Condition three relates to the non-decomposability of an elementary mode and is partly what makes elementary modes interesting. The two other important features as indicated before are pathway uniqueness and thermodynamic plausibility. Decomposition implies that it is possible to represent a mode as a combination of two or more other modes. For example, a mode  might be composed from two other modes,  and :
If a mode can be decomposed, does it mean that the mode is not an elementary mode? Condition three provides a rule to determine whether a decomposition means that a given mode is an elementary mode or not. If it is only possible to decompose a given mode by introducing enzymes that are not used in the mode, then the mode is elementary. That is, is there more than one way to generate a pathway (i.e., something that can sustain a steady state) with the enzymes currently used in the mode? If so, then the mode is not elementary. To illustrate this subtle condition, consider the pathway shown in below.
This pathway represents a stylized rendition of glycolysis. Step three and six are reversible and correspond to triose phosphate isomerase and glycerol 3-phosphate dehydrogenase, respectively.
The network has four elementary flux modes, which are shown in the figure below.
The elementary flux mode vectors are shown below:
Note that it is possible to have negative entries in the set of elementary modes because they will correspond to the reversible steps. Of interest is the observation that the fourth vector,  (where  represents the transpose) can be formed from the sum of the first and second vectors. This suggests that the fourth vector is not an elementary mode.
However, this decomposition only works because we have introduced a new enzyme,  (triose phosphate isomerase) which is not used in . It is, in fact impossible to decompose  into pathways that can sustain a steady state with only the five steps, ,  used in the elementary mode. We conclude therefore that  is an elementary mode.