- The set of four premises $S_{1}$: $\{V → L; L→ B; M → \neg B; V \wedge M\}$ is $inconsistent$ because we can logically derive the contradiction $B \wedge \neg B$.
Given, Set of premises: $S_{1}: \{V → L; L→ B; M → \neg B; V \wedge M\}$
$$\begin{array}{c} 1. V \to L \\ 2. L \to B \\3. M \to \neg B \\ 4. V \wedge M \\ \hline \therefore B \wedge \neg B \\ \hline \end{array}$$
Conclusion: $B \wedge \neg B = F$; Let’s say, $B = T$.
So, we try to make all premises $true$ :
Premises $4. V \wedge M \equiv T$, So, $V \equiv M \equiv T$
Premises $3. M \to \neg B \equiv T \to F \equiv F$
Premises $2. L \to B \equiv T \to T \equiv T$
Premises $1. V \to L \equiv T \to T \equiv T$
So, we can’t make all premises $True$ as premises $3.$ is $false.$
So, we can logically infer that conclusion and the given statement is $true$.
- The set of three premises $S_{2}$: $\{W \to A; A \to L; \neg W \wedge L\}$ is $consistent$ because there exists an interpretation of $W,A$ and $L$ such that under this interpretation all the three premises will be true.
If we can find true sentential interpretation by conjunction of given premises then they will be $consistent.$
$\therefore (W \to A)\wedge (A \to L) \wedge (\neg W \wedge L)$
$=> (W’ + A)(A’ + L)(W’L)$
$=> (W’A’ + W’L + AL)(W’L)$
$=> W’A’L + W’L + W’AL$
$=> W’L + W’AL$
$=> W’L \equiv \neg W \wedge L$
If $W = T$ then $\neg W \wedge L = F \wedge L = F$
If $W = F$ then $\neg W \wedge L = T \wedge L = L$ which can be $T$ also and $F$ also. So, it is $satisfiable.$
So, given set is $consistent.$
$Ans: C$