$15 \cdot 30$ I. Consider the statements: (1) $p \wedge \sim q \wedge r$. (2) $\sim p \wedge \sim q \wedge r$. (3) $p \wedge \sim q \wedge \sim r$. (4) $\sim p \wedge q \wedge r$, where $p, q$, and $r$ are propositions. How many of these can imply that $(p \rightarrow q) \rightarrow r$ is true?
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
or II. Consider the statements: (1) $p$ and $r$ are true and $q$ is false. (2) $r$ is true and $p$ and $q$ are false. (3) $p$ is true and $q$ and $r$ are false. (4) $q$ and $r$ are true and $p$ is false. How many of these can imply that “$r$ can be deduced from ‘$p$ implies $q$’” is true?
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(14th American High School Mathematics Examination, 1963)
[Solution] If “$(p \rightarrow q) \rightarrow r$” is true, its meaning is:
(1) The result $r$ is true, and the premise $p \rightarrow q$ can be true or false;
(2) The result $r$ is false, and the premise is also false.

And “$p \rightarrow q$ is true” means:
(1) The result $q$ is true, and the premise $p$ can be true or false;
(2) The result $q$ is false, and the premise is also false.

Therefore, in (1), if $p \rightarrow q$ is false and $r$ is true, then $(p \rightarrow q) \rightarrow r$ is true; in (2), if $p \rightarrow q$ is true and $r$ is true, then $(p \rightarrow q) \rightarrow r$ is true;
in (3), if $p \rightarrow q$ is false and $r$ is false, then $(p \rightarrow q) \rightarrow r$ is true;
in (4), if $p \rightarrow q$ is true and $r$ is true, then $(p \rightarrow q) \rightarrow r$ is true.
Therefore, the answer is $(E)$.