Predicate logic, models I'm trying to understand predicate logic and models.
I have an old exam question where we are supposed to find a model that shows that the formula does not hold:
$$
\exists x(P(x) \land \neg M(x)), \ \exists y(M(y) \land \neg S(y)) \vdash \exists z (P(z) \land \neg S(z))
$$
Solution: Consider the model $\mathcal{M}$ with universe $A = \{0,1\}$ and  $P^\mathcal{M} = S^\mathcal{M} = \{0\}$ and $M^\mathcal{M}=\{1\}$ and $l$ mapping all variables to $0$. ......more text
I know that $P^\mathcal{M} \subseteq A$. Same with the other predicates. And I suppose $l$ is a lookup-function, but why is it mapping all variables to zero? What is the meaning of that?
When we set $P^\mathcal{M} = \{0\}$, does this mean that $P^\mathcal{M}$ is only true when $x = 0$?   
 A: When we wet $P^\mathcal{M} = \{0\}$, we mean that, according to the model $\mathcal{M}$, $0$ is the only element of the universe that satisfies the predicate $P$. And similarly for the other predicates.
Your exercise asks to show that the formula $\exists z (P(z) \land \lnot S(z))$ is not a consequence of $\exists x (P(x) \land \lnot M(x))$  and $\exists y (M(y) \land \lnot S(y))$. This amounts to show that  there is a model $\mathcal{M}$ such that $\mathcal{M} \models \exists x (P(x) \land \lnot M(x))$ and $\mathcal{M} \models \exists y (M(y) \land \lnot S(y))$ but $\mathcal{M} \not\models \exists z (P(z) \land \lnot S(z))$, i.e., in $\mathcal{M}$, the formulas $\exists x (P(x) \land \lnot M(x))$ and $\exists y (P(y) \land \lnot S(y))$ are true but $\exists z (P(z) \land \lnot S(z))$ is false.
The model $\mathcal{M}$ you described is exactly what you are looking for, because $\mathcal{M} \models P(0) \land \lnot M(0)$ and $\mathcal{M} \models M(1) \land \lnot S(1)$, but $\mathcal{M} \not\models \exists z (P(z) \land \lnot S(z))$ since according to $\mathcal{M}$ all the elements of the universe that satisfy $P$ satisfy $S$ as well. 
Note that the function $l$ mapping all variables to $0$ does not play any role in the exercise, since all the formulas involved in it are closed.
