Logical Fallacy in the negation of P:(∃x∈ℝ)(∀y∈ℝ) x ≼ y ?? Let $P$ be the following proposition.
$P: \ (∃x∈ℝ)(∀y∈ℝ)\, x \leq y$
Let us prove that $P$ is false by a counter example:
Let $y = x-1$ then  $x \leq x-1$ then $0 \leq -1$ (False)
Therefore the proposition $P$ is False.
Let $Q$ be the negation of $P$ ($Q$ is "non $P$")
$Q: \ (∀x∈ℝ)(∃y∈ℝ) \, x > y$
Let us prove by a counter example that $Q$ is false:
Let $x = y-2$ then  $y-2 > y$  then  $-2 > 0$ (False)
Therefore the proposition $Q$ is false.
We know that if a proposition is true, then its negation is false and vice versa. I would like to know where did I make a fallacy. Thanks
 A: You run too fast. Let see see your reasoning in a more "pedantic" way, so as to detect your error.
Since you are talking about a  specific structure (the set of real numbers $\mathbb{R}$ with the usual order $\leq$), proving that a statement is false amounts to prove that its negation holds.
Your proposition $P$
\begin{equation}\tag{$P$}
\exists x \in \mathbb{R}: \ \forall y \in \mathbb{R} , \ x \leq y
\end{equation}
claims that $\mathbb{R}$ has a smallest element with respect to $\leq$.
Clearly, this proposition is false.
To prove that $P$ is false, you have to show its negation holds, i.e. that for every $x \in \mathbb{R}$, there exists a certain $y \in \mathbb{R}$ such that $x \not \leq y$, i.e. $x > y$ (since the order $\leq$ on $\mathbb{R}$ is total).
This is essentially what you correctly did in the first part of your post.
Indeed, let us fix an arbitrary $x \in \mathbb{R}$ and take for instance $y = x -1$; clearly, $y \in \mathbb{R}$ and $x > x- 1 = y$.
So far so good.

Consider now the proposition $Q$, the negation of $P$ (again, we assume that $x \not \leq y$ is equivalent to $x > y$, because $\leq$ is a total order)
\begin{equation}\tag{$Q$}
\forall x \in \mathbb{R}, \ \exists y \in \mathbb{R} : x > y
\end{equation}
Our argument above to show that $P$ is false is actually a proof that the  negation of $P$, i.e. $Q$, holds.
So, for sure we can't prove that $Q$ is false, which amounts to say that you can't prove that the negation of $Q$, i.e. the negation of the negation of $P$, i.e. $P$, holds (unless you discovered that the mathematics of real numbers is contradictory, but this is not the case).
In other words, to prove that $Q$ is false, you have to show that there exists a certain $x \in \mathbb{R}$ such that for every $y \in \mathbb{R}$ we have that $x \not > y$, i.e. $x \leq y$. That is, you have to prove that $P$, i.e. the negation of $Q$, holds. But we have just proved that $Q$ holds!
Therefore, your argument to prove that $Q$ is false contains an error.
Where?
In your argument, you first have to fix a certain $x \in \mathbb{R}$, independently of $y$ or whatever: the order of quantifiers is fundamental! Indeed, writing
$$\exists x \in \mathbb{R}: \ \forall y \in \mathbb{R} ,\dots$$
means that you pick up a $x \in \mathbb{R}$ without knowing anything about $y$. This is because $x \in \mathbb{R}$ comes first (it is on the left) than $\forall y \in \mathbb{R}$.
So, you should be able to talk about $x$ independently of $y$.
The error in your argument is that your $x \in \mathbb{R}$ instead depends on $y$ (you say $x = y-2$) but it doesn't make any sense because $y$ is not fixed: who is $y$?
Said differently, your second argument is not trying to negate $Q$.

Note that your second arguement contains another error: when you set $x = y-2$, you can't say that $y - 2 > y$!
So, what does your second argument actually prove?
It actually shows that, given an arbitrary $y \in \mathbb{R}$, there is a $x \in \mathbb{R}$ (more precisely, $x = y -2$) such that $x < y$, i.e.
\begin{equation}
\forall y \in \mathbb{R}, \ \exists x \in \mathbb{R} : y > x
\end{equation}
which is nothing but $Q$ where you renamed the variables.
Summing up, (the amended version of) your second argument is actually another proof that $Q$, i.e. the negation of $P$, holds!
A: First of all you should fix a set, for example $\mathbb{R}$.
Since your underlying set which you are trying to work, may not have the options that you are using. For example the following fact:
$x \leq y \Rightarrow \forall a; \space  x+a \leq y+a $
is correct in ($\mathbb{R},+,0,\leq )$ but not for any order on any set!
So I assume that you are talking in the real numbers.
Your reasoning about the proposition $Q$ is incorrect.
Q: $\forall x \space \exists y \space x >y$
This means that if you fix $x$ , then someone (Not you) can find a number $y$ such that $x>y$.
So for example if you fix $x$, then one can give you $\space y=x-2$ having the desired property.
In fact your reasoning is trying to assert that the following proposition is false :
$\forall x \space \forall y \space  x>y$
which is clearly not true in $(\mathbb{R},0,1,\leq)$.
Hope it helps...
A: The order of the quantifiers is crucial!
Consider the difference between:
"Everyone likes someone"
and:
"Someone is liked by everyone"
Note that if the second statement is true, then the first statement is true as well: if there is a special person that is liked by everyone, then clearly everyone likes someone ... namely that one very likable person!  However, note that the first statement can also be true without the second statement being true: everyone can like someone ... but it is not the same person for everyone. E.g. One half of the people can like Bob, but not Amanda, while the other half likes Amanda, but not Bob. Then everyone likes someone, but there is not someone liked by everyone.
In logic, these two statements are symbolized as:
$\forall x \exists y \ Likes(x,y)$
and
$\exists y \forall x \ Likes(x,y)$
So note that the only differenece between the two statements is the difference of the order of the quantifiers. And yet, by placing the existential before the universal you apprently get a much stronger statement than if you place the existential after the universal. Hence, the order matters!
The same is true in your case. In particular, if we swap the order of the quantifiers for $Q$, we get:
$\exists y \in \mathbb{R} \forall x \in \mathbb{R} \ x>y$
then we obtain a very strong statement: "there is some real number such that all real numbers are greater than it" ... which is clearly false: no number can be greater than itself, for one!
This is basically how you interpreted statement $Q$ as well ... and your counterexample was to take $y-2$ for any such alleged 'smallest' number $y$. Yes, that also works as a counterexample to this very strong statement.
But note! We just reveresed the order of the quantifiers of $Q$ to get this very strong statement. If we keep the  order of the quantifiers in $Q$ as they are, we get the much weaker statement
$\forall x \in \mathbb{R} \exists y \in \mathbb{R} \ x>y$
which says "for any real number $x$ there is some smaller number $y$" ... and that is in fact true!  Of course, for every number $x$ I can simply pick $y= x-1$.
