# Show that there exists a satisfactory assignment for the unstandard language of arithmetic $\{\textbf{0}, ', <_1\}$

Consider:

$A1: \textbf{0} \not = x'$

$A2: x'=y' \rightarrow x = y$

$A3: \neg x < \textbf{0}$

$A4: x < y' \leftrightarrow (x < y \vee x = y)$

$A5: \textbf{0} < y \leftrightarrow y \not = \textbf{0}$

$A6: x' < y \leftrightarrow (x < y \wedge y \not = x')$

$A7: \forall x,(x = 0 \vee\exists y, x=y')$

Where $\textbf{0}$ denotes $0$, $'$ the successor function, and $m <_1 n$ iff $m$ is odd and $n$ is even, or both $m$,$n$ are the same parity (both odd or even), and $m < n$.

What denotation of $\textbf{0}$ and $'$ will make $A1-A6$ true but $A7$ false?

I was thinking that if we let $\textbf{0} = 1$ and keep the successor function the same, we can obtain the desired result. However, I'm having trouble deciding whether $A4-A6$ are satisfied/true because they contain the $<$ symbol (which is really $<_1$ in the nonstandard language) in them.

It looks like you are trying to interpret everything in $\omega$. The idea would be to make sure the order works before anything else. So let the order be defined as is. Now the order looks like; $1<_{1}3<_{1}5<_{1}7<_{1}...<_{1}2m-1<_{1}2m+1<...<_{1}0<_{1}2<_{1}4<_{1}6<_{1}...<_{1}2n<_{1}2n+2<_{1}...$
Interpret $\bf{0}$ as $1$. Now interpret $m'$ by $m+2$. Now everything you want is satisfied!