Universe of x: all students
Universe of y: all courses
A(y): y is an advanced level course
F(x): x is a first-year student
T(x,y): x is taking y
Translate "No first-year student is taking an advanced level course"
Is my answer correct: $\forall x[F(x) \implies \lnot \forall y [A(y) \implies T(x,y)]]$
Also is this logically equivalent to the above: $ \lnot \exists x [F(x) \implies \land
\forall y[A(y) \implies T(x,y)]]$
Also is there any way to write the above statement with the for all or for each at the beginning? Thanks
The first answer is close ...
It should be:
$\forall x (F(x) \to \neg \exists y (A(y) \land T(x,y)))$
That is: for any first-year student it is not true that there is some advanced level course that they are taking
Now, there are several equivalent expressions. First, we can bring the negation inside to get:
$\forall x (F(x) \to \forall y (A(y) \to \neg T(x,y)))$
We can now also bring the universal out:
$\forall x \forall y (F(x) \to (A(y) \to \neg T(x,y)))$
And now we can do an Exportation to get:
$\forall x \forall y ((F(x) \land A(y)) \to \neg T(x,y))$
Finally, we can bring the negation back out, and get:
$\neg \exists x \exists y (F(x) \land A(y) \land T(x,y))$