Can multiple Pearson permutation tables be reduced or merged?
Claim: A $\pi$ permutation table can be found such that applying it to a message $m$ will yield the same output as applying multiple, different permutations to that message $m$.

i.e. For any number of permutation steps $\phi_n$, $m_{initial}\rightarrow\phi_1\rightarrow\phi_2\rightarrow...\rightarrow\phi_n\rightarrow m_{permuted}$

$\exists$ a single permutation $\pi$ such that: $m_{initial}\rightarrow\pi\rightarrow m_{permuted}$ 

Where both $m_{permuted}$'s are identical.

Proof:

Assume we have two permutation tables, $\phi_1$ and $\phi_2$ s.t. 

$\phi_1=
\begin{bmatrix}
    m_{1} &amp; m_{2} &amp; m_{3} &amp; m_{4} \\
    a &amp; b &amp; c &amp; d
\end{bmatrix} $
$\phi_2=
\begin{bmatrix}
    m_{1\phi_1} &amp; m_{2\phi_1} &amp; m_{3\phi_1} &amp; m_{4\phi_1} \\
    e &amp; f &amp; g &amp; h 
\end{bmatrix} $

Where the first row contains the input broken into bits and the second row contains the positions that each bit will be mapped to. (For $\phi_2$ notice that each bit of $m$ is now $m_{0\phi_1}$, indicating that it is the first bit of the message after being permuted by $\phi_1$). For example:

$
m=3512
$

\begin{bmatrix}
    3 &amp; 5 &amp; 1 &amp; 2 \\
    4 &amp; 2 &amp; 3 &amp; 1 
\end{bmatrix}

yields a $m_{permuted}=2513$, and then this new permuted message can be permuted again (as many times as desired).

Right, so back to our earlier example using $m$, $\phi$, and $a,b,c,...$.

Going through the first permutation, we will have the message $m_{\phi_1}=m_{a}m_{b}m_{c}m_{d}$

Going through the second permutation, we will have the message 
$m_{\phi_2}=m_em_fm_gm_h$

So, it is sufficient to find a permutation matrix that directly maps $m_1,m_2$ etc. to the locations that $m_1,m_2,...$ end up in the final permutation matrix. Finding such a permutation matrix is quite simple, as you can pass a message $m$ through the multiple permutations $\phi_n$, observe where they end up, and create the mapping based on that. This is best illustrated by and example, but proof by example is not acceptable, so I added the other info in before this. Also note that this works for any size of $m$ assuming the permutation has mappings that include all bits in $m$ (So each bit gets assigned a new place). 

Example: 

$m=m_1m_2m_3m_4$

$\rightarrow$$\phi_1=
\begin{bmatrix}
    m_{1} &amp; m_{2} &amp; m_{3} &amp; m_{4} \\
    3 &amp; 2 &amp; 4 &amp; 1
\end{bmatrix} $$\rightarrow m_{\phi_1}=m_3m_2m_4m_1$

$\rightarrow$$\phi_2=
\begin{bmatrix}
    m_{3} &amp; m_{2} &amp; m_{4} &amp; m_{1} \\
    4 &amp; 2 &amp; 3 &amp; 1 
\end{bmatrix} $$\rightarrow m_{\phi_2}=m_1m_2m_4m_3$

$\therefore$ It is sufficient to have a permutation table to complete the mapping $m_1m_2m_3m_4 \rightarrow m_1m_2m_4m_3$

This matrix is
$\pi=
\begin{bmatrix}
    m_{1} &amp; m_{2} &amp; m_{3} &amp; m_{4} \\
    1 &amp; 2 &amp; 4 &amp; 3 
\end{bmatrix} $$\rightarrow \pi(m)=m_1m_2m_4m_3$

And $\phi_2(\phi_1(m))=\pi(m)$