Unable to explain flow of steps in this basic modular expression? Consider the expresion $(13 + 11)· 18 (\mod 7)$:

$(13+11)· 18 ≡ (6+4)· 4 (\mod 7)$  Note the transition from $(13+11)·
> 18$$\implies$ $(6+4)· 4$
$≡ 10 · 4 (\mod 7)$
$≡ 3 · 4 (\mod 7)$ Note the transition from $10 · 4$$\implies$ $3 · 4$
$≡ 12 (\mod 7)$
$≡ 5 (\mod 7)$
$≡ 5$


These 2 transitions involve subtracting 7, but in each case they were either the factor( the $10$ going to $7$ in the 2nd transition) or a component of a factor( the $13$ and $11$ going to $6$ and $4$ in first transition).
I would have understood if they subtracted 7 from the product itself( like the $12$ going to $7$ in the final step) because I can intutively understand that the equivalence that both sides have the same remainder when divided by 7 still holds.
I didn't get how this was possible( there wasn't any law/theorem stating you could do that). Few pages down , I saw this corollary:

$ab ≡ [(a \mod n)(b \mod n)](\mod n)$

Is the transitions some result of the corollary or is there some knowledge I'm lacking entirely to explain those transitions?
 A: Let us define a rule and prove its truth.
Let $a, b, c \in \mathbb{N}$ then:
$a = bn+c \implies a \bmod n = c \bmod n$
Proof:
$a \bmod n = (bn+c) \bmod n = (0 + c) \bmod n = c \bmod n$
Now because this is true, we can say:
\begin{equation*}
\begin{split}
18(13+11) \bmod 7 & ≡ 18(7+6+7+4) \bmod 7 \\ & ≡ 18(6+4) \bmod 7 \\ & ≡ (14(6+4)+4(6+4)) \bmod 7 \\ & ≡ 4(6+4) \bmod 7 \\& ≡ 40 \bmod 7 \\& ≡ (7*5+5) \bmod 7 \\& ≡ 5 \bmod 7
\end{split}
\end{equation*}
Let me know if there is anything I can clarify :)
A: The corollary is indeed the key! In case we have:
$$
a\pmod n=x\\
b\pmod n=y
$$
we can write
$$
a=pn+x\\
b=qn+y
$$
and therefore
$$
a\cdot b=pq\cdot n+(py+qx)n+x\cdot y
$$
showing that
$$
a\cdot b\equiv x\cdot y\pmod n
$$
so this establishes the result, which could also be formulated as

When you do modular arithmetics on products, you can reduce each factor first.


To clarify why partial reductions will also work, let us prove a slightly different result that
$$
a\cdot b\equiv(a+kn)\cdot b\pmod n
$$
This can be seen from
$$
(a+kn)\cdot b=a\cdot b+bk\cdot n
$$
So when $a$ is changed by $kn$ the product is changed by $bk\cdot n$

Changing one factor of a product by a multiple of $n$ changes the product by a multiple of $n$

