For a sentence $\varphi$, we'll define $Spec(\varphi)$ to be the set of all $n\in\mathbb{N}$ for which there is a model $M$ with $|D^M|=n$, such that $M\models\varphi$.
Let $\Sigma=\{P(\cdot),R(\cdot)\}$, where $P,\,R\,$ are relations.
$\boldsymbol{Disprove:} $ There is a sentence $\psi$ above $\Sigma$ such that $Spec(\psi)=\{n\,|\,n\in\mathbb{N}_{odd} \}$.
I have a disproof, but I wonder if part of it is unnecessary.
This is the disproof I have which got all the points (the question taken from an exam):
Assume by contradiction that such $\psi$ exists. Because $1$ is an odd number, there is a model $M$, with $|D^M|=1$ such that $M\models\psi$. Without loss of generality let $D^M=\{a\}$.
Define a new model, $M'$, such that: $D^{M'}=\{a,b\}$ and the next holds:
$\{a\}\in R^M \iff \{a,b\}\in R^{M'}$
$\{a\}\in P^M \iff \{a,b\}\in P^{M'}$
$M'$ is an expansion of $M$. We will define an equivalence relation, $\boldsymbol{\sim}$, with respect to the relations $P,R$. Namely, $\boldsymbol{\sim}$ will have four equivalance classes: as the number of options to be in $P$ or $R$.
By the way $M$ defines $P$ and $R$, there is only one equivalence class, which is: $\{a,b\}$. Let $[\bar{a}]$ be it's representor.
Now we will define an additional model, $N$, as follow: $D^N$ will contain the representatives of the equivalence classes of $\boldsymbol{\sim}$ with respect to $M$. $N$ will define the realtions $P$ and $R$ as follows:
$\{a,b\}\in R^{M'}\iff \{\bar{a}\}\in R^N$
$\{a,b\}\in P^{M'}\iff \{\bar{a}\}\in P^N$
So we actually have:
$M'\models\psi \iff N\models\psi$
From transitivity we have:
$\{a\}\in R^M \iff \{\bar{a}\}\in R^N$
$\{a\}\in R^M \iff \{\bar{a}\}\in P^N$
Namley $N$ is isomorphic to $M$, and becuase $M\models\psi$ we have due to isomorphism $N\models\psi$ which is contradiction, $|D^N|$ is an even number (2).
My question is: wasn't it sufficient to show that $M\models\psi \iff M'\models\psi\,\,$?
Or the whole reason for defining $N$ was because we can't show the above?
I can't really understand what $N$ was needed for.
Would appreciate help!
