Negating an equals sign? If, when negating a statement, and part of that statement is $3y = x$, can you just say $3y$ does not $= x$ by putting a line through the $=$ sign or is there another way to negate the statement? 
The statement was "For all $x$ there exists a $y$, such that $[(y>x) \land (x=3y)]$."
Which I turned into "For all $x$ there exists a $y$ such that [$(y\leq x) \lor (x \neq 3y)]$."
But I'm fairly sure that last bit is wrong, and you have to negate it another way?
 A: No, it is not sufficient to just slash out the equals sign. See the article on universal quantification here: https://en.wikipedia.org/wiki/Universal_quantification#Negation

Generally, then, the negation of a propositional function's universal
  quantification is an existential quantification of that propositional
  function's negation; symbolically:


A: You are trying to negate $\forall x \exists y (y \gt x \wedge x=3y)$  When you negate a quantified sentence, the rule is you change the sense of the quantifier d negate the contents.  So $$\lnot[\forall x \exists y (y \gt x \wedge x=3y)]\\ \Leftrightarrow \exists x \lnot\exists y (y \gt x \wedge x=3y)\\
\Leftrightarrow \exists x \forall y \lnot(y \gt x \wedge x=3y)$$
A: If $P(x)$ is a statement about $x$ then $\neg \forall x P(x)$ is equivalent to $\exists x \neg P(x)$, not $ \forall x \neg P(x)$.  For example, if I say "it is not true that every man is tall" this is not the same as saying "it is true that every man is not tall."  Instead "it is not true that every man is tall" is equivalent to saying "there is a man who is not tall."
Similarly, $\neg \exists x P(x)$ is equivalent to $\forall x \neg P(x)$.  Putting these together, if $P(x,y)$ is a statement about $x$ and $y$ then $$\neg \forall x \exists y P(x,y)$$ is equivalent to $$\exists x \forall y \neg P(x,y).$$  Do you see how to finish your example?
A: Other people have told you how to negate the quantifiers, so you know that
$$\neg\forall x\exists y (y > x \land x = 3y) \iff \exists x \forall y \neg (y > x \land x = 3y).$$
So now how do you negate $(y > x \land x = 3y)$? By applying De Morgan's law, we get
$$\neg (y > x \land x = 3y) \iff \neg (y > x) \lor \neg(x = 3 y)$$
Note that $a \neq b$ means that $(a > b) \lor (a < b)$. So we get
$$\neg (y > x) \lor \neg(x = 3y) \iff (y \leq x) \lor ((x > 3y) \lor (x < 3y)),$$
so taken together we get
$$\neg \forall x \exists y (y > x \land x = 3y) \iff \exists x \forall y ((y \leq x) \lor ((x > 3y) \lor (x < 3y))).$$
A: There is no  problem in the negation of the equality, as indeed
$$\lnot(x=3y)\equiv x\ne3y.$$
The problem is with the use of the quantifiers. As a generalization of De Morgan's laws (see below), you have
$$\lnot(\forall x:P(x))\equiv \exists x:\lnot P(x),\\
\lnot(\exists x:P(x))\equiv \forall x:\lnot P(x).$$
Then
$$\lnot(\forall x:\exists y:y>x\land x=3y)\equiv \\
\exists x:\lnot(\exists y:y>x\land x=3y)\equiv \\
\exists x:\forall y:\lnot(y>x\land x=3y)\equiv \\
\exists x:\forall y:y\le x\lor x\ne 3y.$$

$$\forall x:P(x)$$ can be seen as
$$P(x_0)\land P(x_1)\land P(x_2)\land P(x_3)\cdots$$
Taking the negation,
$$\lnot P(x_0)\lor\lnot P(x_1)\lor\lnot P(x_2)\lor\lnot P(x_3)\cdots$$
which is
$$\exists x:\lnot P(x).$$
(Repeat backward with $Q(x)=\lnot P(x).$)
