5. (1st Grade 11 "Hope Cup" Test 2) $f$ is a one-to-one mapping from $\mathbf{R} \rightarrow \mathbf{R}$, the function $y=f(x)$ is strictly increasing, the solution set of the inequality $x>f(x)$ is $P$, and the solution set of the inequality $x>f[f(x)]$ is $Q$, then
A. $P \varsubsetneqq Q$
B. $P=Q$
C. $Q \mp P$
D. $P \varsubsetneqq Q$ and $Q \varsubsetneqq P$
5. For any $x_{0} \in P, x_{0}>f\left(x_{0}\right)$, we have $x_{0}>f\left[f\left(x_{0}\right)\right]$, so $x_{0} \in Q$, which shows that $P \subseteq Q$; On the other hand, for any $y_{0} \in$ $Q$, then $y_{0}>f\left[f\left(y_{0}\right)\right]$. We will use proof by contradiction to show that $y_{0}>f\left(y_{0}\right)$.

Assume $y_{0} \leqslant f\left(y_{0}\right)$, since the function $y=f(x)$ is strictly increasing, then we have $f\left(y_{0}\right) \leqslant f\left[f\left(y_{0}\right)\right]f\left(y_{0}\right)$, which indicates $y_{0} \in P$, i.e., $Q \subseteq P$.
By $P \subseteq Q$, and $Q \subseteq P$, we know $P=Q$. Choose B.