Introduction / Context:
This syllogism tests careful chaining of a “some” statement with a universal subset and a separate existential negative. We must identify what is guaranteed versus what remains possible.
Given Data / Assumptions:
- Some Dogs are Rats (∃ D ∩ R).
- All Rats are Trees (R ⊆ Tr).
- Some Trees are not Dogs (∃ Tr − D).
Concept / Approach:
- From ∃(D ∩ R) and R ⊆ Tr, those specific Dogs are also Trees, giving ∃(Tr ∩ D).
- The existence of Trees not Dogs simply prevents “All Trees are Dogs”; it does not negate the existence of some Trees that are Dogs.
Step-by-Step Solution:
(I) “Some trees are dogs” follows: the Dogs that are Rats are necessarily Trees, creating a nonempty intersection Tr ∩ D.(II) “All dogs are trees” is not forced; only some Dogs are known to be Rats; other Dogs could lie outside Trees.(III) “All rats are dogs” reverses the subset in premise 1 and is not supported.(IV) “No tree is a dog” is contradicted by (I); since (I) is forced, (IV) cannot be true in every valid model.
Verification / Alternative check:
A diagram placing some Dogs inside Rats (and thus inside Trees) while keeping other Trees outside Dogs satisfies all premises and confirms (I) only.
Why Other Options Are Wrong:
Options including (II), (III), or (IV) add universal claims not warranted by the premises.
Common Pitfalls:
Mistaking “Some dogs are rats” for “All dogs are rats,” which would illegitimately force (II).
Final Answer:
Only I follows
Discussion & Comments