Negate the following statements formally so that **no negation symbol remains:** Negate the following statements formally so that no negation symbol remains: 
(i)  $ \ \ \forall y \exists x (y>0 \to x \leq 0 ) \ $
(ii) $ \ \forall x \forall y \exists z(x <z \leq y ) \ $
Answer:
My approach is as follows:
(i)
The negation of the statements in (i) without negation symbol is 
$ \exists y  \ \ s.t. \ \  \forall x (x>0  \to y \leq  0) \ $ 
(ii) 
Given 
$ \forall x \forall y \ \exists z \ ( x <z  \leq y ) \  \\\sim \forall x \forall y \ \exists z \ ( x <z  \wedge z \leq y ) $ 
The negation is given as  
$ \exists x \exists y \ \ s.t. \ \ \forall z  \ ( x \geq z \wedge  z > y ) $ 
Am I true ?
Is there any help ?
 A: Your answer to (i) is wrong. The correct negation of $A\rightarrow B$ is $A\wedge \neg B$ so in your case, it should be
$\exists y \forall x (y>0) \wedge (x>0)$
which is not equivalent to the answer you have of $\exists y \forall x (x>0 \rightarrow y\leq 0)$. Your sentence is equivalent to $\exists y \forall x ( y\leq 0)\vee(x\leq 0)$, using that $A\rightarrow B$ is equivalent to $B\vee \neg A$.
Your answer to part (ii) is correct.
(EDIT): I misread your answer to (ii), confusing the conjunction with the disjunction. I is wrong, but if you make that change, it will be correct.
A: You did a lot of work correctly, but in the end both of your answers are incorrect.
For (i) you seem to think that the negation of a conditional $P \rightarrow Q$ is $\neg Q \rightarrow \neg P$, but that is not true. The negation of $P \rightarrow Q$ is $P \land \neg Q$
For (ii), you need to change the $\land $ in your final answer to $\lor$, since the negation of $P \land Q$ is $\neg P \lor \neg Q$
A: For both problems we start by using $\neg(\forall xP(x))\equiv\exists x(\neg P(x))$ and then $\neg(\exists xP(x))\equiv\forall x(\neg P(x))$.
(i) Negate $\forall y\exists x(y>0\to x\leq 0)$.
First we get:
$$\exists y\forall x(\neg(y>0\to x\leq 0))$$
Next we note that $A\to B\equiv \neg A\vee B$:
$$\exists y\forall x(\neg(y<0\vee x\leq 0))$$
Then apply DeMorgan's Law to finish:
$$\exists y\forall x(y>0\wedge x\geq 0)$$
(ii) Negate $\forall x\forall y\exists z(x<z\leq y)$.
First we get:
$$\exists x\exists y\forall z(\neg(x<z\leq y))$$
Split the compound inequality:
$$\exists x\exists y\forall z(\neg(x<z\wedge z\leq y))$$
Then apply DeMorgan's Law to finish:
$$\exists x\exists y\forall z(x>z\vee z\geq y)$$
A: You have the quantifier duality okay, and the negations for the order comparators, but not the negation of the predicate's connectives.

(i)  $\forall y\ \exists x\ (y>0\ \to\ x \leq 0 )$

Use $\neg(P\to Q)\iff (P\wedge\neg Q)$
(A conditional is falsified exactly when the antecedant is true but the consequent is false.)

(ii) $ \forall x\ \forall y\ \exists z\ (x <z\ \wedge\ z \leq y ) $

Use $\neg(P\wedge Q)\iff (\neg P\vee\neg Q)$
(deMorgan's Law of Connective Duality)
