I am requested to prove or disprove that $\forall_x$ $(p(x)\lor q(x))$⊨$∀_x p(x)\lor ∀_xq(x)$
I have written the following interpretation bellow. However I have two questions concerning formalization...
1st.) I have written the function of the interpretation as $m(f)=${$0$,if $f=p$ or;$1$,if $f=q$}... However, this seems for me clearly wrong, but I couldn't figure out how to express such a function.
2nd.) It is the overall ending $∀_x(p(x)∨q(x))⊨∀_x p(x)∨∀_x q(x)$ correct? I mean, may I consider the symbol ⊨ as if it was $\rightarrow$? Because it seems to me that I have proved a statement by considering the antecedent true and consequent false of an implication, and so the overall implication is false.
Proof
This statement is false, to show this we choose the domain of interpretation to be {$0,1$}. There are two predicate symbols that occur in their statements: $P$ and $Q$. Let's interpret $P(x)$ as the statement $x=0$, and $Q(x)$ as the statement $x=1$. So in symbols we have a function $m(f)=${$0$,if $f=p$ or;$1$,if $f=q$}.
Now we interpret the statement $∀x(P(x)∨Q(x))$ as the statement "For all $x ∈${$0,1$}, $x=0$ or x=1." This is clearly a true statement.
We also interpret the statement $∀xP(x)$ as "For all $x∈$ {$0,1$}, $x=0$." This is a false statement as $1∈{0,1}$ and $1≠0$.
We also interpret the statement $∀xQ(x)$ as "For all $x∈${$0,1$}, $x=1$." This is a false statement as $0 ∈${$0,1$} and $0≠1$.
Thus, we see that the statement $∀xP(x)∨∀xQ(x)$ is not true in this interpretation, since both $∀xP(x)$ and $∀xQ(x)$ are not true.
In this interpretation, $∀x(P(x)∨Q(x))$ is true, but $∀xP(x)∨∀xQ(x)$ is false. So it is not true that $∀_x(p(x)∨q(x))⊨∀_x p(x)∨∀_x q(x)$.