  3. * ¬ Ax / 2UI x/x
  4. * ¬ Ax ∨Bx / 3Add.
  5. * Ax ⊃Bx / 4MI
  6. * (x)(Ax ⊃Bx) / 5UG
  7. ¬ (∃x)A x ⊃ (x)(Ax ⊃Bx) / 1-6CP
  8. (∃x)A x ∨ (x)(Ax ⊃Bx) / 7MI
  9. (x)(Ax ⊃Bx) ∨(∃x)Ax / 8Comm.
Show: (x)(Ax ⊃Bx) ∨(∃x)Ax
  1. * ¬ [(x)(Ax ⊃Bx) ∨(∃x)Ax] / AIP
  2. * ¬ (x)(Ax ⊃Bx) • ¬ (∃x)Ax / 1DeM
  3. * ¬ (x)(Ax ⊃Bx) / 2Simp.
  4. * (∃x)(Ax • ¬ Bx) / 3CQ
  5. * Am • ¬ Bm / 4EI x/m
  6. * Am / 5Simp.
  7. * ¬ (∃x)Ax / 2Simp.
  8. * (x) ¬ Ax / 7CQ
  9. * ¬ Am / 8UI x/m
  10. * Am • ¬ Am / 6,9Conj.
  11. ¬ ¬ [(x)(Ax ⊃Bx) ∨(∃x)Ax] / 1-10IP
  12. (x)(Ax ⊃Bx) ∨(∃x)Ax / 11DN