When should I use "let" "put" "be" in a proof? From reading some instructions I could find online, I could understand that this isn't universally agreed upon, but in a course I'm taking now the professor insists on a particular connection between the proposition one needs to prove and stating the hypothesis (the opening statement of the proof).  However, the class I'm taking isn't in English, and I'm struggling to properly translate it.
So, for example, my professor says that if the proposition includes a "for all x", then the hypothesis should begin with something that I'd interpret as "let x".  And, if the proposition says "there exists" and the hypothesis should say something like "look at candidate", and I'm not sure how to translate the later.
 A: If I am proposing a particular value for a variable that has already been named in a problem statement, I would use "consider." Especially if I am providing a counterexample do a universal quantified variable.
Prove or disprove: If $a\vert bc$, then $a\vert b$ or $a\vert c$.
Consider $a=4, b=2,c=2$. Then $4\vert 4$, but $4$ does not divide $2$.
Prove or disprove: $f(x) = \frac{d}{dx}\int_0^x f(y)\, dy$ for every integrable function $f$.
Consider $f=\begin{cases}1, &x=0\\0, &x\neq 0.\end{cases}$ We have that...

"Consider" can also be used when you supplying a constructed value for an existence proof:
Theorem: There exists an integer $n$ with $n^2-4=0$.
Proof: Consider $n=2$. By direct computation we have $2^2-4 = 4-4=0,$ as desired.
"Put" or "take" can also be used in this case:
Proof: Put $n=2$. Then by direct computation we have $2^2-4 = 4-4=0,$ as desired.
But neither of these are completely idiomatic, to my ear. I would use a more verbose construction, e.g.
Proof: We verify by direct computation that $n=2$ is a solution: $2^2-4=4-4=0.$

Finally, I would use "let" when I am naming a new quantity that didn't appear in the theorem statement.
Theorem: Every integer greater than 1 has a prime divisor.
Proof: Let $S$ be the set of integers greater than one that do not have a prime divisor. Suppose for contradiction that $S$ is nonempty; then let $d$ be the least element of $S$...
