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Let $W(C)={\sum }_{i\in C}{w}_i$. If the two disjoint winning coalitions exchange players, the aggregate weight is conserved:If both resulting coalitions lost, each weight would be strictly below $q$, making their sum strictly below $2q$. This is impossible. Thus every weighted voting game has swap robustness for disjoint winning coalitions, even when the weights or quota are negative.

The converse is false. The four-player prime-cardinality game from part (a) has swap robustness for disjoint winning coalitions: swapping one player each way preserves each coalition's cardinality, so in fact both winning coalitions remain winning. Part (a) proves it is not a weighted voting game.

Even imposing monotonicity would not repair the converse under the printed disjointness restriction. As a stronger counterexample, use five players and declare a coalition winning exactly when it contains player $1$ and contains either both $2,3$ or both $4,5$. This is a simple cooperative game, and all its winning coalitions contain player $1$, so no two are disjoint; the stated swap condition holds vacuously.

Nevertheless $\{1,2,3\}$ and $\{1,4,5\}$ win, while $\{1,2,4\}$ and $\{1,3,5\}$ lose. Both pairs contain every player the same total number of times. A weighted representation would therefore givea contradiction. This is a trade obstruction to weighted voting and pinpoints how weak the disjoint-only swap requirement is.