$36 \cdot 14$ Given a set $S$ composed of two undefined elements “pib” and “maa”, and the following four axioms: $P_{1}$: Each pib is a set of maa; $P_{2}$: Any two distinct pibs have exactly one common maa; $P_{3}$: Each maa belongs to exactly two pibs; $P_{4}$: There are exactly four pibs.

Consider the three theorems: $T_{1}$: There are exactly 6 maas; $T_{2}$: Each pib has exactly three maas; $T_{3}$: For each maa, there is exactly one other maa that is not in the same pib. Which of the following can be deduced from the axioms?
(A) Only $T_{2}$.
(B) Only $T_{2}$ and $T_{3}$.
(C) Only $T_{1}$ and $T_{2}$.
(D) Only $T_{1}$ and $T_{3}$.
(E) All.
(18th American High School Mathematics Examination, 1967)
[Solution] For convenience, let the digits $1,2,3,4$ represent four different $pib$ (Axiom $P_{4}$); these are all sets of maa (Axiom $P_{1}$).
The only common maa contained in both $pib_{i}$ and $pib_{j}$ is denoted by $ij$ or $ji$ (Axiom $P_{2}$).
Since any maa can be represented in this way (Axiom $P_{3}$), the set of all maa is
$$
\{12,13,14,23,24,34\}
$$

which has exactly $C_{4}^{2}=6$ elements (Theorem $T_{1}$).
$pib_{i}$ contains only three maa, which are $ij (i \neq j)$ (Theorem $T_{2}$). There is exactly one maa that is neither in $pib_{i}$ nor in $pib_{j}$, and thus does not fall into the same $pib$ as $ij$ (Theorem $T_{3}$).
In summary, Theorems $T_{1}, T_{2}, T_{3}$ must be derived using all four axioms.
Therefore, the answer is $(E)$.
Note: Four lines in a plane, no three of which are concurrent and none of which are parallel, provide a geometric model for this problem:

In the figure, the six intersection points (maa) satisfy Axioms $P_{1}, P_{2}, P_{3}, P_{4}$, and the four line segments containing three collinear points are the four $pib$.

In this (finite) geometry: any two lines ($pib$) must intersect at a unique point (maa), but two points (maa) do not necessarily determine a line segment in this system.