I am trying to translate the following English sentence to FOL, however, I am uncertain if my logic in the translation is correct. I will appreciate it very much if anyone can point out my errors.
Question: In addition to the existing predicates in FOL, we can use $Even(x)$, $Prime(x)$, the predicate "$<$", the variables $w,y,z$ and all truth-functional connectives. But we cannot use the predicate "$>$" and the predicate $Square(x)$, but we can express $Square(x)$ with the function symbol "$1x1$", for example the square of 2 is "$2x2$". The domain of discourse is in the set of natural numbers. Translate the following to FOL:
1) All squares are even.
2) No square is prime.
3) The square of some prime is prime.
4) The square of some prime other than 3 is odd.
5) The square of any number greater than 2 is greater than 3.
My Solution:
1) $\forall w Even(wxw)$
2) $\forall w \neg Prime(wxw)$
3) $\exists w (Prime(wxw) \wedge prime(w))$
4) $\exists w (Prime(w) \wedge Prime(wxw) \wedge w \neq 3)$
5) for this question, a more explicit interpretation is: "for any number $w$, if $w$ is greater than 2, then the square of $w$ is greater than 3", so:
$\forall w [(\neg(w < z) \wedge w \neq 2) \rightarrow (\neg ((wxw) < 3) \wedge (wxw) \neq3)] $
I only used one of the three given variables $w,y,z$, so I am not sure if there are any other ways to translate some of the above sentences with the rest of the variables in use. Perhaps I am missing something here?
Thanks
All correct, though I assume that for the last one you meant to write a $2$ rather than a $z$:
$\forall w [(\neg(w < \color{red}{2}) \wedge w \neq 2) \rightarrow (\neg ((w\times w) < 3) \wedge (w\times w) \neq3)] $
But yes, with a $2$ there that's correct.
But that last one can be done a bit more efficient like so:
$\forall w [2 < w \rightarrow 3 < w \times w] $