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What is the first order predicate calculus statement equivalent to the following?

"Every teacher is liked by some student"

1. $∀(x)\left[\text{teacher}\left(x\right) → ∃(y) \left[\text{student}\left(y\right) → \text{likes}\left(y,x\right)\right]\right]$
2. $∀(x)\left[\text{teacher}\left(x\right) → ∃(y) \left[\text{student}\left(y\right) ∧ \text{likes}\left(y,x\right)\right]\right]$
3. $∃(y) ∀(x)\left[\text{teacher}\left(x\right) → \left[\text{student}\left(y\right) ∧ \text{likes}\left(y,x\right)\right]\right]$
4. $∀(x)\left[\text{teacher}\left(x\right) ∧ ∃(y) \left[\text{student}\left(y\right) → \text{likes}\left(y,x\right)\right]\right]$

option C. ∃(y)∀(x)[teacher(x)→[student(y)∧likes(y,x)]]

There is a student y for which every x, if x is a teacher then student y likes teacher x.

option D. ∀(x)[teacher(x)∧∃(y)[student(y)→likes(y,x)]]

For every teacher x and some y, if y is a student then he likes teacher x.
Can the answer to this be "∀x ∃y (teacher (x) ∧ student (y) ∧ likes (y,x))" ?
Use ^ for there exist(∃) and use -> for all(∀).
@rambo1987  No, this says, “every x is a teacher and….”
We need to say, “ if x is a teacher then...”

Answer is (B). In simpler way we can say "If $X$ is a teacher then there exists some $Y$ who is a student and likes $X$".

(A) choice:  If $X$ is a teacher, then there exists a $Y$ such that if $Y$ is a student, then $Y$ likes $X$.
(C) choice: There exist a student who likes all teachers.
(D) choice: Everyone is a teacher and there exists a $Y$ such that if $Y$ is student then $y$ likes $X$. Assuming one cannot be both student and teacher at same time, this just means, everyone is a teacher.

@Rahul Jain25  option A is wrong because with option A, there can arise a chance that even if there is no student, statement can give true. We have at least one student which likes a teacher. in A, we can have a case of no student at all (false -> anything = true )
And in case of B if there is no student then due to AND operation it will give false. Please confirm if my interpretaion is correct? @junk_mayavi.
good question

We can easlily eliminate option A & D as $\exists$ with $\rightarrow$  and  $\forall$ with $\wedge$

now comes the most confusing part option B & C.

Let's take a counter example -

Suppose we have 3 teachers ($x_{1}$, $x_{2}$ & $x_{3}$)  &  10 students ( $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$, $y_{5}$, $y_{6}$, $y_{7}$, $y_{8}$, $y_{9}$, $y_{10}$).

Now in option B outer for loop is x & inner for loop is y i.e for every value of x there will be multiple y. $\exists y$ is fixed in an particular iteration of x, not on all iteration of x.

for all x i.e $x_{1}$ $\rightarrow$ ($y_{2}, y_{3}, y_{4}$)   &   $x_{2}$ $\rightarrow$ ($y_{7}, y_{8}, y_{9}, y_{10}$)   &   $x_{3}$ $\rightarrow$ ($y_{1}, y_{2}, y_{5}, y_{6}$)

So, all of this implies Every Teacher is liked by Some students.( key point is this some is not same for all teacher)

Now in option B outer for loop is y & inner for loop is x, i.e -

there exist some student at the beginning of the statement which says that this is talking about some fixed no. of students, Let's say ($y_{1}, y_{2}, y_{3}$)   &  ($x_{1}, x_{2}, x_{3}$)

now  $y_{1}$ $\rightarrow$ $(x_{1}, x_{2}, x_{3})$,   $y_{2}\rightarrow$ $(x_{1}, x_{2}, x_{3})$,    $y_{3}\rightarrow$ $(x_{1}, x_{2}, x_{3})$

So,this implies Some students like every teacher.

by

best explanation
Thanks. This method is good to differentiate.

(D) states:  Everyone in the class must be a teacher and there should be at least one student who likes that teacher.

This is wrong.

(C) states:  There should be at least one student who likes all the teachers. This is wrong.

(A) might feel right because logically it states that if someone is a teacher there must be at least one student who likes him but due to implies , even if there is no teacher , the result would be true. Hence , this option is also wrong.

because in p-→ q

if p is wrong q’s value doesn't matter.

(B) is right.

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