I was studying discrete mathematics, one of the basic subjects in cs department.
In particular, studying the chapter "Logics", I came to have some trouble.
While solving problem saying
" Let $S(x)$ be the predicate “$x$ is a student”, and $F(x)$ be the predicate “$x$ is a faculty member,” and $A(x, y)$ be the predicate “$x$ has asked $y$ a question”, where the domain consists of all people associated with your school. Use quantifiers to express each of these statements.
e) There is a faculty member who has never been asked a question by a student."
Here, I answered like $∃x(F(x) ∧ ∀y(S(y) → ¬A(y,x)))$.
But some of the solutions I got from Google says: $∃x((F(x) ∧ ∀y(S(y)) → ¬A(y,x))$.
The difference here is the positions of the parentheses implicating the range of antecedents(conditions).
The official solution says that one I thought of was right,
but the problem is that when thinking about the meaning of the second statements,
I can find no errors:(
Wouldn't there be any way to know the difference without using operation rules(i.e. implication into negation and disjunction)?
Thank you for reading this long message.
Consider $a \rightarrow b \equiv \neg a \lor b$, then
$ ∃x((F(x) ∧ ∀y(S(y)) → ¬A(y,x)) \equiv ∃x((\neg F(x) \lor \exists y \neg(S(y)) \lor \neg A(y,x))$
$∃x(F(x) ∧ ∀y(S(y) → ¬A(y,x))) \equiv ∃x(F(x) ∧ ∀y( \neg S(y) \lor ¬A(y,x)))$
Use now universality and existance, and notice that if you make a $\Sigma$ that contain the clauses:
$ ((\neg F(a) \lor \neg(S(b)) \lor \neg A(b,a))$
$(F(a) ∧ \neg S(b) \lor ¬A(b,a))$
$\Sigma_{1} = \{\neg F(a) \lor \neg S(b) \lor \neg A(b,a)\}$
$\Sigma_{2} = \{F(a), \neg S(b) \lor ¬A(b,a)\}$
Notice then the difference of both $\Sigma$, the second one responds to the predicate you said at the beggining