Prove: if $x^2$+x-6 $\ge$ 0, then $x$ $\le$ -3 or $x$ $\ge$ 2.
Typically, I'd just use the factorization method to find that the roots are in fact 2 and 3. But I'm trying to find a proof that satisfies the tautology [$p$ $\Rightarrow$ ($q$ $\lor$ $r$)] $\iff$ [($p$ $\land$ $\sim$$q$) $\Rightarrow$ $r$].
I understand that the antecedent, $p$, is $x^2$+x-6 $\ge$ 0 and that the consequent of the conjunction of $q$ and $r$ is represented by $x$ $\le$ -3 and $x$ $\ge$ 2 respectively. I also know that we can consider $\sim$$q$ to be $x$ $\gt$ -3 (or $x$ $\not\le$ -3)
But I'm not sure how I'd go about showing that the conjunction of $p$ and $\sim$$q$ implies $r$. Please advise.
Assume $p\Rightarrow(q\vee r)$ and $p\wedge\neg q$ where $p:x^2+x-6\geq 0$, $q:x\leq -3$ and $r:x\geq 2$.
From $p\Rightarrow(q\vee r)$ and $p$ follows $q\vee r$. Now argue by case analysis. If $q$ is true then this is in contradiction with $\neg q$, hence $r$ holds. On the other hand, from $r$ follows, clearly $r$. Thus from $q\vee r$ we infer $r$. But $q\vee r$ is true, hence $r$ is also true.