Just so you know, this is a homework-related question.
Let $L=\{0,f\}$ a language, and denote $(f^M)^n(a)=f^M(f^M(...f^M(a))...)$
I must show that there exists a model $M$ that satisfies:
- for every $n\in\mathbb N$ there exists $a\in |M|$ s.t. if $k<n$ then $(f^M)^k(a)\neq 0^M$ and $(f^M)^n(a)=0^M$ .
- for every $a\in\mathbb N$ there exists $n\in \mathbb N$ s.t. $(f^M)^n(a)=0^M$
So I took $M$'s world to be the naturals (with zero) and defined:
$f^{M}(a)=\begin{cases} \underline{0} & a=\underline{0}\\ a-\underline{1} & a>\underline{0} \end{cases} $
So far so good.
Now I must show that there exists a model $M'$ of $T=Th(M)$, and $b\in M'$ s.t. for all $n\in\mathbb N$ we have $(f^{M'})^{n}(b)\neq0^{M'}$. This is where I'm stuck.
I tried to let's say add $\infty$ to the world of some model $M'$ of $T$, and define $f^{M}(\infty)=\infty$, but how do I show it's still a model of $T$?
any help will be appreciated.
You can add a new set of axioms. Take a distinguished element $c$ and add $c \neq 0, c \neq f(0), c \neq f(f(0)) \dots$ Any finite set of them has a model, you just take $c$ large enough. So by compactness, the whole set has a model.