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
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!