Here is a problem from the 2004 USA Math Olympiad.
Suppose that $a_1,\dots,a_n$ are integers whose greatest common divisor is $1$. Let $S$ be a set of integers with the following properties: (i) For $i=1,\dots,n$, $a_i\in S$. (ii) For $i,j=1,\dots,n$ (not necessarily distinct), $a_i-a_j\in S$. (iii) For any integers $x,y\in S$, if $x+y\in S$, then $x-y\in S$. Prove that $S$ must equal the set of all integers.
The solution my book provides first says that if $b_1,b_2,\dots,b_n$ are some integers that "generate" $S$ and satisfy the three conditions in the problem statement, then $b_i-2b_j$ and $2b_i-b_j$ are in $S$ for any indices $i$ and $j$. This part of the argument is easy to follow and just directly uses the three properties.
However, the solution then goes on to say
An $n$-tuple $(b_1,b_2,\dots,b_n)$ as above can be substituted by $(b_1,b_2-b_1,\dots,b_n-b_1)$, which again generates $S$ and, by what we just proved, satisfies (i), (ii), and (iii).
This, I do not understand. Why is it the case that the new $n$-tuple generates the same $S$? In particular, the elements $b_2,b_3,\dots,b_n$ must be in the $S$ generated by the new $n$-tuple (since they are in the $S$ generated by the original $n$-tuple $(b_1,\dots,b_n)$) but I am not sure how they are. Some clarification here would help.
Let $(b_1, b_2, \ldots ) = S$ and $ (b_1, b_2-b_1, \ldots , b_n-b_1) = T$. WTS $ S = T$.
Consider $T$: