Show that the sentence or the negation of a sentence is provable from the theory of discrete total orders

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"Consider the structure $\mathcal{N} = (\mathbb{N}, \leq)$: the natural numbers with the usual ordering (without $0$). Let $\Sigma$ be the set of sentences stating that $\leq$ is a discrete total (i.e., linear) order with an initial point but without a final point. Prove that for every sentence $\varphi$ we have: $\Sigma \vdash \varphi$ or $\Sigma \vdash \neg \varphi$."

Here's what I have so far: I think the set of sentences in $\Sigma$ are:

  • Transitive, reflexive, antisymmetric axioms
  • Linearity (i.e., $\forall x \forall y: x \leq y \vee y \leq x$)
  • There is a smallest element (i.e., $\exists x \forall y: x \leq y$)

My guess is to use Gödel's Completeness Theorem (that is, $\Sigma \vdash \varphi \iff \Sigma \models \varphi$), so we instead have to prove that $\Sigma \models \varphi$ or $\Sigma \models \neg \varphi$. But I'm not sure where to proceed from there.

Thanks in advance for any help!