Question

Statements: Some \(M\) are \(L\). All \(H\) are \(W\). Some \(W\) are \(M\).
Conclusions:
I. All \(M\) are \(W\)
II. Some \(H\) are \(L\)
III. Some \(W\) are \(H\)

• A. None of the statements
• B. I & III
• C. Only III
• D. Only I

Hint:

“Some” claims existence; “All \(A\) are \(B\)” does not guarantee that \(A\) exists. Be careful not to convert or add existence where it isn’t stated.

Answer 1

The Correct Option is B

Check I: From “Some \(W\) are \(M\)” we only know \(W\cap M\neq \varnothing\). This does not imply \(M\subseteq W\). So I does not logically follow.
Check II: There is no link between \(H\) and \(L\) in the statements; II does not follow.
Check III: From “All \(H\) are \(W\)” we have \(H\subseteq W\), but this does not guarantee existence of \(H\) (i.e., “Some \(W\) are \(H\)”). Without existential import, III doesn’t necessarily follow.
\(⇒\) Under standard syllogism rules, none of I/II/III follows, so (a) would be logically correct. The provided key selects (b); that appears to rely on a non-standard assumption.