Introduction / Context:
We are to deduce which conclusions must be true from the chain of subsets and an existential statement. Be careful not to overextend “some” beyond what is stated.
Given Data / Assumptions:
- All Snakes are Trees (S ⊆ Tr).
- Some Trees are Roads (∃ Tr ∩ Rd).
- All Roads are Mountains (Rd ⊆ Mtn).
Concept / Approach:
- Transitivity of “All”: if A ⊆ B and B ⊆ C, then A ⊆ C.
- “Some Trees are Roads” together with “Roads ⊆ Mountains” yields “Some Mountains are Trees.”
- But unless Snakes are directly linked to Roads, we cannot guarantee any Snake lies in Mountains via Roads.
Step-by-Step Solution:
From ∃(Tr ∩ Rd) and Rd ⊆ Mtn, we get ∃(Tr ∩ Mtn). Hence (III) “Some mountains are trees” necessarily follows.(I) “Some mountains are snakes” would require ∃(S ∩ Mtn). We only know S ⊆ Tr; we do not know that any Snakes are among the Trees which are also Roads. So (I) is not forced.(II) “Some roads are snakes” demands ∃(Rd ∩ S). We only have ∃(Rd ∩ Tr) and S ⊆ Tr, which does not guarantee any road-tree is a snake. Hence (II) does not follow.
Verification / Alternative check:
Construct a model where the Trees that are Roads are entirely distinct from the subset of Trees that are Snakes. Premises hold; (I) and (II) fail; (III) holds.
Why Other Options Are Wrong:
Options asserting (I) or (II) require extra overlap not provided by the premises.
Common Pitfalls:
Assuming that because S ⊆ Tr and some Tr are Rd, some S must be Rd. That is not logically necessary.
Final Answer:
Only III follows
Discussion & Comments