= Solution
{c}
For monotonically expanding <scale factor> $R$, $dt=dR/(RH)$, so the elapsed <cosmic time> is
$$
\boxed{t_0-t_i=\int_{R_i}^{R_0}\frac{dR}{R\,H(R)}
=\int_0^{z_i}\frac{dz}{(1+z)H(z)}.}
$$
The second expression uses <cosmological redshift> $1+z=R_0/R$; the <age of the universe> is the limit to its earliest specified state, if that limit is finite.
An <age comparison for an initial vacuum-dominated phase> needs a common normalization. Match the two histories at the end of S1, with $R=R_1$ and $H=H_1>0$, and give them identical subsequent S2 and S3 evolution, including transition radii and today's state. Below $R_1$, the history with S1 has constant $H=H_1$. The comparison history extends S2 backwards, so $\rho\propto R^{-4}$ and $H=H_1(R_1/R)^2$. Starting both at $R_i=\epsilon R_1$, where $0<\epsilon<1$, their early durations are
$$
\Delta t_{\rm S1}=\frac1{H_1}\ln\frac1\epsilon,\qquad
\Delta t_{\rm S2}=\int_{\epsilon R_1}^{R_1}\frac{R\,dR}{H_1R_1^2}
=\frac{1-\epsilon^2}{2H_1}.
$$
The difference is positive because
$$
\boxed{\Delta t_{\rm S1}-\Delta t_{\rm S2}
=\frac1{H_1}\int_\epsilon^1\left(\frac1x-x\right)dx>0.}
$$
The common later evolution adds the same time to both histories. Thus S1 makes this matched universe older. More strongly, as $R_i\to0$, the exact S1 duration diverges logarithmically, whereas a radiation beginning has finite duration $1/(2H_1)$ to the match. The exact <de Sitter spacetime> expanding patch reaches $R=0$ only as $t\to-\infty$ for its <comoving observers>; a finite “age since $R=0$” does not exist in this ideal model. This does not establish past <geodesic completeness> of the patch.
The matching assumption is essential for a finite cutoff or realistic finite period of <cosmic inflation>: arbitrary changes of the present <Hubble parameter> or transition densities prevent a universal age comparison. The literal zero-size limit in this question is instead the infinite-duration idealization just derived.
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