This Number Theory problem follows on from a previous complex number problem with my sincerest apologies, which I posted about here.
The following lemmas follow directly inductively from the equation stated in the linked post, by addition of a third natural number variable:
None the less, I am looking for a simple yet rigorous proof for:
$$((x =y)\lor(x=z)\lor(y=z))\land({\{x,y,z}\} \subset {\{2n-1:n \in \mathbb N}\}) \Rightarrow {\frac {\cos \left( x\pi \,{{\rm e}^{y}}y \right) }{\cos \left( x\pi \, \left( {{\rm e}^{y}}-z \right) y \right) }}=-1$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(C0)}$$
$$\lnot(((x =y)\lor(x=z)\lor(y=z))\land({\{x,y,z}\} \subset {\{2n-1:n \in \mathbb N}\}) )\Rightarrow {\frac {\cos \left( x\pi \,{{\rm e}^{y}}y \right) }{\cos \left( x\pi \, \left( {{\rm e}^{y}}-z \right) y \right) }}=1$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(C1)}$$
Any assistance is very sincerely appreciated,( I realise you may feel like you are enabling an obsessive compulsive personality trait, but I do actually enjoy doing math on this site as opposed to my previous tactic of finding an an equality and announcing QED to myself if it works for the first 10 naturals then moving on) and so a quick and painful minimal counter $(x,y,z)$ is most welcome also.
I think this would probably better articulate the direction I wanted to go with this:
$${\frac {\cos \left( x\pi \,{w}^{y}y \right) }{\cos \left( x\pi \, \left( {w}^{y}-z \right) y \right) }}=-1 \Rightarrow w \in \mathbb W_0 \subset \mathbb R$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(C3)}$$
$${\frac {\cos \left( x\pi \,{w}^{y}y \right) }{\cos \left( x\pi \, \left( {w}^{y}-z \right) y \right) }}=1 \Rightarrow w \in \mathbb W_1 \subset \mathbb R$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(C4)}$$
Problem revised:
Define $\mathbb W_0$ & $\mathbb W_1$ in set builder notation, and show that their union is not the entirity of $\mathbb R$.
an example is if you define $w=k^{\frac{1}{\pi}}$ for some natural $k$, k must neccessarily be equal to $1$ in order for $(C3)$ to be true,
You are complicating this unnecessarily. First of all, these things are a lot easier to understand if you write as much as possible in words. For the first one, I think it would be a lot better to say, "If $x,y,z$ are odd integers, and two of them are equal then, $${\frac {\cos \left( x\pi \,{{\rm e}^{y}}y \right) }{\cos \left( x\pi \, \left( {{\rm e}^{y}}-z \right) y \right) }}=-1\tag{1}$$
Even now, this has a lot of extraneous stuff in it. Write $\theta =x\pi\mathrm{e}^yy$ Then $(1)$ becomes $${\cos{\theta}\over\cos(\theta-xyz\pi)}=-1\tag{2}$$ Now, since $x,y,$ and $z$ are odd, so is $xyz$ and it doesn't matter whether two of them are equal or not. (For example, $1\cdot9\cdot9=1\cdot3\cdot27.$) What this comes down to is $$ \cos\theta = -\cos(\theta-(2n-1)\pi) \text{ if } n\in \mathbf{Z},\tag{3}$$ where I have cleared denominators to avoid division by zero problems. Of course, we know that $(3)$ is true from trigonometry.
Now try to simplify your second statement.