predicate logic formula truth models

Question

Show that the following set of formulas is consistent. (Do it by constructing a model where both formulas are true.)

$\forall x (S(x) \to Q(x))$

$\exists x (Q(x) \land \lnot S(x))$

I know the answer is pretty simple ,but for the life of me I am hitting a blank and can't find a situation where both formulas are true. Any input would be greatly appreciated.

Answer 1

$S$ and $Q$ are unary predicates, i.e. realized as subsets of the domain of a model. As $\forall x(S(x)\rightarrow Q(x))$ ought to be true, we have that the realization of $S$ has to be subset of $Q$.

However, $\exists x(Q(x)\land\neg S(x))$ requires one element(at least) in the domain s.t. it is in the realization of $Q$ but not in the one of $S$. Thus $S$ has to be a proper subset.

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By the means of an example. I assume that your signature is just $(S,Q)$:

Consider the model $$\mathfrak{M}=(\mathbb{Z},S_\mathbb{Z},Q_\mathbb{Z})$$ where $S_\mathbb{Z}=\{x\in\mathbb{Z}\mid x>0\}$ and $Q_\mathbb{Z}=\{x\in\mathbb{Z}\mid x\geq0\}$.

Then, $\mathfrak{M}\models\forall x(S(x)\rightarrow Q(x))$, as for any $x\in\mathbb{Z}$, if $S_\mathbb{Z}(x)$, i.e. $x>0$, then $x\geq 0$, i.e. $Q_\mathbb{Z}(x)$.

However, $\mathfrak{M}\models\exists x(Q(x)\land\neg S(x))$, as for $x=0\in\mathbb{Z}$, we have $x\geq 0$, i.e. $Q_\mathbb{Z}(x)$ but $x\not >0$, i.e. not $S_\mathbb{Z}(x)$.

Answer 2

Let $S(x)$ be $x$ goes in rain without an umbrella, and $Q(x)$ be $x$ gets wet.

Then $S(x)\implies Q(x)$ but there is someone who gets wet without going in rain.