The negation of this proposition Suppose $p(x)$ is a proposition about a real variable $x$ in the interval $(0,1)$.
Let $P$ be the proposition
$$\bigg(\exists x_0\in(0,1), p(x_0)\bigg)\implies\bigg(\forall x\in(0,x_0), p(x)\bigg) .$$
Is the negation of $P$ the proposition
$$\bigg(\exists x\in(0,1), \text{not }p(x)\bigg)\implies\bigg(\forall x_0\in(x,1),  \text{not }p(x_0)\bigg) ?$$
 A: 
OP's above comment:
This is what I mean by $P$:
If there exists $x_0$ between 0 and 1 such that $p(x_0)$ holds, then
$p(x)$ also holds for all $x$ such that  $0<x<x_0$. $$\bigg(\exists x_0\in(0,1), p(x_0)\bigg)\implies\bigg(\forall x\in(0,x_0), p(x)\bigg).$$

This translation is still wrong: as I pointed out, the $x_0$ in the consequent still appears free (and doesn't depend on the $x_0$ in the antecedent). The correct translation is this: $$\forall x_0\in(0,1)\bigg(p(x_0)\implies \forall x\in(0,x_0) \;\; p(x)\bigg)$$ or, equivalently, $$\forall x_0\in(0,1) \;\; \forall x\in(0,x_0)\bigg(p(x_0)\implies p(x)\bigg).$$
And its negation is this: $$\exists x_0\in(0,1)\;\;\exists x\in(0,x_0)\bigg(p(x_0)\;\;\text{and} \;\; \text{not }p(x)\bigg) .$$

Is the negation of $P$ the proposition $$\bigg(\exists x\in(0,1), \text{not }p(x)\bigg)\implies\bigg(\forall x_0\in(x,1),  \text{not }p(x_0)\bigg) ?$$

This proposition does not at all negate the erroneous $P$ (nor the correct $P$). If you require any clarification, leave a comment below.
