I am going back and reviewing some elementary material in logic/set theory. Among the problems in the quantifier section is the following english sentence to logic sentence translation:
Everyone likes Mary, except Mary herself.
Now, my attempt was: (∀x)[(x≠m → L(x,m)) ∧ (x=m → ¬L(x,m))], where m stands for "mary," and L(a,b) stands for "a likes b."
The solution to the exercise in the text was given as: (∀x)(x≠m → L(x,m)). I was a bit skeptical of this and searched the problem online and found another, different solution:
(∀x)(x≠m ↔ L(x,m)).
My question is, which of these three is correct? (or, if they are all/some equivalent, how?)
Ah, that is very subtle. Isn't it true, however, that either Mary likes herself, or Mary does not like herself (since (P OR ¬P) is a tautology)? And since everyone likes mary, EXCEPT Mary herself, it would have to be the case that NOT L(x,m). At least this is what comes to mind initially.
Yes.
That would follow if the statement gave an exhaustive set of people who likes Mary. In that case, it would follow that L(m,m) is false and, by the law of excluded middle (LEM), ~L(m,m). However, it is claimed that the statement says nothing about Mary, neither L(m,m) nor ~L(m,m), so the LEM does not help.
However, saying that some property holds for a smaller set of elements than it actually does is unfair to the reader even if it may be pedantically correct. For example, saying that the set {1, 2, 3} has two elements is misleading. For a real math example, Picard theorem from complex analysis is sometimes formulated as follows: "If a complex function is entire and non-constant, then it assumes all complex points except possibly one." It may be technically correct to say that it assumes all points except one, about which no claim is made, i.e., the function may assume it or it may not. However, the word "possibly" is added to describe the complete image of the function in all cases.
Therefore, I would count (∀x)[(x≠m → L(x,m)) ∧ (x=m → ¬L(x,m))] and (∀x)(x≠m ↔ L(x,m)), which are equivalent, as correct.
Finally, a joke. A passerby noticed that the sheep kept by a certain man were well fed and healthy, so he asked him:
- What do you give these sheep for food?
- Which sheep: the black or the white?
- The white, said the passerby.
- Oh, those eat the grass there.
- What about the black then?
- They eat the same grass.
Puzzled, the passerby asks:
- What about drink then?
- Excuse me, are you asking about the black sheep or the white ones?
- The white, I guess.
- Well, normal water from that river.
- OK, what about the black then?
- Same water.
Frustrated, the passerby yells at the man:
- If everything is the same why do you keep asking me about white and black?
- The white sheep are mine.
- Oh, I'm sorry... And the black sheep?
- They are also mine.
Thank you both for your responses, Plato and emakarov. I finished up the remainder of the exercises in the section and only came upon three others which I was uncertain about:
1) "Anyone who has bought a Rolls Royce with cash must have a rich uncle."
My question with this problem was less with the translation and more with the use of predicates. My translation was: (∀x)[B(x,r) → R(x)], where r stands for "Rolls Royce," B(a,b) stands for "a has bought b with cash," and R(c) stands for "c has a rich uncle." In particular, I was wondering if using either of the predicates B(x) or B(x,r,c) for "x has bought a Rolls Royce with cash" and "x has bought a Rolls Royce (denoted r) with cash (denoted c)," respectively, are acceptable (that is, is what is 'included' in the predicate up to one's discretion)?
2) If nobody failed the test, then everybody who got an A will tutor someone who got a D.
I simply felt unsure about this one for some reason: ¬(∃w)F(w) → (∀x)[A(x) → ((∃y)D(y) → (∃z)T(x,z))], where F(x) stands for "x failed the test," A(y) stands for "y got an A on the test," y(z) stands for "z got a D on the test," and T(a,b) stands for "a tutors b."
Lastly, "x is prime" where the universe of discourse is the natural numbers.
My attempt was: x > 1 ∧ (∀j)(∀k)(x = jk → (j = 1 or j = x))
The solution had a similar statement, except that they stated that there do not exist natural numbers y,z such that x =yz and y and z are both less than x. To me these both seem correct. Is this true?
I will let the logician comment on the first two.
However, I do agree with you on the last.
You were clever in forcing k to be 1 or x.
A prime number is a integer with exactly two divisors.
That gets around the bogus definition that one sees in print (math-ed material) "a prime number is an integer is divisible only by itself and 1".
But of course by that definition 1 is prime.
I agree with 1).
This depends on the problem requirement or other external circumstances. At one extreme, you can denote the whole proposition by one letter. I think, a reasonable and slightly mode detailed formalization is (∀x)[B(x,r) → (∃y)(U(x,y) ∧ R(y))] where U(x,y) means "y is x's uncle" and R(y) means "y is rich."
Should be
¬(∃w)F(w) → (∀x)[A(x) → (∃y)(D(y) ∧ T(x,y))]
As it is, the person x who got an A tutors some z who did not necessarily got a D. Also, if some people got As but nobody got either a D or an F, then the original English statement is false while the formula in the quote is true since the premise (∃y)D(y) is false.