Let
- Student$(x): x$ is a student
- Loves$(x,\text{Bill}):x\; \text{loves Bill}$
"No student loves Bill"
Domain of discourse: All the people
We can interpret like this way.
- If Raju is a student, then he does not love Bill.
- If Rani is a student, then she does not love Bill.
- If Rakesh is a student, then he does not love Bill.
- If Vani is a student, then she does not love Bill.
- and, so on..
Now, we try to convert same statements, in terms of variables.
- If $x_{1}$ is a student, then $x_{1}$ does not love Bill.
- If $x_{2}$ is a student, then $x_{2}$ does not love Bill.
- If $x_{3}$ is a student, then $x_{3}$ does not love Bill.
- ---------------------------------------------------------------
- If $x_{n}$ is a student, then $x_{n}$ does not love Bill.
Now, try to conclude, the above statements.
- $\text{Student}(x_{1}) \rightarrow \neg \text{Loves}(x_{1},\text{Bill})$
- $\text{Student}(x_{2}) \rightarrow \neg \text{Loves}(x_{2},\text{Bill})$
- $\text{Student}(x_{3}) \rightarrow \neg \text{Loves}(x_{3},\text{Bill})$
- ------------------------------------------------------------
- $\text{Student}(x_{n}) \rightarrow \neg \text{Loves}(x_{n},\text{Bill})$
We , can write above statements using universal quantifier $(\forall)$.
- $\forall x(\text{Student}(x) \rightarrow \neg \text{Loves}(x,\text{Bill}))$
- $\forall x(\neg \text{Student}(x) \vee \neg \text{Loves}(x,\text{Bill}))\quad [\because P \rightarrow Q \equiv \neg P \vee Q]$
Demorgan's laws:
- $\neg \forall x (P(x) \vee Q(x)) \equiv \exists x (\neg P(x) \wedge \neg Q(x))$
- $\neg \exists x (P(x) \wedge Q(x)) \equiv \forall x (\neg P(x) \vee \neg Q(x))$
Now, $\forall x(\neg \text{Student}(x) \vee \neg \text{Loves}(x,\text{Bill})) \equiv \forall x[\neg (\text{Student}(x) \wedge \text{Loves}(x,\text{Bill}))]\equiv \neg \exists x[\text{Student}(x) \wedge \text{Loves}(x,\text{Bill})]$
So, the correct answer is $A;B$