A model for which the following is false, $\forall xR(x)\lor \forall xQ(x)$

Question

Does the following model cause the formula below to equate to false?

$A = \{a,b\}$

$R^m = \{(a)\}$

$Q^m =\{(b)\}$

$\forall xR(x)\lor \forall xQ(x)$

How about $\forall x(R(x)\lor Q(x))$?

I am trying to get a better understanding of how the quantifiers qualify predicate symbols. According to my understanding, the model provided here should cause the first formula to equate to false but the latter to true.

Answer 1

\begin{equation}\tag{1} \forall x R(x)\lor\forall x Q(x) \end{equation}

\begin{equation}\tag{2} \forall x (R(x)\lor Q(x)) \end{equation}

To construct a model (also known as an interpretation) where $(1)$ and $(2)$ are false, you need only find a truth assignment that makes them false. For example, you can assert that every instance of $R(x)$ and $Q(x)$ are false, namely $\forall x\lnot R(x)$ and $\forall x\lnot Q(x)$.

A simple model that makes (2) true but (1) false is the following: $R(a)=1$, $R(b)=0$, $Q(a)=0$, $Q(b)=1$. This means that there is an R(x) that is false, and a Q(x) that is too, so $(1)$ is false. However, since $Q(x)$ is true when $R(x)$ is false, and vice versa, $(2)$ is true.

I think this is what you were going for in your question, if it was, then your attempt is correct.

Answer 2

You did that correctly.

Here is an intuitive one that works as well: Take as the domain all natural numbers (or integers), and interpret $R(x)$ as '$x$ is even' and $Q(x)$ as '$x$ is odd'