Let us Understand this by taking a Simple Example...
let x∈{Set of students}
y∈{Set of courses offered by a University to their students},
and, Predicate P(x, y)="x takes course y".
Now, Let's see the options for this question.
(A) If we translate this FOL into the equivalent English statement it becomes
LHS= "All students take all courses"
RHS= "All courses are taken by all students"
so we see there are same thing (LHS=RHS).
so, A is TRUE.
(B) Here LHS and RHS are equivalent FOL expressions, just move ~ sign inside the braces and get LHS.
(C) Similarly, If we translate this FOL into the equivalent English statement
LHS= "Some student take some course".
RHS= "Some course is taken by Some student"
Both LHS and RHS Meaning is same ('you eat fruit' or 'fruit is eaten by you' is same meaning),so we see there are equivalent(LHS=RHS).
so, C is also TRUE.
(D) Similarly, for this option,
LHS= "Some student takes all courses".
RHS= "All courses are taken by Some student"
here LHS and RHS look similar but understand this with an another example...
Consider, x and y both ∈ {Set of real Numbers}
and, P(x, y)= x + y=2
now, LHS says " There exist a value of x, for all value of y such that, x + y=2 is true" (FALSE statement)
so LHS is FALSE, and FALSE → (anything)== always TRUE
so, D is also TRUE.
(E) BUT now, for this option,
LHS= "All students takes some course".
RHS= "Some course is taken by All students"
here also, LHS and RHS look similar
Consider, x and y both ∈ {Set of real Numbers}
and, P(x, y)= x + y = 2
LHS= "For all values of x, there exist a value of y such that , x + y=2 is true". (ALWAYS TRUE).
but........
RHS= "There exist a value of y, for all value of x such that, x + y =2 is true" (FLASE statement)
and TRUE(LHS) → FALSE(RHS) == FALSE
so, E is FALSE.
SO, ANSWER -- E