First-order logic: How to check this sentences are true?

Question

Let $T$ be arbitrary theory and $\varphi$, $\psi$ arbitrary sentences. Consider following statements:

1. If $T \models \varphi \land \psi$, then $T \models \varphi$ and $T \models \psi$.
2. If $T \models \varphi \lor \psi$, then $T \models \varphi$ or $T \models \psi$.
3. If $T \models \varphi$ and $T \models \psi$, then $T \models \varphi \land \psi$.
4. If $T \models \varphi$ or $T \models \psi$, then $T \models \varphi \lor \psi$.

Which ones are true/false? What's step-by-step process to decide?

Answer 1

1.True. For any model $\mathcal{M}\models T$, if $\mathcal{M}\models \varphi\land\psi$ then by the definition of evaluation of closed formulas (sentences), $\mathcal{M}\models \varphi$ and $\mathcal{M}\models \psi$. Therefore $T\models\varphi$ and $T\models \psi$.

2.False. Consider language contains equality symbol and the theory $T=\{\forall x(x=x)\}$. Then $T\models\forall x\forall y(x=y)\lor\forall x\exists y(\lnot x=y)$, (because two sentences are contradicting each other) but if we take $T$-model $\mathcal{M}=\{a\}$ then $\mathcal{M}\not\models\forall x\exists y(\lnot x=y)$, and if we take $\mathcal{N}=\{a,b\}$ then $\mathcal{N}\not\models\forall x\forall y(x=y).$ Therefore $T\not\models\forall x\forall y(x=y)$ and $T\not\models \forall x\exists y(\lnot x=y)$.

3.True. For all model $\mathcal{M}\models T$, by the hypothesis $\mathcal{M}\models \varphi$ and $\mathcal{M}\models \psi$ hence $\mathcal{M}\models \varphi\land\psi$.

4.True, similar to 3.