For composable couple of morphisms $g\circ f$ let

$T(g, f): F(g)\circ F(f) \Rightarrow F(g\circ f)$ the canonical cell.

For a triple of composable $h\circ g\circ f$ let $T_r(h, g, f):=T(T(h, g),f)$ (A short way to write $T(hg, f)\ast T(h, g)f$)

and $T_l(h,g,f):= T(h,T(g, f))$ for the axiom of coherence $T'_3=T''_3$.

If we have composable $n$ morphisms $f_n\circ \ldots\circ f_1$ for the various (associativity) coherent disposition of parenthesis (as in a no necessarily associative binary composition) we have a cell $F(f_n)\circ \ldots F(f_1) \to F(f_n\circ \ldots f_1)$, we want to prove that all these cells are equal, for $n=3$ this is true. For induction we suppose the assert for 3, ..., n and indicate with $T^k$ ($3\leq n\leq n$) the unique composition cell $F(f_k)\circ \ldots F(f_1) \to F(f_k\circ \ldots f_1)$. Let $T_{n+1}: F(f_{n+1})\circ \ldots F(f_1)\xrightarrow{1\circ T^n} F(f_{n+1})\circ F(f_n\circ \ldots f_1)\xrightarrow{T} F(f_{n+1}\circ f_n\circ \ldots f_1) $, let $T': F(f_{n+1})\circ \ldots F(f_1)\to F(f_{n+1}\circ f_n\circ \ldots f_1) $ associated to some coherent parenthesis disposition. Now if $f_{n+1}$ is not inside a parenthesis (do not considering the total parenthesis on the whole string) we have that $T'(f_{n+1}\circ \ldots f_1): F(f_{n+1})\circ F(f_n)\circ \ldots F(f_1)\xrightarrow{1\circ T} F(f_{n+1})\circ F(f_n\circ \ldots f_1)$ and this is $T_{n+1}$, if $f_{n+1}$ is inside a parenthesis then

$T'(f_{n+1}\circ \ldots f_1): T^3(f_{n+1}, T^{n-i}(f_n,\ldots, f_{i+1}), T^{i}(f_i\circ \ldots f_1))=$

$T_l(f_{n+1}, T^{n-i}(f_n,\ldots, f_{i+1}), T^i(f_i\circ \ldots f_1))=$

$T_{n+1}(f_{n+1}\circ \ldots f_1) $.

This is only a traslation from Boubaki Algebra

For composable couple of morphisms $g\circ f$ let

$T(g, f): F(g)\circ F(f) \Rightarrow F(g\circ f)$ the canonical cell.

For a triple of composable $h\circ g\circ f$ let $T_r(h, g, f):=T(T(h, g),f)$ (A short way to write $T(hg, f)\ast T(h, g)f$)

and $T_l(h,g,f):= T(h,T(g, f))$ for the axiom of coherence $T'_3=T''_3$.

If we have composable $n$ morphisms $f_n\circ \ldots\circ f_1$ for the various (associativity) coherent disposition of parenthesis (as in a no necessarily associative binary composition) we have a cell $F(f_n)\circ \ldots F(f_1) \to F(f_n\circ \ldots f_1)$, we want to prove that all these cells are equal, for $n=3$ this is true. For induction we suppose the assert for 3, ..., n and indicate with $T^k$ ($3\leq n\leq n$) the unique composition cell $F(f_k)\circ \ldots F(f_1) \to F(f_k\circ \ldots f_1)$. Let $T_{n+1}: F(f_{n+1})\circ \ldots F(f_1)\xrightarrow{1\circ T^n} F(f_{n+1})\circ F(f_n\circ \ldots f_1)\xrightarrow{T} F(f_{n+1}\circ f_n\circ \ldots f_1) $, let $T': F(f_{n+1})\circ \ldots F(f_1)\to F(f_{n+1}\circ f_n\circ \ldots f_1) $ associated to some coherent parenthesis disposition. Now if $f_{n+1}$ is not inside a parenthesis (do not allow the total parenthesis on the whole string) we have that $T'(f_{n+1}, \ldots f_1): F(f_{n+1})\circ F(f_n)\circ \ldots F(f_1)\xrightarrow{1\circ T} F(f_{n+1})\circ F(f_n\circ \ldots f_1)$ and this is $T_{n+1}$, if $f_{n+1}$ is inside a parenthesis consider the maximal parenthesis containing $f_{n+1}$ and let $f_{n+1},\ldots f_{i+1}$ the part of string contained in this parenthesis, the case $i=n-1$ follow as above,

if $i< n-1$ then

$T'(f_{n+1}, \ldots f_1)= T(T^{n-i}(f_{n+1},\ldots f_i), T^{i}(f_i,\ldots f_1))=$

$T(T((f_{n+1},T^{n-i-1}(f_n, \ldots f_i)), T^{i}(f_i,\ldots f_1))=$

$T((f_{n+1},T^n(f_n, \ldots f_1))=$

$T_{n+1}(f_{n+1}\circ \ldots f_1) $.

This is only a traslation from the classical algebra theorem.