Gnarly equality proof? Or not? I have this gnarly equality which Mathematica's Reduce says it doesn't have the chops to handle:
$\left.k\in \mathbb{Z}\land \left\lceil \frac{k-2}{2}\right\rceil  \left(\left\lceil \frac{k-2}{2}\right\rceil -((k-2) \bmod 2)+2\right)+k+1=\left\lceil \frac{k}{2}\right\rceil  \left(\left\lceil \frac{k}{2}\right\rceil -(k \bmod 2)+2\right)\right]$,
so I simplified to this: 
$k\in \mathbb{Z}\land (k \bmod 2)+k=2 \left\lceil \frac{k}{2}\right\rceil$, which Mathematica reduces to: $c_1\in \mathbb{Z}\land k=c_1$.
So, can I say that this proves the equality for $k\geq1$? Or is something subtle going on? Do I need to do more?

Rad, dude. You are correct on all counts. The equality is too gnarly for Mathematica; your reduction is correct; and your interpretation of Mathematica's final answer is good. Way to go.
You can actually go further: Mathematica's answer is good for $k \in \mathbb{Z}$, not just $k \in \mathbb{Z}^{\geq 1}$.
