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.

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41 votes

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

"Every teacher is liked by some student"

- $∀(x)\left[\text{teacher}\left(x\right) → ∃(y) \left[\text{student}\left(y\right) → \text{likes}\left(y,x\right)\right]\right]$
- $∀(x)\left[\text{teacher}\left(x\right) → ∃(y) \left[\text{student}\left(y\right) ∧ \text{likes}\left(y,x\right)\right]\right]$
- $∃(y) ∀(x)\left[\text{teacher}\left(x\right) → \left[\text{student}\left(y\right) ∧ \text{likes}\left(y,x\right)\right]\right]$
- $∀(x)\left[\text{teacher}\left(x\right) ∧ ∃(y) \left[\text{student}\left(y\right) → \text{likes}\left(y,x\right)\right]\right]$

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59 votes

Best answer

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.

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5 votes

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.**

0 votes

**(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) **i**s right.**