Syllogism — two “some” premises with a shared middle term Statements: • Some tables are TVs. • Some TVs are radios. Conclusions to evaluate: I. Some tables are radios. II. Some radios are tables. III. All radios are TVs. IV. All TVs are tables.

Difficulty: Easy

Correct Answer: None follows

Explanation:


Introduction / Context:
This question probes your grasp of non-transitivity of some-statements. Although both conclusions mention the outer classes (tables and radios), the premises only guarantee overlaps with the middle class (TVs). Without more information, no necessary relationship among the outer classes can be deduced.



Given Data / Assumptions:

  • Tb = tables, Tv = TVs, R = radios.
  • Some Tb are Tv.
  • Some Tv are R.
  • Nothing is stated about Tb and R directly.


Concept / Approach:
Two some-type premises sharing a middle term do not allow you to infer an overlap of the two outer terms. In addition, universal conclusions like all radios are TVs or all TVs are tables cannot be justified from such weak premises. We must check each conclusion against a countermodel consistent with the statements.



Step-by-Step Solution:

Build a model where Tb ∩ Tv uses one portion of Tv, and Tv ∩ R uses a different, disjoint portion of Tv. Then Tb and R remain disjoint; thus I and II fail.Premise strength is insufficient to support III (All radios are TVs). Some radios may be outside the overlapping subset of Tv; the statement does not restrict the rest of R.Similarly, IV (All TVs are tables) is the illicit converse of Some Tb are Tv and is not warranted.


Verification / Alternative check:
Concrete example: let Tv = {a, b}, Tb = {a}, R = {b}. Premises hold. I–IV are all false, proving that none of the listed conclusions is necessary.



Why Other Options Are Wrong:

  • All follow: clearly incorrect as shown by the counterexample.
  • Only I and III or only II and IV: each includes conclusions that can fail in valid models.
  • None of these: option A already captures the correct outcome.


Common Pitfalls:
Assuming transitivity for some; converting some A are B into all B are A.



Final Answer:
None follows

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