A method for comparing analytical proofs Is there a method for checking the validity of a mathematical/analytical proof based on another, different proof that we know is right?
When it comes to mere calculatory problems, even though we can follow different paths to attain the correct answer or even express this same thing differently, at the end one can always operate on the expressions to check if they are actually the same. Nevertheless, it doesn't seems to work analogously when dealing with proofs to mathematical assertions. There are often multiple ways of proving them, and they don't seem to be mutually translatable, as calculatory answers are. 
Therefore, is there a method to compare mathematical proofs?
What makes Stack Exchange so great is the concreteness we impose to questions, I know and agree, but my question cannot be more specific and still I feel this is the community I should refer to.
 A: I feel like there are a few main cases I can think of


*

*The proofs abuse a common underlying property. So there is a way to connect the proofs to one another, as long as you can find out how each proof relates mathematically to that property

*The proofs take different viewpoints (for example, one proof is topological in nature and one proof is not) and abuse a common underlying property.  In this case, it may not even be obvious how to reconcile the proofs intuitively (never mind rigorously), or maybe it's not possible because the contexts are just too different

*The proofs rely on different underlying properties.  In this case the proofs are most likely fundamentally different, unless you can connect them by connecting the properties.

Example of Case #2: Proving $0.999... = 1$

Proof 1 (Classic analytic proof): We know for real $|\alpha|<1$, this infinite sum converges, $\sum_{n=0}^{\infty} \alpha^n = \frac{1}{1-\alpha}$.  Therefore, $0.999... = 9(0.111...)= 9 \sum_{n=1}^{\infty} \left(\frac{1}{10} \right)^n  = 9\left(\frac{1}{1-\frac{1}{10}} - 1\right) = 1$
Proof 2 (Topological proof): For real numbers $x$ and $y$ in the usually topology, we can say $x\neq y \iff \mbox{ there exists } \alpha \mbox{ such that } x<\alpha<y$ (this means that there is an open interval around $x$ not containing $y$, so they must be different).  So let's try to find such an $\alpha$ strictly between $0.999...$ and $1$.  It is not possible, since we would need $\alpha$ to have a decimal expansion of the form $0.a_1a_2a_3...$, where each $a_i \geq 9$. So $a_i = 9$.
My personal commentary: To someone who hasn't studied topology, this proof appears to come at the problem from 2 different angles, and the connection between the proofs is not rigorously obvious, maybe not even intuitively obvious.  BUT to someone who has studied topology, these proofs may seem equivalent because limits in topology are defined using open intervals
