Simplification of inequalities I am trying to simplify the following conditions 
$$\beta\leq1 $$
$$ 1-\alpha-\beta \leq 0$$
Can I say that the above two conditions are equivalent to saying that
$$1- \beta \geq 0$$
$$\alpha \geq 0$$
Proof: $$\beta\leq1 \rightarrow -\beta\geq -1\rightarrow 1- \beta \geq 0 $$
Then the only way for $$ 1-\alpha-\beta \leq 0$$ to be satisfied is if $$\alpha \geq 0$$
Is my reasoning correct?

$$\beta\leq 1 $$ is equivalent to $$1- \beta \geq 0$$
But then, if you plug this into
$$ 1-\alpha-\beta \leq 0$$ or
$$ (1-\beta)-\alpha \leq 0$$ or
$$ (1-\beta) \leq \alpha$$ we see that
$$\alpha \geq 1- \beta \geq 0$$
Using the transitivity, we indeed conclude that
$$\alpha \geq 0$$
So, yes, $\alpha$ must be greater/equal than zero. But that is not enough.
Every solution to $\alpha \geq 1- \beta$ is also a solution for  $\alpha \geq 0$. But not the other way around.
For example, if $\beta = 1/6$ then $ (1-\beta) = 5/6$, so not only must $\alpha \geq 0$, but it must also be greater/equal to $5/6$
