Finite Interpretations of Logical Independence

Question

Is the sentence $a$ independent of the set $A$?

$a=\{$All philosophers are either obscure or shallow$\}$

$A=\{$All philosophers are either rationalists or empiricists; some rationalists are obscure; some empiricists are shallow$\}$

$A=\{ \forall x\, [Px \rightarrow (Rx \vee Ex)]$; $\exists x\, [Rx\ \&\ Ox]$; $\exists x\, [Ex\ \&\ Sx] \}$

$a=\{$ $\forall$x $[$Px $\rightarrow$ (Ox $\vee$ Sx)$]$ $\}$

Answer 1

$A \cup a$ must be of the form $T \cup F$

Domain=$\{$1,2,3$\}$: P=$\{$1,3$\}$; R=$\{$1,3$\}$; E=$\{$1$\}$; O=$\{$1,2$\}$; S=$\{$1,2$\}$.

$\forall$x $[$Px $\rightarrow$ (Rx $\vee$ Ex)$]$ $\Rightarrow$ True as all values in P are either in R or E.

$\exists$x $[$Rx & Ox$]$ $\Rightarrow$ True as some values in R and also in O.

$\exists$x $[$Ex & Sx$]$ $\Rightarrow$ True as some values in E are also in S.

$\forall$x $[$Px $\rightarrow$ (Ox $\vee$ Sx)$]$ $\Rightarrow$ False as 3 is not in O or S.

With true premise and a false conclusion, the inference of $A$ to $a$ is invalid.

$A \cup \neg a$ must be of the form $T \cup T$

Domain=$\{$1$\}$: P=$\{$1$\}$; R=$\{$1$\}$; E=$\{$1$\}$; O=$\{$1$\}$; S=$\{$1$\}$.

$\forall$x $[$Px $\rightarrow$ (Rx $\vee$ Ex)$]$ $\Rightarrow$ True as all values in P are either in R or E.

$\exists$x $[$Rx & Ox$]$ $\Rightarrow$ True as some values in R and also in O.

$\exists$x $[$Ex & Sx$]$ $\Rightarrow$ True as some values in E are also in S.

$\forall$x $[$Px $\rightarrow$ (Ox $\vee$ Sx)$]$ $\Rightarrow$ True as all values of P are in either O or S.

With true premise and a true conclusion, negating $a$, the inference of $A$ to $\neg a$ is invalid.

Since $A \cup a$ and $A \cup \neg a$ are both invalid, the the sentence $a$ is thus independent from the set $A$.