besides, except

Inclusion and exclusion are critical for reasoning, and these two words are supposed to make it easier to factor things in or out. It is not always so. The prepositions ‘besides’ and ‘except’ present both grammatical and logical challenges.

‘Besides’ adds, ‘except’ subtracts:

Everyone besides the vicar gambles.
Everyone except the vicar gambles.

This distinction seems to break down in negative E-type sentences where both words exclude:

No one besides / except the vicar gambles.

Various other combinations of ‘besides’ and ‘except’ with A, E, I, and O-type sentences produce more or less felicitous outcomes, including anomalies such as:

Someone except for the vicar gambles.
Some people except for the vicar gamble.

Logically speaking, that is when we take logic to be common sense, the first sentence is odd. The second sentence holds up, especially when we use intonation to dramatic effect (successful replacement candidates for ‘except’ would be ‘but not’ or ‘other than’). On the strict understanding of ‘some’ in logic though, both sentences are OK. Each says: there is at least one person who gambles and that person is not the vicar.

If we give the vicar a name, for example: Some people except for Michael gamble, then the translation into FOL comes out as:

(∃x)[¬(x = m) • Gx] • ¬ Gm

This says: Some people other than Michael gamble, and Michael doesn’t gamble. We need the part: ¬ Gm, to show that ‘m’ has the subject property but not the predicate property (we could have fleshed out our symbolization by saying that ‘x’ is a person, or ‘Px’ if we wanted to – it would have set off the meaning but crowded the formula).

We ought to be able to show something similar for:

Some people except for the vicar gamble.

The difference this time is that we have to accommodate a definite description. It would be very awkward to maneuver a definite description into our sentence so that ‘some people’ would have scope over it. My suggestion is: Some people who are not a vicar gamble, and the vicar doesn’t gamble. Again, breaking it down this way shows that ‘x’ has the subject property but does not have the predicate property. The truth value is also preserved. Net loss: we are saying that some people who are not a or any vicar, rather than the vicar from our example, gamble.

(∃x)( ¬Vx • Gx) • (∃x)[Vx • (y)(Vy ⊃y = x) • ¬ Gy]

The corresponding sentence for inclusion: Some people besides the vicar gamble, would then be:

(∃x)( ¬Vx • Gx) • (∃x)[Vx • (y)(Vy ⊃y = x) • Gy]

A few more examples:

Everyone besides the vicar gambles.

(x)(¬ Vx ⊃Gx) • (∃x)[Vx • (y)(Vy ⊃y = x) • Gy]

Everyone except the vicar gambles.

(x)(¬ Vx ⊃Gx) • (∃x)[Vx • (y)(Vy ⊃y = x) • ¬ Gy]

No one besides / except the vicar gambles.

(∃x){Gx • (y)(Gy ⊃ y = x) • (∃z)[Vz • (w)(Vw ⊃ w = z) • z = x]}

The last statement is of course equivalent to: The only person who gambles is the vicar.

Deductive Logic, Warren Goldfarb, Hackett, 2003, Part IV, Ex. 13(a-b)

Russell's way of expressing definite descriptions was: (∃x)[Fx • (y)(Fy ⊃y = x)]; Quine preferred (∃x)(y)(Fy ≡ x = y). They are equivalent. We show this by first working one way and then the other (line 26). The equivalence is expressed as a conjunction of two conditionals (line2), then each conditional is proved separately. There are other ways of proving this equivalence - for one we could have condensed lines 28-34 using pre-nex rules rather than my cumbersome instantiation, then generalization, then instantiation again, but I have done it to show that it is possible. We can generalize within an assumption as long as the variable we generalize from is not free on the first line of the assumption.

1. (∃x)[Fx • (y)(Fy ⊃y = x)] ≡ (∃x)(y)(Fy ≡ x = y)
2. {(∃x)[Fx • (y)(Fy ⊃y = x)] ⊃(∃x)(y)(Fy ≡ x = y)} • {(∃x)(y)(Fy ≡ x = y) ⊃ (∃x)[Fx • (y)(Fy ⊃y = x)]} / 1BE
3. (∃x)[Fx • (y)(Fy ⊃y = x)] ⊃ (∃x)(y)(Fy ≡ x = y) / 2Simp.
4. * (∃x)[Fx • (y)(Fy ⊃y = x)] / ACP
5. * * ¬ (∃x)(y)(Fy ≡ x = y) / AIP
6. * * (x)(∃y) ¬ (Fy ≡ x = y) / 5CQ
7. * * (x)(∃y) ¬ [(Fy ⊃x = y) • (x = y ⊃Fy)] / 6BE
8. * * Fa • (y)(Fy ⊃y = a) / 4EI x/a
9. * * Fa / 8Simp.
10. * * (y)(Fy ⊃y = a) / 8Simp.
11. * * (∃y) ¬ [(Fy ⊃a = y) • (a = y ⊃Fy)] / 7UI x/a
12. * * ¬ [(Fm ⊃a = m) • (a = m ⊃Fm)] / 11EI y/m
13. * * ¬ (Fm ⊃a = m) ∨¬ (a = m ⊃Fm)] / 12DeM
14. * * Fm ⊃m = a / 10UI y/m
15. * * Fm ⊃a = m / 14Id
16. * * ¬ (a = m ⊃Fm) / 15,14DS
17. * * ¬ [¬ (a = m) ∨Fm] / 16MI
18. * * a = m • ¬ Fm / 17DeM
19. * * ¬ Fm / 18Simp.
20. * * ¬ (a = m) / 9,19Id
21. * * a = m / 18 Simp.
22. * * a = m • ¬ (a = m) / 20,21Conj.
23. * ¬ ¬ (∃x)(y)(Fy ≡ x = y) / 5 - 22IP
24. * (∃x)(y)(Fy ≡ x = y) / 23DN
25. (∃x)[Fx • (y)(Fy ⊃y = x)] ⊃ (∃x)(y)(Fy ≡ x = y) / 4 - 24CP
26. (∃x)(y)(Fy ≡ x = y) ⊃ (∃x)[Fx • (y)(Fy ⊃y = x)] / 2Simp.
27. * (∃x)(y)(Fy ≡ x = y) / ACP
28. * (y)(Fy ≡ a = y) / 27EI x/a
29. * Fy ≡ a = y / 28UI y/y
30. * (Fy ⊃ a = y) • (a = y ⊃Fy) / 29BE
31. * Fy ⊃ a = y / 30Simp.
32. * (y)(Fy ⊃ a = y) / 31UG
33. * a = y ⊃Fy / 30Simp.
34. * (y)(a = y ⊃Fy) / 33UG
35. * a = a ⊃Fa / 34UI y/a
36. * a = a / Id
37. * Fa / 36,35MP
38. * Fa • (y)(Fy ⊃a = y) / 37,32Conj.
39. * (∃x)[Fx • (y)(Fy ⊃x = y)] / 36EG
40. (∃x)(y)(Fy ≡ x = y) ⊃ (∃x)[Fx • (y)(Fy ⊃y = x)] / 27 - 39CP
41. {(∃x)[Fx • (y)(Fy ⊃y = x)] ⊃(∃x)(y)(Fy ≡ x = y)} • {(∃x)(y)(Fy ≡ x = y) ⊃ (∃x)[Fx • (y)(Fy ⊃y = x)]} / 25,40Conj.
42. (∃x)[Fx • (y)(Fy ⊃y = x)] ≡ (∃x)(y)(Fy ≡ x = y) / 41BE

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 4th edition, 2004, Ex. 10.5, 5(f)

We are asked to show that a set of sentences is inconsistent. The solution comes down to showing that we can derive a sentence and its negation within the scope of only the primary assumptions. In instantiating line 3 by Existential Instantiation we have to choose a constant other than 'a', as that already occurs in our premises. Working through the set, we derive 'Kam' on line 12 and '¬ Kam' on line 15, which shows that our sentences are indeed inconsistent.

1. (x)[(Sx • Bxx) ⊃Kax] / Premise
2. (x)(Hx ⊃Bxx) / Premise
3. (∃x)(Sx • Hx) / Premise
4. (x) ¬ (Kax • Hx) / Premise
5. Sm • Hm / 3EI x/m
6. Hm ⊃Bmm / 2UI x/m
7. Hm / 5Simp.
8. Bmm / 7,6MP
9. (Sm • Bmm) ⊃Kam / 1UI x/m
10. Sm / 5Simp.
11. Sm • Bmm / 8,10Conj.
12. Kam / 11,9MP
13. ¬ (Kam • Hm) / 4UI x/m
14. ¬ Kam ∨¬ Hm / 13DeM
15. ¬ Kam / 7,14DS

'because' - a non-truth-functional operator

The temptation to treat the operator because as truth functional is a good enough reason to remind ourselves periodically why it is not. The fact that because is not truth-functional follows directly from the definition of a function.

A function is a relation which assigns a member of the domain x one and only one member of the range f(x). A function may take more than one input x to produce the same output, but this much is certain: for each input there is only one output. How does that bear on because?

For comparison, let us start with the conjunction and, as in: p • q. We let p be ‘It is cloudy’ and q ‘it is raining’. The outcome: 'It is cloudy and it is raining', is true iff p is true and q is true. Our output sentence is true irrespective of whether there is a connection between clouds and rain. Thus, two truths produce a truth if the connective is and. If we let p be ‘An isosceles triangle has two equal sides’, which is true, and q ‘A prime number is a whole number larger than 1 that is divisible only by 1 and itself’, which is also true, our reason for uttering: 'An isosceles triangle has two equal sides and a prime number is a whole number larger than 1 that is divisible only by 1 and itself’ may not be immediately clear, but there is nothing illegitimate about the sentence. In both cases, the input is two T’s (two true sentences), and in both cases the final sentence comes out true.

It is easy to see that this is not the case with because. Let p ‘The apple dropped down’ be true, and q ‘the apple was overripe’ be true too. Our output sentence is: ‘The apple dropped down because it was overripe,’ is true, and we attribute the truth to the connection between the apple falling and its being overripe. Two true sentences produce one true one. But what about the next pair of sentences above? To say that ‘An isosceles triangle has two equal sides because a prime number is a whole number larger than 1 that is divisible only by 1 and itself’ is definitely not true. One thing has nothing to do with the other. Thus, two T’s (two true sentences) do not always produce a true output sentence when connected via because.

This violates our definition of a function, which assigns one and only one member of the range to each member of the domain. The examples above show that because assigns once T (true) once F (false) to the same value of the input. The truth value of the compound is not completely determined by the truth values of the component sentences alone. We look to our knowledge of the external world for the answer to whether the compound sentence is true or not.

The operator because is variously handled in first order logic by and or if, and the sentence it introduces is absorbed into a subordinate clause.

Deductive Logic, Warren Goldfarb, Hackett, 2003, Part IV, Ex. 5(c)

The proof of this argument is not particularly challenging but it is notable for the application of the first law of identity which says that everything is identical with itself, and which we can introduce at any point without justification (line 5). Since we are asked to show an equivalence, we have to work both ways.

1. (x)(y)(x = y ⊃Gxy)
2. ∴(x)Gxx
3. (y)(x = y ⊃Gxy) / 1UI x/x
4. x = x ⊃Gxx / 3UI y/x
5. (x)(x = x) / Id
6. x = x / 5UI x/x
7. Gxx / 6,4 MP
8. (x)Gxx / 7UG

1. (x)Gxx
2. ∴(x)(y)(x = y ⊃Gxy)
3. * x = y / ACP
4. * Gxx / 1UI x/x
5. * Gxy / 3,4Id
6. x = y ⊃Gxy / 3-5CP
7. (y)(x = y ⊃Gxy) / 6UG
8. (x)(y)(x = y ⊃Gxy) 7UG

Deductive Logic, Warren Goldfarb, Hackett, 2003, Part IV, Ex. 5(b)

We set out to show that the following schema is valid using the laws of identity. Once an assumption has been made for indirect proof, we set up Modus Tollens to derive the identity 'a = m' on line 13. Next we continue instantiating line 5, which we stopped on line 7. Since the Convention says that the instantiating variable must be free or the instantiating constant must occur in the instantiating statement in at least all those places where the variable is free in the 'decapitated' general statement (see Suber, Copi), we are allowed to instantiate (z)(Gaz ⊃¬ Gzm) into Gaa ⊃¬ Gam. Then it is just a matter of replacing 'm' with 'a' on line 16 based on the previously derived identity.

1. (x)Gxx
2. (x)(y)[¬(x = y) ⊃(∃z)(Gxz • Gzy)
3. ∴(x)(y)(∃z)(Gxz • Gzy)
4. * ¬(x)(y)(∃z)(Gxz • Gzy) / AIP
5. * (∃x)(∃y)(z)(Gxz ⊃¬ Gzy) / 4CQ
6. * (∃y)(z)(Gaz ⊃¬ Gzy) / 5EI x/a
7. * (z)(Gaz ⊃¬ Gzm) / 6EI y/m
8. * (z)(¬Gaz ∨¬ Gzm) / 7MI
9. * (z) ¬ (Gaz • Gzm) / 8DeM
10. * ¬ (∃z)(Gaz • Gzm) / 9CQ
11. * (y)[¬(a = y) ⊃(∃z)(Gaz • Gzy) / 2UI x/a
12. * ¬ (a = m) ⊃(∃z)(Gaz • Gzm) / 11UI y/m
13. * a = m / 10,12MT
14. * Gaa / 1UI x/a
15. * Gaa ⊃¬ Gam / 7UI z/a
16. * ¬ Gam / 14,15MP
17. * ¬ Gaa / 13,16Id
18. * Gaa • ¬ Gaa / 14,17Conj.
19. ¬ ¬ (x)(y)(∃z)(Gxz • Gzy) / 4 - 18IP
20. (x)(y)(∃z)(Gxz • Gzy) / 19DN

Material conditional

I can’t pretend to have any special way of explaining the truth values of the material conditional, so I am only going to say which of the explanations I’ve seen works best for me. First, a list of some of the explanations and examples I’ve come across which are meant to throw some light on the workings of the material conditional:

- Use a conditional that expresses a promise, for example: If you get an A, you will pass the exam. What if I don't get an A? I didn't say you won't pass if you don't get an A, so the sentence is true, and I wouldn't have lied if you didn't get an A and didn't pass.

- The last two lines of the truth table are irrelevant anyway, because we do not deliberately use a conditional when we know that the antecedent is false. So we may as well make conditionals with false antecedents true.

- If x < 2 then x < 4. Substitute '5' for 'x'. What do you get? A true sentence, with a false antecedent and a false consequent.

- The majority of people, when asked, would say that a conditional is true as long as the consequent is true.

- Antecedent and consequent are deliberately unrelated and often absurd, as in this classic example: If the earth is flat, the moon is made of green cheese. Turning this conditional around by contraposition, which preserves its truth value, shows that it is true in fact: If the moon is not made of green cheese, then the earth is not flat.

The explanation I find most convincing is the one which exploits the distinction between what is asserted and what is implied, and is in fact similar to the first listed (conditional promise). For example:

(1) If it is after 5 pm, the shop is closed.

I can’t in the same breath say that if it is after 5 pm indeed, the shop is closed and is not closed. One of the statements must be false. If I accept that ‘the shop is closed’ is true, then ‘the shop is not closed’ is false. That’s the first two lines of the truth table done.

It’s a quarter to five and the shop is closed – is this true or false? I have never said in my original statement that if it is before 5 pm the shop is not closed. If you got that impression, you have overinterpreted what I have said. You have read too much into it. It could well be that the shop is closed before and after 5 pm. The sentences are not contradictory.

It’s a quarter to five and the shop is not closed – true or false? Again, there is nothing in this sentence that contradicts my original assertion. The sentence is true. You may think that this, the 4th line of the truth table, excludes the preceding situation, but we are not comparing these two! We are relating every line to the original assertion.

Deductive Logic, Warren Goldfarb, Hackett, 2003, Part IV, Ex. 5(a)

We are to show by deduction, using the laws of identity, that the following formula: (∃x)Fxy • (x) ¬ Fxz implies: ¬ (z = y). If we first assume z = y by indirect proof, then drop the quantifiers in the premise, we will obtain Fay and ¬ Faz. Substituting 'y' for 'z', or the other way round, we will obtain a contradiction, which will prove that ¬ (z = y) does indeed follow from the original premise. However, if we choose not to use the laws of identity (below), we must be careful to generalize by UG the free variables 'y' and 'z' first before we use indirect proof. Doing things in reverse order will mean we have to generalize them within the assumption, which will violate the rule which says that we can't generalize within the scope of an assumption if the instantiating variable occurs free in the first line of the sequence.

1. (∃x)Fxy • (x) ¬ Fxz
2. ∴¬ (z = y)
3. (∃x)Fxy / 1Simp.
4. (y)(∃x)Fxy / 3UG y/y
5. (x) ¬ Fxz / 1Simp.
6. (z)(x) ¬ Fxz / 5UG z/z
7. * z = y / AIP
8. * (∃x)Fxm / 4UI y/m
9. * (x) ¬ Fxm / 6UI z/m
10. * Fam / 8EI x/a
11. * ¬ Fam / 9EI x/a
12. * Fam • ¬ Fam / 10,11Conj.
13. ¬ (z = y) / 7 - 12IP