Set inclusion puzzle: ‘‘All tubes are handles’’ and ‘‘All cups are handles’’ — decide whether it logically follows that all cups are tubes and whether some handles are not cups.

Question

Solve this multiple-choice question and choose the correct option.

Curioustab · Accepted Answer

We have two universal inclusions into the same superset and must judge two claims. Premise 1: All tubes are handles, so Tubes ⊆ Handles. Premise 2: All cups are handles, so Cups ⊆ Handles. Conclusions: I. All cups are tubes. II. Some handles are not cups. Concept/ApproachTwo distinct subclasses of the same parent class need not overlap, and nothing here provides existence of handles outside cups.Testing conclusion IFrom Cups ⊆ Handles and Tubes ⊆ Handles we cannot infer Cups ⊆ Tubes. The two subclasses could be disjoint inside Handles. Thus I does not follow.Testing conclusion IITo claim Some handles are not cups, we need at least one handle that is not a cup. The premises allow Handles to equal Cups as a special case, so the existence claim is not forced. Thus II does not follow.Verification/AlternativeModel: Handles = {a, b}, Cups = {a, b}, Tubes = ∅. Both premises hold, yet I is false and II is false.Common pitfallsAssuming subclasses of the same superclass must intersect, or assuming existence without explicit support.Final AnswerNeither I nor II follows.