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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. $∀(x)\left[\text{teacher}\left(x\right) ∧ ∃(y) \left[\text{student}\left(y\right) → \text{likes}\left(y,x\right)\right]\right]$
asked in Mathematical Logic by Boss (18.3k points)
retagged by | 2.1k views
+2
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.
0
Can the answer to this be "∀x ∃y (teacher (x) ∧ student (y) ∧ likes (y,x))" ?

1 Answer

+30 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. 

answered by Active (2.6k points)
edited by
+13
Yes. Option C is wrong because it says that all teachers are liked by the SAME student.
+1
please provide the statement equivalent to option D
+1
Added...
+1
sir , i did not understand one thing how to think that " when to use "implication" and when to use "AND" in above type of question...?
+12
When we want to say something "EXIST" we must not use implication but AND, because implication is true even if nothing exist.
0
* note - gateforum booklet have provided answer c which is wrong.
+1
awesome explanation.. superb.
–1
what if X is not a teacher .. then it is also like by student . Then
0
Sir, I have an argument,

A) is not answer bcoz it says that if y is a student than all y "for which"

student(y) is true must like all teacher. Is this valid argument for rejecting A)???
0
@rahul jain25 ya right
0
∀(x)[teacher(x)→∃(y)[student(y)∧likes(y,x)]]=∀(x)∃(y)[teacher(x)→[student(y)∧likes(y,x)]]

and option c->the given statement  and option b <-> the given statement ..
+1
Let's say there are 30 students in a class and roll no. 10 likes every teacher, now Roll no. 10 is "some" student of that class and he/she likes every teacher so by that thought process isn't Option C also correct @Arjun Sir ?

or does the question say like if there are 3 teachers then Teacher1 is liked by roll no. 5, Teacher2 is liked by Roll No. 10 and Teacher3 is liked by Roll No. 15

or Option C is also correct but Option B is stronger/better answer ?
0

"Every teacher is liked by some student" is equivalent to If (for all people ) person is the teacher then liked by some student.

Option B say exactly that. But Option D says (for all people ) person is the teacher and liked by some student.

+2
@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 )
Answer:

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