Q:

Prove that $(t$ mod $N) $ mod $ p = (t$ mod $p)$

Let $t \in \mathbb{Z}$. Let $p$ and $q$ be primes, and let $N = p \cdot q$.
Prove that $(t$ mod $N) $ mod $ p = (t$ mod $p)$ 
I found this question on some website. Intuitively I can easily understand that it's true and also explain it in words, but I can't find a formal proof for it. Help would be appreciated!

A:

Hint: let $t = a p +b$ with $0 \le b \le p-1\,$, so that $t \bmod p = b\,$. Then let $a = cq+d$ with $0 \le d \le q-1\,$, so that $t = cpq+dp+b\,$. It follows that $t \bmod pq = dp+b\,$ (why?), so in the end $(t \bmod pq) \bmod p = (dp+b) \bmod p = b $.

