Let’s assume for simplicity:
- $x$ and $y$ are from a set of names of students
- $P(x)$ reads “$x$ is a girl”
- $Q(x,y)$ reads “$y$ likes $x$”
Then the given logic:
S: $\forall x\bigl(P(x)\implies\exists y(x,y)\bigr)$ reads as:
“For all students ($x$), if $x$ is girl then there exists at least one student ($y$), such that $y$ likes $x$”
Option A: $\exists x\bigl(P(x) \land \forall y Q(x,y)\bigr)$ reads as:
“There exists at least one student ($x$) such that $x$ is a girl and for all students ($y$) $y$ likes $x$”
This can be true if all students like just one girl but S reads that for every girl there is a student who likes her.
So A doesn’t imply S
Option B: $\forall x \forall y Q(x,y)$ reads as:
“For all students ($x$) there are all students ($y$) such that $y$ likes $x$”
Everybody likes everybody. This implies S because then $Q(x,y)$ becomes always true or RHS of S becomes always true
If you have trouble understanding then think like this, try to impose statement B on S and check whether S is still true.
B can be reframed as: For all students ($x$) irrespective of whether $x$ is a girl or not, all students ($y$) like $x$
So B implies S
Option C: $\exists y \forall x \bigl( P(x) \implies Q (x,y) \bigr)$ reads as:
“There exists at least one student ($y$) for all students ($x$) such that if $x$ is a girl then $y$ likes $x$”
For all students ($x$) if $x$ is a girl then there exists at least one student ($y$) who likes all girls
So C implies S
Option D: $\exists x\bigl(P(x) \land \exists y Q(x,y)\bigr)$ reads as:
“There exists at least one student ($x$) such that $x$ is a girl and there exists at least one student ($y$) such that $y$ likes $x$”
This can be true if for only one girl there is a student who likes her but S reads for all girls there is a student who likes them.
So D doesn’t imply S
