This is not possible. pot is a pure function of inc and exc: the set of undecided transactions, picked from high feerate to low feerate ignoring topology, until the feerate stops increasing.
Presumably in your case inc=\{a\}, exc=\{\}, so und = \{b,c\}. pot is computed as follows:
- Start with pot = inc = \{a\}.
- See if adding b (the highest feerate transaction in und \setminus pot) improves it. And indeed, \{a,b\} is better than \{a\}, so b \in pot.
- See if adding c improves it. It does not: \{a,b,c\} is worse than \{a,b\}, so c \not\in pot.
Thus, pot in this case is \{a,b\}.