Home » Logical Reasoning » Logical Problems

Syllogism validation with set inclusion: if all trees in the park are flowering and some trees are dogwoods, assess the claim 'All dogwoods in the park are flowering trees'

Difficulty: Easy

Correct Answer: True

Explanation:


Given data

  • All Trees_in_park ⊆ Flowering.
  • Some Trees_in_park ∩ Dogwoods ≠ ∅.


Concept/Approach (why this method)

Universal inclusion: every member of the 'trees in the park' set is flowering; any subclass (e.g., dogwoods in the park) inherits this property.


Step-by-Step deduction
1) Let D = {dogwoods in the park}. By premise, D ⊆ Trees_in_park.2) Since Trees_in_park ⊆ Flowering, transitivity gives D ⊆ Flowering.


Verification/Alternative

Venn diagram: place 'Trees' circle fully inside 'Flowering'; 'Dogwoods in park' is a sub-region within 'Trees', hence also inside 'Flowering'.


Common pitfalls

  • Confusing 'some trees are dogwoods' with 'some dogwoods are flowering'—here all park trees are flowering, so all park dogwoods are flowering.


Final Answer
True

← Previous Question Next Question→

More Questions from Logical Problems

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion