Translating English into First Order Logic Translate the following into a formula of first-order logic. "A language L that is regular will have
the following property: there will be some number N (that depends on L) such that if s is a string
in L (a string is a sequence of characters) whose length is at least N then s can be written as $xyz$
where y is not the empty string and $xy^i
z$ is in the language L for every nonnegative integer i."
can anyone help me with this? This is what I came up with so far and it's definitely not right...
My Guess: Universe of Discourse: Language
N(x)= x is some number depending on L
I(x)= x is non negative integer
S(x)= x is string in L
for existential quantifier ill use "bE" and for universal quantifier ill use "bA"
Guess starts here: bEx(N(x) ^ (S(x) > N(x)) --> bEx bEy bEz((S(x)=xyz)^y does not equal S element empty set)) ^ bAx(I(x)
This is probably completely wrong; I don't really get it. Thanks for any input/help.

Before putting everything into symbols, try to rewrite the sentence in a more logical way. For example:
For all $L$, if $L$ is a language and $L$ is regular, then there exist $N$ and $s$ such that if


*

*$N \in \mathbb{N}$, and

*$s \in L$, and

*$\mathrm{length}(s) \geq N,$
then there exist $x$, $y$ and $z$ such that 


*

*$s=xyz$

*$x,y$ and $z$ are strings

*$y$ is not the empty string

*for all $i,$ if $i \in \mathbb{Z}$ and $i \geq 0$, then $xy^iz \in L.$
By the way, a good universe of discourse would be the universe of sets.

Edit. Here's one possible symbolization. If it looks hideous (which it does), try drawing it as a tree diagram and it will be better (no parantheses!). By asterisk I mean: next line!


*

*$\forall L(\mathrm{language}(L) \wedge \mathrm{regular}(L) \rightarrow *)$

*$\exists N \exists s(\mathrm{natural}(N) \wedge s \in L \wedge \mathrm{length}(s) \geq N \rightarrow *)$

*$\exists x \exists y\exists z(xyz=s \wedge \mathrm{string}(x) \wedge \mathrm{string}(y) \wedge \mathrm{string}(z) \wedge y \neq \mathrm{TheEmptyString} \wedge \forall i(\mathrm{natural}(i) \rightarrow xy^iz \in L))$

