Truth of a sentence in predicate logic

Question

I'm working through the exercises in this Logic problem set and I'm struggling to understand ex. 7.5:

Determine whether the following sentence is logically true in predicate logic:

There is someone such that, if he or she is asleep, everyone is asleep.

The solutions section suggests this is True and gives a formal 'Natural Deduction' proof that I don't really understand (it probably doesn't help that I don't have a copy of the book the solutions keep referring to).

Is there another way to approach this?

Greg Nisbet · Answer 1 · 2021-03-16T21:29:13.743

This is a classic example of the mismatch between the meaning of if in classical logic and its ordinary intuitive meaning. It is called the drinker's paradox.

Let's use the fact that $\varphi \to \psi$ is equivalent to $\lnot \varphi \lor \psi$ in classical logic to paraphrase the sentence in natural language and avoid the troublesome if.

There exists a person w such that w is not asleep or everyone is asleep.

Let $c$ be an arbitrary person in our group of people.

Suppose everyone is asleep, then we pick $w$ to be $c$. Since everyone is asleep, this value of $w$ satisfies the condition.

Suppose at least one person is awake, call them $a$. Then we pick $a$ as $w$.

I think the reason why this is counterintuitive is that the choice of $w$ depends on who happens to be asleep right now or not, but the phrasing of the question in natural language implies that a fixed $w$ has some special property that someone causally determines which configurations of awake and asleep people are possible.

The paradox goes away completely if we give the conjunction or wide scope over exists. We can do this unambiguously in natural language by reordering the dijsuncts. Reordering the dijsuncts, and changing the order of the disjunction and existential quantifier in this specific case are both allowed in classical logic.

Everyone is asleep or there's a person who is not asleep.

This is a really helpful answer, especially identifying it as the drinker's 'paradox'. Would it be fair to say that this is analogous to the distinction in real analysis between continuity and uniform continuity? We might intuitively interpret the sentence to mean 'there exists a person such that for all sleep states, if the person is asleep everyone is asleep', when the correct formulation is 'for all sleep states, there exists a person [who may depend on the sleep state] such that...'? — JCW, Mar 16 '21 at 21:49
That's a good way of thinking about it, but you need to be careful. You can analyze the sentence as having an implicit universal quantification over states of the world, but in a first order logic setting that quantification is part of the semantics, you can't reach out and touch it directly and can't reorder it with a "real quantifier". I asked a similar question two years ago that some knowledgeable folks answered. FWIW, I was thinking of the sleep state as "constant but unknown" in my answer. — Greg Nisbet, Mar 16 '21 at 23:14

score 2 · Answer 2 · answered Mar 16 '21 at 20:56

2

Case 1: There is someone who is not asleep. In that case, let $p$ be the person who is not asleep.

Then we see the following implication holds: if $p$ is asleep, then everyone is asleep. This is vacuously true, since $p$ is not asleep.

Then there indeed exists some person $p$, such that if $p$ is asleep, everyone is asleep.

Case 2: There is not anyone who is not asleep. In that case, everyone is asleep. Let $p$ be some person.

Then we see that the following implication holds: if $p$ is asleep, then everyone is asleep. This is true because everyone is, in fact, asleep.

Then there indeed exists some person $p$, such that if $p$ is asleep, everyone is asleep.

answered Mar 16 '21 at 20:56

Mark Saving

33,541

That would have been my argument. It seems entirely trivial by comparison to the long chain of formal manipulations given as the solution, so I assumed it had to be faulty. Is there any advantage that such an opaque argument can offer that an elegant one like this doesn't? – JCW Mar 16 '21 at 21:05
1

@JCW A proof in formal logic can be viewed as just a way of precisely encoding an informal proof. I could, for instance, take my above proof and turn it into a purely formal proof. Formal proofs have the benefit that they can (in principle) be checked by a machine. Informal English proofs cannot be checked by a computer. But for human comprehensibility, it's usually better to give an informal proof. – Mark Saving Mar 16 '21 at 21:09

score 2 · Answer 3 · answered Mar 16 '21 at 21:02

This is a sometimes odd-sounding feature of existentially quantified implications.

If we express the statement formally it's: $$\exists x (A(x) \to \forall y\, A(y)).$$ Consider two cases:

everyone is asleep, or
not everyone is asleep.

In case 1, $\forall y\, A(y)$ holds, and hence the implication $A(x) \to \forall y\, A(y)$ would hold (without worrying about whether $x$ is asleep or not),

In case 2, we can find someone so that $A(x)$ is false, hence again the implication must be true.

score 1 · Answer 4 · answered Mar 17 '21 at 10:21

[...] It seems entirely trivial by comparison to the long chain of formal manipulations given as the solution, so I assumed it had to be faulty. [...]

The problem with your assumption is that formal proofs from scratch need to prove every single bit of the reasoning involved. In particular, how exactly do you justify that either "there is someone who is not asleep" or "there is not anyone who is not asleep"? Depending on your specific chosen formal system, this may be an axiom called LEM (law of excluded middle) or require a few lines of proof. Similarly, to get from the latter to "everyone is asleep" again may be allowed in one inference step or require some more lines of proof.

Here is a completely formal proof using a Fitch-style natural deduction system with LEM. The key to get an intuitive proof is to use LEM, so if the system does not support LEM then you simply prove the instance of LEM that you need! Shorter proofs in a formal system may be less intuitive.

If ∃x∈People:  [✰]
  ∃y∈People ( ¬Asleep(y) ) ∨ ¬∃y∈People ( ¬Asleep(y) ).  [LEM]
  If ∃y∈People ( ¬Asleep(y) ):
    Let c∈People such that ¬Asleep(c).
    If Asleep(c):
      Contradiction.
      ∀x∈People ( Asleep(x) ).
    Asleep(c) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  If ¬∃y∈People ( ¬Asleep(y) ):
    Let d∈People.  [from ✰]
    If Asleep(d):
      Given any x∈People:
        If ¬Asleep(x):
          ∃y∈People ( ¬Asleep(y) ).
          Contradiction.
        Asleep(x).
      ∀x∈People ( Asleep(x) ).
    Asleep(d) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).

Important note: The statement in your question is actually not a logical tautology. As you should see from the above proof, it depends crucially on the assumption that there exists some person, otherwise it is trivially false.

~ ~ ~ ~ ~ ~ ~

Also, contrary to popular belief, the apparent paradox here is not really due to any mismatch between the English "if" and logical "if"! Rather, it is due to incorrect translation of the English sentence, due to the use of the generic present tense! Notice that the following sentence is actually true when interpreted according to standard English:

There is someone such that, if he or she is asleep at midnight on 16 Mar 2021, then everyone is asleep at midnight on 16 Mar 2021.

The problem with the original sentence was that because it used the present tense without a context specifying a single time, the time ended up being quantified outside the "if", and so it ended up being interpreted as:

There is a person A such that, at every time t, if A is asleep at time t, then everyone is asleep at time t.

Such implicit quantifiers do not go past explicit quantifiers, which is why it did not go all the way to the outside. Note that the logical tautology corresponds precisely to that:

At every time t, there is a person A such that, if A is asleep at time t, then everyone is asleep at time t.

I hope this partly linguistic explanation satisfies your inquiry into the apparent paradox, and yes as you guessed it is a quantifier swap, though not quite involving sleep-states but involving time. On the other hand, be aware that in general it is difficult to set down any rules to translate English into logical form. After all, there are still many people who insist on saying "All that glitters is not gold." instead of "Not all that glitters is gold."...

Is time here not just another way of allowing for transitions between what I called 'sleep states'? — JCW, Mar 17 '21 at 10:39
I find the initial part of this answer a bit confusing, because we prove relatively much more important things, like almost all theorems in mathematics, without trying to formalise them in this way - if we believe this formalisation is necessary to be sure an argument correct, why do we restrict it to such relatively trivial examples? — JCW, Mar 17 '21 at 10:41
@JCW: In this special case, it's pretty much the same whether you think about it as involving time or sleep-states. My point was simply that if you look at other apparent paradoxes of the same kind, it won't be about "sleep" anymore, but the key issue will still be the tense. As for your second question, here we are talking about formal proofs. From the logic viewpoint we simply don't have formal proofs to most of mathematics (yet). That is why there have been some instances of wrong proofs, sometimes even of false 'theorems', even accepted for years. — user21820, Mar 17 '21 at 10:45

Dan Christensen · Answer 5 · 2021-03-17T00:44:19.280

0

Assume we have a non-empty universe of people, i.e. there exists at least one person. Using a simple form of natural deduction, we have the following proof by contradiction (lines 2 -13) where