$15 \cdot 29$ (I) Consider the statements: (1) $p \wedge q$. (2) $p \wedge \sim q$. (3) $\sim p \wedge q$. (4) $\sim p \wedge \sim q$. Where $p$ and $q$ are statements, and $\wedge$ represents "and," and $\sim$ represents the negation of the statement. Among these statements, how many can lead to $\sim(p \wedge q)$ being true?
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
or (II) Consider the statements: (1) $p$ and $q$ are both true. (2) $p$ is true and $q$ is false. (3) $p$ is false and $q$ is true. (4) $p$ is false and $q$ is false. How many of these statements can lead to the negation of "both $p$ and $q$ are true"?
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(13th American High School Mathematics Examination, 1962)
[Solution] For type (I), we have:
Since $\sim(p \wedge q)=\sim p \wedge \sim q$, then statements (2), (3), (4) are correct.
For type (II), we have:
The negation of the statement “$p$ and $q$ are both true” means: “$p$ is true and $q$ is false” or “$p$ is false and $q$ is true” or “$p$ and $q$ are both false”. Hence, statements (2), (3), (4) are correct.
Therefore, the answer is $(D)$.