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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.

Difficulty: Medium

Correct Answer: Neither I nor II follows

Explanation:

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/Approach
Two distinct subclasses of the same parent class need not overlap, and nothing here provides existence of handles outside cups.
Testing conclusion I
From 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 II
To 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/Alternative
Model: Handles = {a, b}, Cups = {a, b}, Tubes = ∅. Both premises hold, yet I is false and II is false.
Common pitfalls
Assuming subclasses of the same superclass must intersect, or assuming existence without explicit support.
Final Answer
Neither I nor II follows.
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