Syllogism — Chain a particular statement through two universal inclusions Statements: • Some radios are telephones. • All telephones are mirrors. • All mirrors are desks. Conclusions: I) Some radios are desks. II) Some radios are mirrors. III) Some desks are telephones. Choose the conclusion set that must be true.

Introduction / Context:
Here, one existential premise (“some radios are telephones”) is combined with two universal inclusions to test whether you can correctly push membership forward and derive the necessary existential consequences at each stage.

Given Data / Assumptions:

• There exists r with r ∈ Radios and r ∈ Telephones.
• Telephones ⊆ Mirrors.
• Mirrors ⊆ Desks.

Concept / Approach:
If an element is in Telephones and all Telephones are Mirrors, that same element is in Mirrors. If all Mirrors are Desks, it is also in Desks. Therefore, the very same radio-telephone element witnesses both “Some radios are mirrors” and “Some radios are desks.” Moreover, since at least one Telephone exists (from the first premise), and all Telephones are Desks, it follows that some Desks are Telephones.

Step-by-Step Solution:

Verification / Alternative check:
Diagram the chain: Radios —(some overlap)→ Telephones —(subset)→ Mirrors —(subset)→ Desks. The original witness element carries through the chain, and the existence of at least one Telephone gives III.

Why Other Options Are Wrong:
Any option omitting one of I–III ignores a consequence forced by pushing the same element through the subset chain.

Common Pitfalls:
Losing track of the same individual as it moves through universal inclusions; forgetting that one “some” can propagate forward through “all.”

Final Answer:
All follow.