Starting from the definition of $\mathbf{x}\mathbf{\equiv }\mathbf{ymodN}$(namely, that $\mathbf{N}$divides $\mathbf{x}\mathbf{-}\mathbf{y}$), justify the substitution rule $\mathbf{x}\mathbf{\equiv }\mathbf{x}\mathbf{\text{'}}\mathbf{modN}\mathbf{,}\mathbf{y}\mathbf{\equiv }\mathbf{y}\mathbf{\text{'}}\mathbf{modNx}\mathbf{+}\mathbf{y}\mathbf{\equiv }\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{modN}$,and also the corresponding rule for multiplication.

The primary distinction between addition reactions and substitution reactions is that the former entail the joining of two or more atoms or functional groups, whilst the latter involve the displacement of an atom or functional group by another functional group.

Assume that, $\mathrm{x}\mathrm{mod}\mathrm{N}=\mathrm{p}\mathrm{and}\mathrm{y}\mathrm{modN}=\mathrm{q}$.

Then,

$\mathrm{x}+\mathrm{y}\mathrm{modN}=\mathrm{p}+\mathrm{q}$

Now,

$p$ can be written as $\mathrm{p}=\mathrm{x}\text{'}\mathrm{modN}$, since we have $\mathrm{x}=\mathrm{x}\text{'}\left(\mathrm{modN}\right)$.

In the same way,

$q$can be written as$\mathrm{q}=\mathrm{y}\text{'}\mathrm{modN}$ , since we have $\mathrm{y}=\mathrm{y}\text{'}\left(\mathrm{modN}\right)$.

By using the above equations,

$(\mathrm{x}+\mathrm{y})\mathrm{modN}=(\mathrm{x}\text{'}+\mathrm{y}\text{'})\mathrm{modN}$

Therefore,

$\mathrm{x}+\mathrm{y}=\mathrm{x}\text{'}+\mathrm{y}\text{'}\left(\mathrm{modN}\right)$

For Multiplication rule we have to prove $\mathrm{x}*\mathrm{y}\equiv \mathrm{x}\text{'}*\mathrm{y}\text{'}\left(\mathrm{mod}\mathrm{N}\right)$

Assume that,

$\begin{array}{l}\mathrm{x}\mathrm{mod}\mathrm{N}=\mathrm{p}\\ \mathrm{y}\mathrm{mod}\mathrm{N}=\mathrm{q}\end{array}$

By multiplying $\mathrm{p}\mathrm{and}\mathrm{q}$.

$\mathrm{x}\times \mathrm{y}\left(\mathrm{modN}\right)=\mathrm{a}\times \mathrm{b}$

Now,

$p$can be written as$\mathrm{p}=\mathrm{x}\text{'}\mathrm{modN}$ since we have$\mathrm{x}=\mathrm{x}\text{'}\left(\mathrm{modN}\right)$ .

In the same way,

$q$can be written as$\mathrm{q}=\mathrm{y}\text{'}\mathrm{modN}$ since we have $\mathrm{y}=\mathrm{y}\text{'}\left(\mathrm{modN}\right)$

By using the above equation,

$(\mathrm{x}\times \mathrm{y})\mathrm{modN}=(\mathrm{x}\text{'}\times \mathrm{y}\text{'})\mathrm{modN}$

The above expression is,

$\mathrm{x}\times \mathrm{y}=\mathrm{x}\text{'}\times \mathrm{y}\text{'}\left(\mathrm{modN}\right)$

Therefore, the solution is$\mathrm{x}\times \mathrm{y}=\mathrm{x}\text{'}\times \mathrm{y}\text{'}\left(\mathrm{modN}\right)$ .