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For any modulus $m > 1$, if $[a] = [b]$ and $[c] = [d]$ then $[a]+[c]=[b]+[d]$ and $[a][c]=[b][d]$

If $[a]=[b]$ then $a \equiv b ({mod}\ m)$

Also

If $[c]=[d]$ then $c \equiv d ({mod}\ m)$

So it follows that $a-b=m(x)$ and $c-d=m(y)$ for some integers $x$ and $y$. So then $a=m(x)+b$ and $c=m(y)+d$

Thus you have $a+c=m(x+y)+b+d$ then $(a+c)-(b+d)=m(x+y)$ so you have that $(a+c) \equiv (b+d) ({mod}\ m)$ then you have that $[a+c]=[b+d]$ or $[a]+[c]=[b]+[d]$

then you can do a similar argument for the $[a][c]=[b][d]$

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    I think you are using the result in your proof. How does $a\equiv b(\textnormal{ mod } m)$ imply the next line? – 2017-02-21
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    sorry I fixed it now I had my notes at first – 2017-02-21

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To prove that $[a][c] = [b][d]$, you will find it easier to work with the following: $$a \equiv b \mod m \iff a = b + km$$ where $k \in \mathbb{Z}$.