As the title said, what is the best translation for "All students have already done the exam for all course except course X"? Me and my fellow students debated for quite a long time about this question. We use x is a domain of students and y is a domain of the courses, and Predicate P(x,y) is x already done exam course X. My answer is $$\forall x \forall y (P(x,y) \iff (y \neq Course X))$$ One of my friend answer is $$\forall x \forall y(P(x,y) \implies (y \neq Course X))$$ and the other answer is $$\forall x \forall y (P(x,y) \implies (y \neq Course X)) \land \exists x(\lnot P(x, Course X))$$ Which one do you think is right? And if none are right, what is the translation and why?
Quantifier Logic : Translation of sentence "All students have already done the exam for all course except course X"
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On
$\forall x,y (P(x,y) \iff (y \neq Course X))$ can also be accepted, meaning no one has done exam for $X$ but all did other exams
EDIT
$$\forall x \forall y ((y \neq Course X) \implies P(x,y)) \land \exists x(\lnot P(x, Course X))$$ So the other answer you provided is close to being right only it has to switch places in the implication. This means each exam not being $X$ has been done but also fact is that not everyone has done $X$, so there is someone who didn't do it
On
Definitions:
variables: [($s$: student),($c$: course)]
constant: [$X$: course X]
relation: [$T(s,c)$: student $s$ has taken the exam for course $c$]
Interpretation 1:
$\forall s\forall c((c\neq X)\Rightarrow T(s,c))$.
In words: If a course, $c$, is not course $X$ then students have taken the exam for course $c$.
Interpretation 2(contrapositive of 1):
$\forall s\forall c(\neg T(s,c)\Rightarrow(c=X))$.
In words: If students have not taken the exam for course $c$, then course $c$ is course $X$.
On
Actually, all the existing answers do not accurately convey the same meaning as the original sentence (ignoring the grammatical errors), as interpreted by a native English speaker.
The actual meaning is:
Every student has taken the exam for every course $C$ that is not course $X$, but there is some student who has not taken the exam for course $X$ (otherwise we would not say "except $X$").
Which translates to:
$\forall s \in Students\ \forall C \in Courses\ ( C \ne X \to ( \text{$s$ has taken the exam for $C$} ) ) \\\ \land \exists s \in Students\ ( \text{$s$ has not taken the exam for $X$} )$
And I'll leave it to you to finish translating it to pure first-order logic.
Consider $S$ be the domain of students.
And $C:$ be the domain of course tests.
Let $\text{X-Course}$ be the "name" of the course to be excluded (err, the course that no student has yet taken a test in).
Let $P(x, y)$ denote "$x$ has taken $y$."
$$\forall x \in S, \forall y \in C \Big((y \neq \text{X-Course})\to P(x, y) \Big)$$
Or we can say, $$\forall x \in S, \forall y \in (C\setminus \{\text{X-Course}\}) \Big(P(x, y)\Big)$$
where $C\setminus \{\text{X-Course}\}$ is the set of all course tests, except for the test for Course $X$.
$\Huge($ And if you insist on the fact that we are to assume that the domain for $x$ is "All students", and the domain for $y$ is "All courses", with $P(x, y)$ meaning $x$ has taken the test in course $y$, and $X$ the name for the course in which no test has yet been taken,
Then at its simplest, we have: $$\forall x, \forall y\Big(\big(y\neq X\big) \ \rightarrow P(x, y)\Big)\Huge)$$
Alternatively, let the domain $D$ of x, y, consist of all persons, places, things.
Let $S(x)$ denote that element $x\in D$ is a student.
Let $C(x)$ denote that element $x \in D$ is a test for an academic course.
Let $P(x, y)$ denote "x has taken y".
Let $\text{X-Course }$ name the "not-yet-if-ever tested academic course X"
Then the sentence you are given can be translated as follows:
$$ \forall x, \forall y\Big(\big(S(x) \land C(y) \land (y \neq \text{X-course})\big) \to P(x, y)\Big)$$