If a model $\mathcal{M}$ is s.t. $\mathcal{M} \models \Delta$ and $\mathcal{M} \models \Gamma$, then $\mathcal{M} \models \Delta \cup \Gamma$?

Question

If a model $\mathcal{M}$ is s.t. $\mathcal{M} \models \Delta$ and $\mathcal{M} \models \Gamma$, then $\mathcal{M} \models \Delta \cup \Gamma$?

I'm pretty sure this is trivially true. But I'm not sure how I would prove it.

Something along the lines of $\mathcal{M}$ makes every member of $\Delta$ true and every member of $\Gamma$ true. So $\mathcal{M}$ makes the members in either $\Delta$ or $\Gamma$ true. Therefore since the definition of $\Delta \cup\Gamma$ is s.t. every sentence $\varphi \in \Delta \cup\Gamma$ iff $\varphi \in \Delta$ or $\varphi \in \Gamma$. We have $\mathcal{M}$ makes every member of $\Delta \cup \Gamma$ true.

Answer 1

This is just to move this off the unanswered queue, since the OP's argument is correct; I've made this CW to avoid reputation gain.

Yup, that's right! In unnecessary detail:

Suppose $\mathcal{M}\models\Delta$ and $\mathcal{M}\models\Gamma$. To show $\mathcal{M}\models\Delta\cup\Gamma$, fix $\theta\in\Delta\cup\Gamma$; we want to show that $\mathcal{M}\models\theta$.

We argue by cases:

• $\theta\in\Delta$: then $\mathcal{M}\models\theta$ since $\mathcal{M}\models\Delta$.

• $\theta\in\Gamma$: then $\mathcal{M}\models\theta$ since $\mathcal{M}\models\Gamma$.

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Really the only subtle issue with $\models$ and Boolean operations is intersection on the left: we can have $\Gamma\models\theta$ and $\Delta\models\theta$ but $\Gamma\cap\Delta\not\models\theta$. (Remember that "$\Theta\models\eta$" is shorthand for "If $\mathcal{M}\models\Theta$ then $\mathcal{M}\models\eta$.")