Here is a very simple method using which we can solve this type of problems easily.

Given **F:** $\forall$x($\exists$yR(x,y))

Let the Domain of x be (1,2) and y be (1,2)

Then above expression F can be rewritten as

F : ( (R(1,1) v R(1,2) ) ^ (R(2,1) v R(2,2) ) for all x there must exist some y such that R(x,y) is true.

Now questions asks for which of the expressions are implied by this means given an expression E, is F$\rightarrow$E a tautology ?

Let's consider options one by one.

(I)$\exists$y ($\exists$xR(x,y))

Considering above domains of x and y we can rewrite them as

( (R(1,1) v R(2,1) ) v ( R(1,2) v R(2,2) )

Now is this expression implied by F? means

is ( (R(1,1) v R(1,2) ) ^ (R(2,1) v R(2,2) ) $\rightarrow$ ( (R(1,1) v R(2,1) ) v ( R(1,2) v R(2,2) ) a tautology?

Yes, it is absolutely a tautology because this implication will only be false when the antecedent is true and the consequent is false and you can check the above expression, it will never happen. So this is **implied expression.**

(II)$\exists$y($\forall$R(x,y))

This can be rewritten as

( (R(1,1) ^ R(2,1) ) v ( R(1,2) ^ R(2,2) ) ) for some y there must exist all x such that R(x,y) is true.

is this implied by F?

( (R(1,1) v R(1,2) ) ^ (R(2,1) v R(2,2) ) $\rightarrow$ ( (R(1,1) ^ R(2,1) ) v ( R(1,2) ^ R(2,2) ) ) a tautology?

This implication can be false when R(1,1) is true, R(2,2) is true and R(1,2) is false and R(2,1) is false.So, this is not implied expression.

(III)$\forall$y($\exists$xR(x,y)) can be rewritten as

( (R(1,1) v R(2,1) ) ^ ( R(1,2) v R(2,2) ) ) for all y there must exist some x such that R(x,y) is true.

is ( (R(1,1) v R(1,2) ) ^ (R(2,1) v R(2,2) ) $\rightarrow$ ( (R(1,1) v R(2,1) ) ^ ( R(1,2) v R(2,2) ) ) a tautology?

This implication can also become false when only R(1,1) and R(2,1) are true. So, this is also not implied expression.

(IV) $\sim\boldsymbol{ \exists x(\forall y \sim R(x,y))}$ The expression in bold is just the complement of the Given F. and so this option looks like $\sim$( $\sim$ F) which is equal to F. and so this is trivially implied by F.

So, **answer**** is (I) and (IV).**