= Solution
A <proportional hazards family> has a common <baseline hazard> $h_0$ and positive constants $a_z$ such that $h_z(t)=a_z h_0(t)$. Thus its <hazard ratios> are constant in time, and its <survivor functions> satisfy $S_z(t)=S_0(t)^{a_z}$. An <accelerated life family> instead has positive constants $c_z$ such that $T_z$ has the distribution of $c_zT_0$. Equivalently, $S_z(t)=S_0(t/c_z)$: all time <quantiles> are multiplied by the same factor. These two properties are generally different.
Integrating the printed <Weibull distribution> density, or differentiating the following expression and checking its initial value, gives
$$
S_{p,\lambda}(t)=\exp[-(\lambda t)^p],\qquad
H_{p,\lambda}(t)=(\lambda t)^p,\qquad
h_{p,\lambda}(t)=p\lambda^p t^{p-1}.
$$
Here $\lambda$ is an inverse-time parameter: the coefficient of $t^p$ in the <cumulative hazard> is $\lambda^p$, not $\lambda$. This matters when comparing different parameterizations of a <Weibull distribution>.
With common shape $p$, the <hazard ratio> is
$$
\boxed{\frac{h_{p,\lambda_1}(t)}{h_{p,\lambda_2}(t)}
=\left(\frac{\lambda_1}{\lambda_2}\right)^p.}
$$
It does not depend on $t$, proving membership in one <proportional hazards family>. For the other property,
$$
S_{p,\lambda_1}(t)=S_{p,\lambda_2}\left(\frac{\lambda_1}{\lambda_2}t\right),
\qquad
\boxed{T_1\ \overset{d}{=}\ \frac{\lambda_2}{\lambda_1}T_2.}
$$
Consequently they also belong to one <accelerated life family>. A larger inverse-time parameter gives shorter survival times and larger hazards, consistently in both descriptions. This is the common-shape case of <Weibull accelerated-life and proportional-hazards families>.
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