Subject or predicate

The sentence:

A fox is a scavenger.

represents no cognitive challenge, with the ‘A fox’ being the subject of the sentence and ‘is a scavenger’ being the predicate. The subject is at least one or else all members of a class (particular v universal) which the predicate tells us something about. There are also singular statements where an assertion is made about a specifically named person or thing: Mr. Tod is a fox. Reversing the subject and predicate does pose a cognitive challenge:

A scavenger is a fox.

The problem here is that we have defined a broader term with a narrower one, which jars immediately. Needless to say, it is the freak sentences that keep the mind buzzing.

The sentence: Mr. Tod is a fox, translates into the language of FOL (first order logic) simply as: Fj. The sentence: A fox is Mr. Tod, requires an existential quantifier and goes into symbols as: (∃x)(Fx • x = t), where ‘t’ stands for Mr. Tod. This sentence is very, very weak. It says that there exists an individual such that it is a fox and it is called Mr. Tod. An example of a credible sentence is wanting, and any suggestions are welcome.

The definite description: The fox is Mr. Tod, requires a more robust translation: (∃x)]Fx • (y)(Fy ⊃ y = x) • x = t], and says that there is exactly one fox that is called Mr. Tod. As an example of immediate application, I can, for example, ask a friend about the fellow who came to last night’s fancy dress party dressed as a fox, and be told that the fox was Mr. Tod.

This frivolous what-is-what dilemma was prompted by a sentence I heard on the news:

The interests of the country are the priority.

When compared with: The priority is the interests of the country, the difference is as between: Mark Twain was Samuel Clemens, and Samuel Clemens was Mark Twain, there being no apparent difference at all. Identity is commutative, so the sentences are equivalent.

Identity aside, what would it take for the A scavenger is a fox sentence not to cause a dissonance? It seems to me that when we put the extensional rather than intentional spin on a sentence, we can convince ourselves that it sounds OK. The intentional definition of a scavenger is ‘an animal that eats anything it can find’. The extensional definition lists examples: foxes, rats, crows, and so on. Thus:

A scavenger is a fox, for example.

is perfectly well-formed. On paper (on screen) the trick was accomplished by tacking on ‘for example’, but in the head the ‘for examples’ can be dispensed with. At any rate, the mind resets itself from intentional to extensional mode even before the ‘for example’ is added. That the mind can reset itself like this can be gleaned from saying the sentence in quick succession many times: … is a fox is a scavenger is a fox is a scavenger is a fox is a scavenger is a fox is a scavenger is a … As the mind wanders from one to the other, the subject and predicate are in a state of indeterminacy.

Understanding Symbolic Logic, Virginia Klenk, Prentice Hall, 2008, 5th edition, Unit 20, 1(l), p. 381

There are, I'm sure, other ways of coming to grips with the proof of this argument, exploiting perhaps some of the disjunctions, but the one below suggested itself to me first and here it is.

There are exactly three composers in the room. Exactly one of the composers in the room is a pianist. Any composer in the room who is not a pianist is an opera singer. Therefore, there are at least two opera singers in the room.

1. (∃x)(∃y)(∃z){Cx • Cy • Cz • Rx • Ry • Rz • ¬ (x = y) • ¬ (x = z) • ¬ (y = z) • (w)[(Cw • Rw) ⊃(w = x ∨ w = y ∨ w = z)]}
2. (∃x){Cx • Rx • (y)[(Cy • Ry) ⊃y = x] • Px}
3. (x)[(Cx • Rx • ¬ Px) ⊃Ox]
4. ∴(∃x)(∃y)[Ox • Oy • Rx • Ry • ¬ (x = y)]
5. * ¬ (∃x)(∃y)[Ox • Oy • Rx • Ry • ¬ (x = y)] ......... AIP
6. * (x)(y)[(Ox • Oy • Rx • Ry) ⊃ (x = y)] ......... 5 QC
7. * (∃y)(∃z){Ca • Cy • Cz • Ra • Ry • Rz • ¬ (a = y) • ¬ (a = z) • ¬ (y = z) • (w)[(Cw • Rw) ⊃(w = a ∨ w = y ∨ w = z)]} ......... 1 EI x/a
8. * (∃z){Ca • Cm • Cz • Ra • Rm • Rz • ¬ (a = m) • ¬ (a = z) • ¬ (m = z) • (w)[(Cw • Rw) ⊃(w = a ∨ w = m ∨ w = z)]} ......... 7 EI y/m
9. * Ca • Cm • Cr • Ra • Rm • Rr • ¬ (a = m) • ¬ (a = r) • ¬ (m = r) • (w)[(Cw • Rw) ⊃(w = a ∨ w = m ∨ w = r)] ......... 8 EI z/r
10. * ¬ (a = m) ......... 9 Simp.
11. * Ci • Ri • (y)[(Cy • Ry) ⊃y = i] • Pi ......... 2 EI x/i
12. * (y)[(Cy • Ry) ⊃y = i] ......... 11 Simp.
13. * (Cm • Rm) ⊃m = i ......... 12 UI y/m
14. * Cm • Rm ......... 9 Simp.
15. * m = i ......... 14,13 MP
16. * ¬ (a = i) ......... 15,10 Id
17. * (Ca • Ra) ⊃a = i ......... 12 UI y/a
18. * ¬ (Ca • Ra) ......... 16,17 MT
19. * ¬ Ca ∨¬ Ra ......... 18 DeM
20. * Ca ......... 9 Simp.
21. * ¬ Ra ......... 20,19 DS
22. * Ra ......... 9 Simp.
23. * Ra • ¬ Ra ......... 21,22 Conj.
24. ¬ ¬ (∃x)(∃y)[Ox • Oy • Rx • Ry • ¬ (x = y)] ......... 5-23 IP
25. (∃x)(∃y)[Ox • Oy • Rx • Ry • ¬ (x = y)] ......... 24 DN

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 5th edition, 2008, Unit 20, Ex. 1(k), p. 381

Here is an argument which works equally well on the intuitive and on the symbolic level, as 'at least' and 'at most' in relation to the same quantity are simply the long way of saying 'exactly'. Perhaps the trickiest thing here is to realize that we have to instantiate line 2 more than once, first into constants, then into variables, and then again into constants, to free up the desired predicates.

There are at least two pianists in the room. All the pianists in the room are composers. There are at most two composers in the room. Therefore, there are exactly two pianists in the room.

1. (∃x)(∃y)[Px • Py • Rx • Ry • ¬ (x = y)]
2. (x)[(Px • Rx) ⊃ Cx]
3. (x)(y)(z)[(Cx • Cy • Cz • Rx • Ry • Rz) ⊃(z = x ∨z = y ∨y = x)]
4. ∴(∃x)(∃y){Px • Py • Rx • Ry • ¬ (x = y) • (z)[(Pz • Rz) ⊃(z = x ∨z = y)]}
5. (∃y)[Pa • Py • Ra • Ry • ¬ (a = y)] ......... 1 EI x/a
6. Pa • Pm • Ra • Rm • ¬ (a = m) ......... 5 EI y/m
7. (Pa • Ra) ⊃ Ca ......... 2 UI x/a
8. Pa • Ra ......... 6 Simp.
9. Ca ......... 8, 7 MP
10. * Pz • Rz ......... ACP
11. * (y)(z)[(Ca • Cy • Cz • Ra • Ry • Rz) ⊃(z = a ∨z = y ∨y = a)] ......... 3 UI x/a
12. * (z)[(Ca • Cm • Cz • Ra • Rm • Rz) ⊃(z = a ∨z = m ∨m = a)] ......... 11 UI y/m
13. * (Ca • Cm • Cz • Ra • Rm • Rz) ⊃(z = a ∨z = m ∨m = a) ......... 12 UI z/z
14. * (Pz • Rz) ⊃ Cz ........ 2 UI x/z
15. * Cz ......... 10, 14 MP
16. * Pm • Rm ......... 6 Simp.
17. * (Pm • Rm) ⊃ Cm ......... 2 UI x/m
18. * Cm ......... 16, 17 MP
19. * Ra ......... 8 Simp.
20. * Rm ......... 16 Simp.
21. * Rz ......... 10 Simp.
22. * Ca • Cm • Cz • Ra • Rm • Rz ......... 9, 15, 18, 19, 20, 21 Conj.
23. * z = a ∨z = m ∨m = a ......... 13, 22 MP
24. * ¬ (a = m) ......... 6 Simp.
25. * ¬ (m = a) ......... 24 Comm.
26. * z = a ∨z = m ......... 23, 25 DS
27. (Pz • Rz) ⊃ z = a ∨z = m ......... 10-26 CP
28. (z)[(Pz • Rz) ⊃ z = a ∨z = m ] ......... 27 UG
29. Pa • Pm • Ra • Rm • ¬ (a = m) • (z)[(Pz • Rz) ⊃ z = a ∨z = m ] ......... 6, 28 Conj.
30. (∃y){Pa • Py • Ra • Ry • ¬ (a = y) • (z)[(Pz • Rz) ⊃ z = a ∨z = y ]} ......... 29 EG m/y
31. (∃x)(∃y){Px • Py • Rx • Ry • ¬ (x = y) • (z)[(Pz • Rz) ⊃ z = x ∨z = y ]} ......... 30 EG a/x

Work and time problems

Snatches of conversation about work planning in the office drift in my direction during my lightning visits to print teaching materials and make photocopies. Person A, working alone, can translate a document in 4 hours – I overhear – while person B, working alone, can do the same job in 6 hours. So, if we get them to share the job, we’ll have the job done in 5 hours (1/2 times 4 + 1/2 times 6).

That is not exactly true. Although the answer is much less, and although lots of other factors may distort the calculation significantly in real life, and although it does not really matter that much after all, here are a couple of examples how to get it exactly right.

Example 1

Person A can do a job in 4 hours, while person B can do the same job in 6 hours. How long would it take both, working together, to do the same job?

Problems of this kind often involve figuring with the principle of 1, where 1 represents the entire job (one job). It is clear that we must somehow add the times of A and B to get this entire job. But how? Let t be the number of hours it takes A and B to complete the job working together. Person A can do the job faster, or more of it in the same time, or, a fourth of the job against a sixth of the job which person B can do in the same time (1/4 is greater than 1/6). Our equation will thus be:

t(1/4) + t(1/6) = 1

By solving the equation for t, we get t = 2.4 hours

Problem 2

We can reverse the situation while at the same time running a check on the previous problem.

It takes persons B 2 hr longer to do a job than it takes person A. If they work together, they can do the job in 2.4 hours. How long would it take each, working alone, to do the job?

Let 2.4 be the number of hours it takes A and B to complete the job working together. Person A can do 1/t of the job while person B can do 1/(t + 2) of it in the same time. So:

2.4(1/t) + 2.4[1/(t + 2)] = 1

We will get two answers, one of which is negative, hence meaningless, while the other is t = 4. This is the time it takes A to do the job. B will do it in 4 + 2 hours.

Understanding Symbolic Logic, Virginia Klenk, Pearson Prentice Hall, 5th edition, 2008, Unit 20, Ex. 1(j), p. 381

Proving the conclusion in this argument requires some footwork but the path eventually suggests itself. The first premise is written here as a single sentence, but since 'and' is a coordinate conjunction, I have written the second clause on a separate line for clarity. There may well be a quicker route which I have missed, but I blame it on the late hour.

Adam and Eve were the only people in the Garden of Eden, and they were tempted. Anyone in the Garden of Eden who was tempted succumbed to temptation. Anyone who succumbed to temptation was kicked out of the Garden of Eden. Therefore, everyone in the Garden of Eden was kicked out.

1. (∃x)(∃y){Px • Py • Ex • Ey • (z)[(Pz • Ez) ⊃(z = x ∨z = y)] • x = a • y = e}
2. Ta • Te
3. (x)[(Px • Ex • Tx) ⊃Sx]
4. (x)[(Px • Sx) ⊃Kx]
5. ∴(x)[(Px • Ex) ⊃Kx]
6. * ¬ (x)[(Px • Ex) ⊃Kx] ......... AIP
7. * (∃x)(Px • Ex • ¬ Kx] ......... 6 QC
8. * Pm • Em • ¬ Km ......... 7 EI x/m
9. * (∃y){Pn • Py • En • Ey • (z)[(Pz • Ez) ⊃(z = n ∨z = y)] • n = a • y = e} ...... 1 EI x/n
10. * Pn • Pr • En • Er • (z)[(Pz • Ez) ⊃(z = n ∨z = r)] • n = a • r = e ......... 9 EI y/r
11. * Pn • En ......... 10 Simp.
12. * n = a ......... 10 Simp.
13. * Ta ......... 2 Simp.
14. * Tn ......... 13,12 Id
15. * Pn • En • Tn ......... 11,14 Conj.
16. * (Pn • En • Tn) ⊃Sn ......... 3 UI x/n
17. * Sn ......... 15,16 MP
18. * Pn ......... 11 Simp.
19. * Pn • Sn ......... 17,18 Conj.
20. * (Pn • Sn) ⊃Kn ......... 4 UI x/n
21. * Kn ......... 19,20 MP
22. * (z)[(Pz • Ez) ⊃(z = n ∨z = r)] ......... 10 Simp.
23. * (Pm • Em) ⊃(m = n ∨m = r) ......... 22 UI z/m
24. * Pm • Em ......... 8 Simp.
25. * m = n ∨m = r ......... 23,14 MP
26. * ¬ Km ......... 8 Simp.
27. * ¬ (m = n) ......... 21, 26 Id
28. * m = r ......... 27, 25 DS
29. * r = e ......... 10 Simp.
30. * m = e ......... 28, 29 Id
31. * Pm ......... 24 Simp.
32. * Pe ......... 30,31 Id
33. * Em ......... 24 Simp.
34. * Ee ......... 30,31 Id
35. * Te ......... 2 Simp.
36. * Pe • Ee • Te ......... 32,34,35 Conj.
37. * (Pe • Ee • Te) ⊃Se
38. * Se ......... 36,37 MP
39. * Pe • Se ......... 32,38 Conj.
40. * (Pe • Se) ⊃Ke ......... 4 UI x/e
41. * Ke ......... 39,40 MP
42. * Km ......... 30, 41 Id
43. * Km • ¬ Km ......... 26,42 Conj.
44. ¬ ¬ (x)[(Px • Ex) ⊃Kx] ......... 6-43 IP
45. (x)[(Px • Ex) ⊃Kx) ......... 44 DN

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

Symbolize and prove the following argument. The second sentence can be written as either one or two premises.

All John's pets are Siamese cats. John has at least one pet and at most one Siamese cat. Therefore, John has exactly one pet, which is a Siamese cat.

1. (x)[(Px • Hjx) ⊃Sx]
2. (∃x){Px • Hjx • (y)(z)[(Sy • Hjy • Sz • Hjz) ⊃y = z]}
3. ∴(∃x){Px • Hjx • (y)[(Py • Hjy) ⊃y = x] • Sx}
4. Pa • Hja • (y)(z)[(Sy • Hjy • Sz • Hjz) ⊃y = z] ......... 2EI x/a
5. Pa • Hja ......... 4Simp.
6. (Pa • Hja) ⊃Sa ......... 1UI x/a
7. Sa ......... 5,6MP
8. * Py • Hjy ......... ACP
9. * ( Py • Hjy) ⊃Sy ......... 1UI x/y
10. * Sy ......... 8,9MP
11. * (y)(z)[(Sy • Hjy • Sz • Hjz) ⊃y = z] ......... 4Simp.
12. * (z)[(Sy • Hjy • Sz • Hjz) ⊃y = z] ......... 11UI y/y
13. * (Sy • Hjy • Sa • Hja) ⊃y = a ......... 12UI z/a
14. * Hjy ......... 8Simp.
15. * Hja ......... 5Simp.
16. * Sy • Hjy • Sa • Hja ......... 10,14,7,15Conj.
17. * y = a
18. Py • Hjy ⊃y = a ......... 8-17Cp
19. (y)[(Py • Hjy) ⊃y = a] ......... 18UG y/y
20. Pa ......... 5Simp.
21. Pa • Hja • (y)[(Py • Hjy) ⊃y = a] • Sa ......... 20,15,19,7Conj.
22. (∃x){Px • Hjx • (y)[(Py • Hjy) ⊃y = x] • Sx} ......... 21EG a/x