$15 \cdot 24$ Given two assumptions: ( I ) At least one Mems is not an Ens. ( II ) No Ens is a Vees, we can infer
(A) At least one Mems is not a Vees.
(B) At least one Vees is not a Mems.
(C) No Mems is a Vees.
(D) At least one Mems is a Vees.
(E) We cannot infer $(A)$ or $(B)$ or $(C)$ or $(D)$ from the given assumptions.
(18th American High School Mathematics Examination, 1967)
[Solution] Let $M, E, V$ denote the sets of Mems, Ens, and Vees, respectively.
From the problem, we know there is at least one $m \in M$, and $m \notin E$, and $E \cap V=\varnothing$. If $M=V$, then $(A),(B),(C)$ do not hold;
if $M \cap V=\varnothing$, then $(D)$ does not hold.
Therefore, the answer is $(E)$.