To prove (1), suppose by way of contradiction that \(df = 0\) for some nonzero \(f \in A\text{.}\) Since \(A\) is \(p\)-torsion-free and \(p\)-adically separated, we may divide \(f\) by a suitable power of \(p\) to reduce to the case where \(f \notin pA\text{.}\) Now
\begin{equation*}
0 = \delta(df) = f^p \delta(d) + \delta(f) \phi(d).
\end{equation*}
Multiplying by \(\phi(f)\) and using that \(\phi\) is a ring homomorphism, we obtain
\begin{equation*}
0 = f^p \phi(f) \delta(d) + \delta(f) \phi(df) = f^p \phi(f) \delta(d).
\end{equation*}
Since \(A\) is \((p,d)\)-local, \(\delta(d)\) is a unit in \(A\text{,}\) so \(f^p \phi(f) = 0\text{.}\) Reducing modulo \(p\text{,}\) we obtain \(f^{2p} \equiv 0 \pmod{p}\text{.}\) Since \(A/p\) is reduced, this implies \(f \equiv 0 \pmod{p}\text{,}\) contradicting our earlier choice of \(f\) and thus proving the claim.
To prove (2), it is enough to show that
\((A/d)[p^2] = (A/d)[p]\text{.}\) That is, given
\(f,g \in A\) with
\(p^2 f = gd\text{,}\) we must have
\(pf \in dA\text{.}\) Since
\(gd \in p^2 A\text{,}\) we have
\(\delta(gd) \in pA\) and hence
\(\phi(g) \delta(gd) \in pA\text{.}\) Rewriting this as
\(\delta(d) g^p \phi(g) + \delta(g) \phi(gd)\text{,}\) we see that
\(\delta(d) g^p \phi(g) \in pA\text{.}\) Since
\(A\) is
\((p,d)\)-local,
\(\delta(d)\) is a unit in
\(A\text{,}\) so
\(g^p \phi(g) \in pA\) and so
\(g^{2p} \in pA\text{.}\) Because
\(A/p\) is reduced, this implies
\(g \in pA\text{;}\) since
\(A\) is
\(p\)-torsion-free, this implies that
\(pf \in dA\) as desired. (Compare
[25], Lemma 2.34.)