First we need to review the definition of "implies".
Suppose $P,Q$ are two sentences. $P\implies Q$ is exactly the case that if $P$ is true then $Q$ is also true.
In your example, $x\ge 1$ implies $x\ge 0$ because $1\ge 0$ and $\ge$ is transitive, that is if $x\ge y$ and $y\ge z$ then $x\ge z$.
Suppose now that you have $\mathcal P=\{P_1,\ldots\}$ a set of sentences (not necessarily finite), and $Q$ another sentence. We say that $\{P_1,\ldots\}\implies Q$ exactly if all the sentences in $\mathcal P$ are true, then $Q$ is also true.
An example is:
$$\{"x\ge2", "x\text{ is odd}"\}\implies x\ge 3$$
Neither $x\ge 2$ nor $x\text{ is odd}$ imply $x\ge 3$ on their own, however combine the two facts and you have that $x\ge 3$.
This can be extended into replacing $Q$ by some $\mathcal Q=\{Q_1,\ldots\}$ another set of sentences, in which case we say that $\mathcal P\implies\mathcal Q$ exactly whenever all the sentences in $\mathcal P$ are true, then all the sentences in $\mathcal Q$ are true as well.
For example:
$$\{"x\ge 1", "x\text{ is even}", "x\text{ can be divided by } 6"\}\implies\{"x\text{ can be divided by } 3", "x\ge 6"\}$$
There is no strict requirement that a "big" sets have to imply the truth of some "smaller" or "bigger" set than itself. One sentence can imply many sentences, and it may be the case that several sentences are required to imply a single sentence.
A mistake in your post is that $\implies$ is not a partial order, but rather a quasi-order. That means reflexive and transitive. It is not antisymmetric.
For example: $$"x<1 \text{ and } x>-1"\implies "x=0"$$ as well: $$"x=0"\implies "x<1 \text{ and } x>-1"$$
However formally speaking, $x=0$ is not the same sentence as $"x<1 \text{ and } x>-1"$.