Recall that divisibility of integers is transitive, meaning:
Proposition 0. If $a \mid b$ and $b \mid c$, then $a \mid c$.
This can be generalized slightly by involving modular arithmetic:
Proposition 1. If $a \mid b$ and $c \equiv c' \mod b,$ then $c \equiv c' \mod a.$
Proposition 0 is obtained in the case $c' = 0$.
Question. Are there any further generalizations of this where e.g. $a \mid b$ is replaced by a more general condition and the necessary changes are made to the conclusion?
The idea of Proposition 0, as I read it, is just an associativity of products: $$ m_1(m_2m_3)=m_1m_2m_3=(m_1m_2)m_3 $$ for integers $m_1,m_2,m_3$. Indeed: with $m_1=c/b$, $m_2=b/a$, and $m_3=a$, we have: $$ c=\frac{c}{b}\left(\frac{b}{a}a\right)=\left(\frac{c}{b}\frac{b}{a}\right)a $$ giving Proposition 0. Letting $m_1=(c-c')/b$ gives Proposition 1. So you can generalize in a few ways: $$ m_1(m_2\ldots m_N)=(m_1\ldots m_{N-1})m_N $$ yielding for example $$ d_N=\frac{d_N}{d_{N-1}}\left[\left(\prod_{i=1}^{N-2}\frac{d_{i+1}}{d_i}\right)\cdot d_1\right]=\left[\prod_{i=1}^{N-1}\frac{d_{i+1}}{d_i}\right]d_1. $$ And you can replace $d_i$ with $d_i-d_i'$ of course for some or all of the $i$'s.