The pentagon axiom and the triangle axiom are independent

In $\mathcal{Z}$, every integer is invertible for composition, so any chosen integers can serve as the components of $\alpha ,\lambda ,\rho$. The tensor prescription is a bifunctor because $((m^{\mathrm{\prime }}\otimes n^{\mathrm{\prime }})\circ (m\otimes n))=(m^{\mathrm{\prime }}+n^{\mathrm{\prime }})+(m+n)=(m^{\mathrm{\prime }}+m)+(n^{\mathrm{\prime }}+n)=((m^{\mathrm{\prime }}\circ m)\otimes (n^{\mathrm{\prime }}\circ n))$. Thus only the pentagon and triangle equations need to be checked.

First choose $\alpha =0$, $\lambda =1$, and $\rho =0$. Then the pentagon reads $0+0=0+0+0$, so it holds, while the triangle becomes $0+1=0$, namely $1=0$, so it fails. Hence the pentagon does not imply the triangle.

Now choose $\alpha =1$, $\lambda =0$, and $\rho =1$. Then the triangle reads $1+0=1$, so it holds, while the pentagon becomes $1+1=1+1+1$, namely $2=3$, so it fails. Hence the triangle does not imply the pentagon.

The two witnesses prove that neither axiom is a consequence of the other.