Usually I look at a category as a digraph with an associative (partial) binary function $\circ$ on its arrows.1 When regarding commutative diagrams I look at a category as a digraph with an equivalence relation $\simeq$ on its paths. Just like the function $\circ$ has to be associative, the equivalence relation $\simeq$ has to fulfill a condition – to be considered "categorical" or "commutative".
I guess (but I may be wrong) that this condition simply is:
An equivalence relation $\simeq$ on the paths of a digraph $G$ is commutative when there is an associative binary function $\circ$ on its arrows such that2
$ \qquad G \models p_1 + \dots + p_n \simeq q_1 + \dots + q_m$
$ \qquad\qquad\qquad$if and only if
$ \qquad G \models p_1 \circ \dots\circ p_n = q_1 \circ \dots \circ q_m$
i.e. when $\simeq$ is (naturally?) induced by an associative function $\circ$.
What I'd like to know:
How can commutative equivalence relations on the paths of a digraph be characterized, i.e. which of their necessary properties are sufficient?
Consider the following properties that commutative equivalence relations – as defined above – must have by definition:3
(1) $ \qquad |p|=|q|=1 \qquad\qquad\qquad\quad\ \ \Rightarrow \big(\exists\ r\ : |r| = 1 \wedge p + q \simeq r\big)$
(2) $ \qquad |p|=|q|=1 \qquad\qquad\qquad\quad\ \ \Rightarrow \big( p \simeq q \Leftrightarrow p = q \big)$
(3) $ \qquad |p|=|q|=|r|=1 \qquad\qquad\quad \Rightarrow \big( p \simeq q \Rightarrow p + r \simeq q + r \big)$
(3') $ \qquad |p|=|q|=|r|=1 \qquad\qquad\quad \Rightarrow \big( p \simeq q \Rightarrow r + p \simeq r + q\big)$
(4) $ \qquad |p|=|q|=|q'| = |r|= |r'|= 1 \Rightarrow \big( q + p \simeq r \wedge p + q' \simeq r' \Rightarrow q + r' \simeq r + q'\big)$
Note that (4) is an explicit expression of the associativity of $\circ$ in terms of $\simeq$.
So my main question is:
Are the properties (1) to (4) – eventually – sufficient to characterize commutative equivalence relations?
If not so: what is missing?
More of a side question is:
(Under which conditions) can the preconditions of properties (3) to (4) – requiring that the paths involved be arrows – be dropped, i.e., the conclusions hold for all paths, not only for arrows (which seems true at blue-eyed first sight)?
1 The bookkeeping details concerning source and target of both arrows and paths and their composites are tacitly assumed.
2 Let $+$ be the (associative) path concatenation operator.
3 Let $|p|$ be the length of the path $p$, so $|p|=1$ means that $p$ is an arrow.