X implies Y (X->Y)means "Whenever X is true Y has to be True".
Here X is F:∀x(∃yR(x,y)), now we have to find out what are the options will be definitely true if F:∀x(∃yR(x,y)) is true.
Option I. will be obviously true if F:∀x(∃yR(x,y)) is true
Option II can be false when F:∀x(∃yR(x,y)) is true, for example let's say domain of x is {1,2} and domain of y is {a,b,c} and R(1,a) and R(2,b) is true and R(1,b),R(2,a),R(1,c),R(2,c ) are false, hence ∀x(∃yR(x,y)) is true ,but ∃y(∀xR(x,y)) is false
Option III. is also false, if we take the last example there is no x exists such that R(x,c) is true
Option IV is true, if we negate a statement twice then we will get back the original statement.(¬¬F=F).
We will negate F twice and then apply de morgan law in order to get option IV
∀x(∃yR(x,y))= ¬¬∀x(∃yR(x,y)) , now if we apply simple de morgan's law then we will get the option IV (De morgan law=>¬∀xP(x)=∃x¬P(x) and ¬∃xP(x)=∀x¬P(x) )
¬¬∀x(∃yR(x,y))= ¬∃x¬(∃yR(x,y)) = ¬∃x(∀y¬R(x,y))
Answer B