Prove that a given encryption scheme is perfectly secret
For every key $k \in \mathcal K$ $$Dec(Enc(m)) \overset{\underset{\mathrm{def}}{}}{=} (Enc(m) - k) \bmod 4,$$ where $$Enc(m) \overset{\underset{\mathrm{def}}{}}{=} (m + k) \bmod 4,$$ so, combining it, we have $$Dec(Enc(m)) = (((m + k) \bmod 4) - k) \bmod 4,$$ which, by properties of modular arithmetic (or, more precisely, congruences), is equivalent to $$(m \bmod 4) + (k \bmod 4) - (k \bmod 4) = m + k - k = m$$
This is actually a great example, as it demonstrates the ultimate elegance of the mathematical definition of security. Encryption scheme $(E, D)$ over $(\mathcal K, \mathcal M, \mathcal C)$ is perfectly secret, if for all $m_i, m_j \in \mathcal M, c \in \mathcal C$ $$Pr[E(k, m_i) = c] = Pr[E(k, m_j) = c],$$  where $k$ is uniform in $\mathcal K$. Here is a table of encryptions of $m$ with key $k$ for all $m \in \{0, 1, 2, 3\}$ and $k \in \{0, 1, 2, 3\}$:
$$
\begin{array}{c|lc}
m/k &amp; 0 &amp; 1 &amp; 2 &amp; 3 \\
\hline
0 &amp; 0 &amp; 1 &amp; 2 &amp; 3 \\
1 &amp; 1 &amp; 2 &amp; 3 &amp; 0 \\
2 &amp; 2 &amp; 3 &amp; 0 &amp; 1 \\
3 &amp; 3 &amp; 0 &amp; 1 &amp; 2
\end{array}$$
As you can see, for every $m$, $Pr[Enc(m) = c] = 1/4$, thus given encryption scheme is perfectly secure.