What is the correct translation of the following statement into mathematical logic? “Every student who walks talks”

(I) ∀x ((student(x) & walk (x)) → talk (x)))

(II) ∀x (student(x) → (walk (x) → talk (x)))

(III) ¬ ∃x ((student(x) & walk (x)) & ¬(talk (x))))

# Query on mathematical logic

See here you need to frame your statement for solving this

- Every X who is a student and walking then he talk
- Every X who is a student then who is walking then talk
- There not exists a X who is a student and walking and not talking.

So the answer must be 2 and 3