I need some help for one equality in the following proposition. It was a hint to conclude that $\sup\{f(\cdot):f\in\Phi\}$ is additive. I highlighted it blue. Ultimately I am interested in proving the proposition, but the main question is a hint with how to obtain the blue equality $\overset{☹}{=}$.
Proposition. Let $(X,+,0)$ be an abelian monoid. Let $(M,\preceq, +,0)$ be an ordered monoid. Fix $\Phi\subseteq M^X$. Assume the following.
(i) $(\forall f\in\Phi)[f:(X,+,0)\to (M,+,0)]$
(ii) $(\forall f_1,f_2\in\Phi)(\exists f_3\in\Phi)(\forall x\in X)[f_1(x)\preceq f_3(x)\text{ and }f_2(x)\preceq f_3(x)]$
(iii) $(\forall x\in X)[\sup\{f(x):f\in\Phi\}\text{ exists in }M]$
Then $\sup\{f(\cdot):f\in\Phi\}$ is additive.
hint provided from source material: For all $x,y\in X$, $$\begin{align} \color{blue}{\sup\{f(x):f\in\Phi\}+\sup\{f(y):f\in\Phi\}}&\overset{\color{blue}☹}=\color{blue}{\sup\{f_1(x)+f_2(y):f_1,f_2\in\Phi\}}\\ &\preceq \sup\{f(x)+f(y):f\in\Phi\}\\ &=\sup\{f(x+y):f\in\Phi\}\\ &\preceq\sup\{f(x):f\in\Phi\}+\sup\{f(y):f\in\Phi\} \end{align}$$
I can easily show $\succeq$ of $\overset{☹}{=}$, but not the other direction. I believe I am missing something subtle with the assumptions because with ordered monoids one has $$\sup S+\sup T\succeq \sup(S+T)\tag{1}$$ when the suprema all exist. In particular, if one considers $(-\infty,+\infty]$ adjoined with $\{n\cdot +\infty\}_{n\in\mathbb{N}}$ we have an ordered monoid where if $S=T=\mathbb{R}$, then we get a strict inequality of $(1)$. If instead one considers ordered groups, then one can show equality in $(1)$. It is because of this that I somehow believe the proposition should be considering ordered groups instead of ordered monoids, but, again, I may be missing something subtle with the assumptions.
I also want to add I already know how to prove the proposition if I have the blue equality at hand.