Prove that ⊨ is not symmetric

Question

Another question from a beginner:

We want to prove that ⊨ is not symmetric by finidng concrete formulas φ and ψ for which we can show that φ ⊨ ψ and ψ ⊭ φ.

Thank you for your help!

Answer 1

If $\models$ means classical consequence counterexamples are easy to find. Let $R, S$ be one place relation symbols. We have $\exists x (Rx \wedge Sx) \models \exists x Rx \wedge \exists x Sx$, but not vice versa.

For the first claim let $\mathfrak{A} = \langle D, I\rangle$ be a first-order model and $\beta$ an assignment in $\mathfrak{A}$. $\mathfrak{A}, \beta \models \exists x (Rx \wedge Sx) \Rightarrow~ \text{There is some}~a \in D~\text{with}~ \mathfrak{A}, \beta \frac{a}{x} \models Rx ~\text{and}~\mathfrak{A}, \beta \frac{a}{x} \models Sx \Rightarrow~\text{There is some}~a \in D~\text{with}~\mathfrak{A}, \beta \frac{a}{x} \models Rx ~\text{and}~\text{there is some}~a \in D~\text{with}~\mathfrak{A}, \beta \frac{a}{x} \models Sx \Rightarrow \mathfrak{A}, \beta \models\exists x Rx \wedge \exists x Sx$. (Notation: $\beta \frac{a}{x}$ is that assignment that differs form $\beta$ at most in that it assigns $a$ to $x$)

The following model shows that $\exists x Rx \wedge \exists x Sx \not \models \exists x (Rx \wedge Sx)$: $\mathfrak{A} = \langle \mathbb{N}, I \rangle$, such that $I(R) = \lbrace n \in \mathbb{N}: n = m \cdot 2~ \text{for some}~ m \in \mathbb{N} \rbrace, I(S) = \mathbb{N} \setminus I(R)$. Since obviously $I(R), I(S)$ are non-empty we have (for some assignment $\beta$) $\mathfrak{A}, \beta \models \exists x Rx \wedge \exists x Sx$. But because $I(R) \cap I(S) = \emptyset$ we also have that $\mathfrak{A}, \beta \not \models \exists x (Rx \wedge Sx)$.

Answer 2

We have that :

$p \land q \vDash p$

but :

$p \nvDash p \land q$.

Use truth-table to verify it.