6. Given that the age order of $a$, $b$, and $c$ satisfies:
(1) If $b$ is not the oldest, then $a$ is the youngest;
(2) If $c$ is not the youngest, then $a$ is the oldest.
Then the ages of these three people from oldest to youngest are ( , .
(A) bac
(B) $c b a$
(C) $a b c$
(D) $a c b$
6. (A).

Age is represented from oldest to youngest using $1,2,3$, and $a_{i}$ indicates that $a$'s age ranking is $i(1 \leqslant i \leqslant 3)$, with $b_{i}, c_{i}$ having the same meaning.

From condition (1), if $b$ is not the oldest, then $\bar{b}_{1} a_{3}$; or if $b$ is exactly the oldest, then $a$ can take any order $b_{1} a_{1}+b_{1} a_{2}+b_{1} a_{3}=b_{1}$. If 0 represents the proposition is false and 1 represents the proposition is true, we get
$$
\bar{b}_{1} a_{3}+b_{1}=1 \text {. }
$$

Similarly, from condition (2) we have
$$
\bar{c}_{3} a_{1}+c_{3}=1 \text {. }
$$
(1) $\times$ (2) gives
$$
\left(\vec{b}_{1} a_{3}+b_{1}\right)\left(\bar{c}_{3} a_{1}+c_{3} i=1,\right.
$$

which simplifies to $\bar{b}_{1} a_{3} \bar{c}_{3} a_{1}+b_{1} a_{3} c_{3}+b_{1} \bar{c}_{3} a_{1}+b_{1} c_{3}=1$.
Noting that one person cannot have two ranks, and one rank cannot be shared by two people, we have
$$
\begin{array}{l}
\bar{b}_{1} a_{3} \bar{c}_{3} a_{1}=0, \\
\bar{b}_{1} a_{3} c_{3}=0, \\
b_{1} c_{3} a_{1}=0 .
\end{array}
$$

Therefore, $b_{1} c_{3}=1$.
Thus, $b_{1}=1, c_{3}=1$, and consequently, $a_{2}=1$.
So, $b$ is the oldest, $a$ is the second oldest, and $c$ is the youngest.