First-Order Logic: semantic entailment and disjunction

Question

In First-Order Logic, we consider that $\Sigma$ is a set of well-formed formulas (wff), and $\alpha$, $\beta$ are such wff. I would like to prove/disprove the following two statements:

1. If either $\Sigma\models\alpha$ or $\Sigma\models\beta$, then $\Sigma\models\left(\alpha\lor\beta\right)$
2. If $\Sigma\models\left(\alpha\lor\beta\right)$, then either $\Sigma\models\alpha$ or $\Sigma\models\beta$

Any idea?

Answer 1

It is Exercise 5 [page 27] of Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001).

For assertion 1, we have to apply the definition :

$\Sigma \vDash \varphi$ iff every truth assignment $v$ which satisfy every formula in $\Sigma$ (i.e. such that $v(\sigma)=T$ for every $\sigma \in \Sigma$) satisfy also $\varphi$ (i.e. $v(\varphi$)=T).

Thus :

if either $Σ \vDash (α)$ or $Σ \vDash (β)$

we have that for every truth assignment $v$ such that $v(\sigma)=T$ for every $\sigma \in \Sigma$, then either $v(α)=T$ or $v(β)=T$.

Thus, we have that for every truth assignment $v$, if $v(\sigma)=T$ for every $\sigma \in \Sigma$, then either $v(α \lor β)=T$, i.e.

$Σ \vDash (α \lor β)$.

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For 2, we can find a counter-example showing that it does not hold.

We will try to apply it with $\Sigma = \emptyset$, in which case $\emptyset \vDash \varphi$ (abbreviated as : $\vDash \varphi$) amount to saying that $\varphi$ is a tautology.

Consider the sentential letter $p_0$; clearly, $p_0 \lor \lnot p_0$ is a tautology.

Thus :

$\vDash (p_0 \lor \lnot p_0)$.

Assuming assertion 2 : if $\vDash(α∨β)$, then either $\vDash α$ or $\vdash β$, we may conclude with :

either $\vDash p_0$ or $\vDash \lnot p_0$.

But no propositional letter or negation of a propositional letter is a tautology.

Thus, we have shown that assertion 2 does not hold.