In my assignment I have the following question: Find a model $M$ with domain $\{a, b, c, d\}$ so that:
$M\vDash R(\overline{a},\overline{b})\land R(\overline{b},\overline{a})\land\neg Q(\overline{d})$
$M\vDash\forall x(P(x)\to Q(x)),$
$M\vDash\forall x(R(x,x)\to P(x))$,
$M\vDash\forall x\exists y (R(x,y))$
where $R^M$ is a transitive relation.
What I struggle to understand here is how the negation of $Q$ can be true with the input $\overline{d}$, however later it is expected that for all $x$, if $P(x)$ is true, it implies that $Q(x)$ too shall be true (according to the truth table for implication).
How I interpret this is that $x$ can be everything in the domain, so either $a, b, c\text{ or }d$, but since the NEGATION of $Q$ with $\overline{d}$ as input is true, isn't there a contradiction here?
How is the negation (of this relation $Q$) meant to be understood here?
There is no contradiction. The idea is that $P$ must be false for $d$.
Consider the following interpretation in the domain $\{a,b,c,d\}$: \begin{align} R^M &= \{(a,b), (b,a), (a,a), (b,b), (c,c), (d,c)\} \\ P^M &= \{a,b,c\} \\ Q^M &= \{a,b,c\} \\ \end{align} It is immediate to prove that $R^M$ is transitive and that the structure $M$ (where each $x \in \{a,b,c,d\}$ is the interpretation of the constant $\overline{x}$) satisfies the four axioms.