Give the first order predicate calculus of the following statement. “Some boys like every girl”

Question

A). ∀(x) [ girl(x) -> ∃ (y) [ boy(y) Λ likes(y,x) ] ]

B). ∀(x) [ girl(x)  Λ ∃ (y) [ boy(y) -> likes(y,x) ]

C) ∀(x) [ girl(x) -> ∃ (y) [ boy(y) -> likes(y,x) ] ]

D) None of these

Answer 1

Option D is correct.

A - Every girl is liked by some boy.

B - Everything is a girl, and is either liked by some boy, or something is not a boy. (I know, sounds funny.)

C - Every girl is liked by some boy, or there is something which is not a boy.

Comments

Some boys like every girl

Does this mean a boy like every girl or every girl is liked by some boy? I guess it is a bit ambiguous especially the usage of "boys" and not "boy".

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Some boys  There exists a boy

So, the statement "Some boys like every girl" is equivalent to the following two statements.

To support my claim, consider the following:

Some tigers are man-eaters.

The above statement clearly means that there is atleast one tiger which is a man-eater.

Some tigers like to eat all animals.

The above statement means that there is atleast one tiger, who enjoys eating an animal irrespective of which animal we consider.

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I agree, got confused with "boys". I should check my grammar knowledge :)

Answer 2

A) for all x, X is a girl implies there is a y, y is a boy and y likes x.

Comments

It sounds like, Every girl is liked by somebody, as Pragy Agarwal also mentioned. Isn'it?