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