W(x): x is in the world
B(x): x is bad
From these two I form the statement, 'Something in the world is bad', given by $\exists x(W(x) \wedge B(x) )$
Now I want to write the negation of this statement, but i'm confused which one would be correct.
Would the negation be $\forall x \neg (W(x) \wedge B(x) )$ or would it be $\forall x W(x) \implies \neg B(x)$ ?
Indeed. Well, almost. Do not drop the bracketting.
$$\begin{align}&\neg \exists x~\big(W(x)\wedge B(x)\big)&&\text{There is not something in the world that is bad.}\\\equiv&&&\text{(via quantifier duality)}\\&\forall x~\neg\big(W(x)\wedge B(x)\big)&&\text{Everything is not both in the world and bad.}\\\equiv&&&\text{(via de Morgan's rule)}\\ & \forall x~\big(\neg W(x)\vee\neg B(x)\big)&&\text{Everyting is not in the world or it is not bad.} \\\equiv&&&\text{(via implication equivalence)}\\&\forall x~\big(W(x){\implies}\neg B(x)\big)&&\text{Everything, that is in the world, is not bad.} \end{align}$$