I have a few logic statements and am unsure how to formulate them

  1. A student is excellent if he can solve all the logic questions
  2. No student can solve all the logic questions
  3. there is a logic problem no student can solve
  4. No student is excellent

I was thinking of formulating it the following way:

  1. forall_yS(x,y)->E(x) (for all logic questions y if solve student x questions, then x is excellent)
  2. ~exsit_xforall_yS(x,y) (not exist student x for all questions y student solve questions)
  3. exist_yforall_x~S(x,y) (exist question y for all student x student not solve question)
  4. ~exist_xE(x) (not exist x such that x is excellent)

How can i be sure if i'm wrong or right ?


One way to check your translations from English to logic is to read back to yourself strictly what you have symbolized. For example, a strict reading of your second sentence would be

$$\lnot \exists x\forall y(S(x, y))$$ "There does not exist a something such that for all things, it is not the case that the something solves all the things." Now, you may know what you meant to say, but the bottom line is you failed to specify the kind of thing that does not exist, and the kind of things that are solved.

Correcting for that we'll create a translation key to cover all the bases. You should always do this, so you and others will understand the translation, and reproduce, more or less, the original statement.

$E(x):\;$ "$x$ is excellent."

$S_t(x):\;$ "$x$ is a student."

$Q(x):\;$ "$x$ is a logic question."

$S_o(x, y):\;$ "$x$ solves y".

Now, we translate "No student can solve all the logic questions."

$$\lnot \exists x\Big(S_t(x) \land \forall y(Q(y) \rightarrow S_o(x, y))\Big)$$

Now, using a translation key similar to the one here, try to rewrite your translations to capture all that the sentences declare.

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  1. You should always give your translation key in answering this sort of exercise. What is the domain of quantification? That needs to be made clear, as Mauro points out. And what does $S$ mean? Does $S(x, y)$ say e.g. that (a) $x$ can solve $y$, or that (b) $x$ is a student and $y$ is a logic problem and $x$ can solve $y$, for example? This sort of thing should be made explicit too.

  2. The first sentence is intended to be read as generalization, but the translation has a dangling free variable. You want at least

$$\forall x(\forall yS(x,y) \to E(x))$$

or maybe something more like

$$\forall x([T(x) \land \{\forall y(Ly \to S(x,y)\}] \to E(x))$$

(with $T(x)$ saying that $x$ is a student, etc.), depending whether you are giving interpretation (a) or (b) to $S$.

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  • $\begingroup$ what about the other statements ? $\endgroup$ – Lena Bru Jun 9 '14 at 7:40
  • $\begingroup$ How do you make the statements look good ? $\endgroup$ – Lena Bru Jun 9 '14 at 7:41
  • 1
    $\begingroup$ Just rudely demanding answers is not exactly the best way of eliciting friendly responses here. $\endgroup$ – Peter Smith Jun 9 '14 at 11:43
  • $\begingroup$ @LenaBru I suspect that "How do you make the statements look good?" is referring to the latex/mathjax "typesetting" used on this site. You can see the way it was done by clicking "edit" under a question or an answer and viewing the source. There is also info on the math.stackexchange help page. $\endgroup$ – joeA Jun 11 '14 at 22:35

It might be useful to collectivize some of the logical properties into sets, such as "student", "logic question", "excellent student". So you'd have

  1. $\forall x\in X,\,\forall y\in Q,\,S(x,y)\implies x\in E$
  2. $\forall x\in X,\,\exists y\in Q,\,¬S(x,y)$
  3. $\exists y\in Q,\,\forall x\in X,\,¬S(x,y)$
  4. $\forall x\in X,\,x\notin E$

where $X$ stands for the set of students, $Q$ for that of logic questions and $E$ for the (sub)set of excellent students. Note that if you define $E$ to be a subset of $X$ the last sentence could be read $E=\varnothing$ ($E$ is empty). I've given alternative representations of the middle sentences for you to compare, yours are correct.

Also, be careful if you're taking the first sentence as a definition, for they're commonly understood as of the 'iff' class even when you don't write it down.

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You may have some problems with the domain of discourse.

In your problem, you have to "speak of" students and questions; thus, your domain of discourse must contain both "types" of objects.

When you translate :

4) No student is excellent

with :

$\lnot \exists x E(x)$

where the predicate $E$ means "... is excellent", are we sure that there are no questions which are excellent ?

If we have a question $q$ such that $E(q)$ holds, $\lnot \exists x E(x)$ will be false. In order to avoid this problems, we must add two more predicates : $Q$ for "... is a question" and $St$ for " ... is a student".

Thus we may translate 4) as :

$\lnot \exists x (St(x) \land E(x))$.

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