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Consider the statement

No-one is right all the time

I have a question which requires me to express this in terms of predicate logic. I have attempted this by first defining the following predicates:

\begin{align*} &\underline{P}(x) : \text{ $x$ is a person} \\ &\underline{T}(x) : \text{ $t$ is a time} \\ &\underline{R}(x,t) : \text{ $x$ is right at time $t$} \end{align*} I have then attempted to write the argument in terns of this predicates, in the following form:

$$ \neg (\exists x)(\underline{P}(x) \wedge (\forall t) (\underline{T}(t) \rightarrow \underline{R}(x,t))) $$

However, I have the solution to this question and, although I have defined the predicates correctly, the expression should be $$ (\forall x) \underline{P}(x) \rightarrow \neg \forall t(\underline{T}(t) \vee \underline{R}(x,t)) $$ Are these expressions equivalent, or is my expression incorrect? If my attempt is not correct, why not?

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  • $\begingroup$ Your formula has seven open parentheses and six close parentheses. Could you fix it? $\endgroup$ – Fabio Somenzi May 21 '17 at 20:29
  • $\begingroup$ I have amended it. $\endgroup$ – M Smith May 21 '17 at 20:30
  • $\begingroup$ Thanks. Note that for your sentence to be true there should be no $x$ such that $P(x)$ holds. You need to change the $\vee$ into a $\wedge$ to get a sentence equivalent to the given solution. $\endgroup$ – Fabio Somenzi May 21 '17 at 20:36
  • $\begingroup$ Sorry, that was also a typo. $\endgroup$ – M Smith May 21 '17 at 21:16
  • $\begingroup$ It looks OK now. $\endgroup$ – Fabio Somenzi May 22 '17 at 0:45
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While your expression is a correct symbolization of the English sentence, the sentence you indicate as the 'solution' is not!

The problem is that in the 'solution' $x$ is a free variable in the term $R(x,t)$. Fortunately, that's fixed easily enough by adding some parentheses (maybe you forgot to type those in?):

$$\forall x (P(x) \rightarrow \neg \forall t (T(t) \lor R(x,t)))$$

OK, but is that equivalent to your sentence? Again, I can immediately see that your sentence is correct, but this sentence I have a hard time understanding unless we do some algebra. In particular, let's bring in the negation, so we get:

$$\forall x (P(x) \rightarrow \exists t \neg (T(t) \lor R(x,t)))$$

DeMorgan gets us:

$$\forall x (P(x) \rightarrow \exists t (\neg T(t) \land \neg R(x,t)))$$

OK, ... That's definitely not right, as this says 'for everyone there is something that is not a time, and that they are not right at (... that time ... But it isn't a time. ... Definitely something wrong there already ...). In fact, that statement can be made true trivially by picking for every person themselves: since a person is not time, there is indeed such a thing. But clearly 'no one is right all the time say something interesting, not trivial, so the 'solution' is still not correct!

To correct for this second mistake, we can add a $\neg$ in front of the $T(t)$ in the original sentence, so that we get:

$$\forall x (P(x) \rightarrow \neg \forall t (\neg T(t) \lor R(x,t)))=$$

$$\forall x (P(x) \rightarrow \exists t \neg (\neg T(t) \lor R(x,t)))=$$

$$\forall x (P(x) \rightarrow \exists t (T(t) \land R(x,t)))$$

Ah! Now, this is something we can easily translate back into an actully meaningful English sentence: 'Everyone is not right at some time'. In fact, that is (at least intuitively) indeed equivalent to 'No one is right all the time'

To formally show that equivalence, we can do some more algebra:

Bringing in the negation in your sentence, we get:

$$\forall x \neg (P(x) \land \forall t (T(t) \rightarrow R(x,t)))$$

DeMorgan:

$$\forall x (\neg P(x) \lor \neg \forall t (T(t) \rightarrow R(x,t)))$$

Implication:

$$\forall x (P(x) \rightarrow \neg \forall t (T(t) \rightarrow R(x,t)))$$

Implication:

$$\forall x (P(x) \rightarrow \neg \forall t (\neg T(t) \lor R(x,t)))$$

And yes, that is indeed the same as what I proposed we coul crrect the mistaken 'solution' sentenc into.

But long story short: Your sentence is correct, and it is superior to the 'solution' in three ways:

  1. The 'solution' is not correct!

  2. Even if we make some small changes to the 'solution' so that it does become correct, it is hard to see that it indeed correctly captures the intended English sentence, given its 'weird' interplay between the negations and the quantifiers (put differently: the original sentence is not in 'aristotelean' format, where as your sentence is

  3. Even if we put the corrected 'solution' sentence in more readable Atistotelean format, it would still translate a little differently from the oiginal English sentence - 'everyone is incorrect at least sometimes' is equivalent to 'No one is right all the time', but again your sentence is a direct translation of 'no one is right all the time'

So: job well done on your part! And job not well done with the provided 'solution'!

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