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

Analysis I: Given “Some \(W\) are \(M\)”, we only know \(W\cap M\neq \varnothing\). This doesn't imply \(M\subseteq W\). Thus, I is not logically sound.
Analysis II: The statements provide no connection between \(H\) and \(L\); II is invalid.
Analysis III: From “All \(H\) are \(W\)”, we have \(H\subseteq W\), but this doesn't guarantee the existence of \(H\) (i.e., “Some \(W\) are \(H\)”). Without existential import, III is not necessarily valid.
\(\Rightarrow\) Under standard syllogism rules, none of I/II/III are valid, so (a) is logically correct. The provided key selects (b); this seems to depend on a non-standard assumption.