I am trying to analyze the following statement, which I believe is true, but I would appreciate guidance on proving it (or counterexamples if it is false):
$$ \forall n, m \in \mathbb{N}^{\geq 2}, n \neq m \implies \forall a, b \in \mathbb{Z}, (a \equiv_n \circ \equiv_m b) $$
Here, $ \equiv_n \circ \equiv_m $ is a relation on $ \mathbb{Z} $, defined as:
$$ \equiv_n \circ \equiv_m = \{(a, b) \in \mathbb{Z}^2 \mid \exists c \in \mathbb{Z}, a \equiv_n c \text{ and } c \equiv_m b \}. $$
Where $ \equiv_n $ represents equivalence modulo $ n $, given by:
$$ \equiv_n = \{(a, b) \in \mathbb{Z}^2 \mid \exists k \in \mathbb{Z}, a = kn + b\}. $$
My question is as follows:
- Is the composition $ \equiv_n \circ \equiv_m $ well-defined for all $ n \neq m $?
- If so, how can this be proven? If not, could you provide a counterexample or explain why?
Any help in clarifying or proving this would be greatly appreciated!
