We first reduce to the Noetherian case. By Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} there exists a finite type $\mathbf{Z}$ subalgebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ such that $A = A_0 \otimes_{R_0} R$. By Lemma \ref{lemma-relative-dualizing-noetherian} there exists a relative dualizing complex $K_0 \in D(A_0)$. Thus if we show (2) for $K_0$, then we find that $K_0 \otimes_{A_0}^\mathbf{L} A$ is a dualizing complex for $R \to A$ and that it also satisfies (2) by transitivity of derived base change. The uniqueness of relative dualizing complexes (Lemma \ref{lemma-uniqueness-relative-dualizing}) then shows that this holds for any relative dualizing complex. \medskip\noindent Assume $R$ Noetherian and let $K$ be a relative dualizing complex for $R \to A$. Given a ring map $R \to R'$ set $A' = A \otimes_R R'$ and $K' = K \otimes_A^\mathbf{L} A'$. To finish the proof we have to show that $K'$ is a relative dualizing complex for $R' \to A'$. By More on Algebra, Lemma \ref{more-algebra-lemma-base-change-relatively-perfect} we see that $K'$ is $R'$-perfect in all cases. By Lemmas \ref{lemma-base-change-relative-algebraic} and \ref{lemma-relative-dualizing-noetherian} if $R'$ is Noetherian, then $K'$ is a relative dualizing complex for $R' \to A'$ (in either sense). Transitivity of derived tensor product shows that $K \otimes_A^\mathbf{L} (A \otimes_R A) \otimes_{A \otimes_R A}^\mathbf{L} (A' \otimes_{R'} A') = K' \otimes_{A'}^\mathbf{L} (A' \otimes_{R'} A')$. Flatness of $R \to A$ guarantees that $A \otimes_{A \otimes_R A}^\mathbf{L} (A' \otimes_{R'} A') = A'$; namely $A \otimes_R A$ and $R'$ are tor independent over $R$ so we can apply More on Algebra, Lemma \ref{more-algebra-lemma-base-change-comparison}. Finally, $A$ is pseudo-coherent as an $A \otimes_R A$-module by More on Algebra, Lemma \ref{more-algebra-lemma-more-relative-pseudo-coherent-is-moot}. Thus we have checked all the assumptions of More on Algebra, Lemma \ref{more-algebra-lemma-compute-RHom-relatively-perfect}. We find there exists a bounded below complex $E^\bullet$ of $R$-flat finitely presented $A \otimes_R A$-modules such that $E^\bullet \otimes_R R'$ represents $R\Hom_{A' \otimes_{R'} A'}(A', K' \otimes_{A'}^\mathbf{L} (A' \otimes_{R'} A'))$ and these identifications are compatible with derived base change. Let $n \in \mathbf{Z}$, $n \not = 0$. Define $Q^n$ by the sequence $$ E^{n - 1} \to E^n \to Q^n \to 0 $$ Since $\kappa(\mathfrak p)$ is a Noetherian ring, we know that $H^n(E^\bullet \otimes_R \kappa(\mathfrak p)) = 0$, see remarks above. Chasing diagrams this means that $$ Q^n \otimes_R \kappa(\mathfrak p) \to E^{n + 1} \otimes_R \kappa(\mathfrak p) $$ is injective. Hence for a prime $\mathfrak q$ of $A \otimes_R A$ lying over $\mathfrak p$ we have $Q^n_\mathfrak q$ is $R_\mathfrak p$-flat and $Q^n_\mathfrak p \to E^{n + 1}_\mathfrak q$ is $R_\mathfrak p$-universally injective, see Algebra, Lemma \ref{algebra-lemma-mod-injective}. Since this holds for all primes, we conclude that $Q^n$ is $R$-flat and $Q^n \to E^{n + 1}$ is $R$-universally injective. In particular $H^n(E^\bullet \otimes_R R') = 0$ for any ring map $R \to R'$. Let $Z^0 = \Ker(E^0 \to E^1)$. Since there is an exact sequence $0 \to Z^0 \to E^0 \to E^1 \to Q^1 \to 0$ we see that $Z^0$ is $R$-flat and that $Z^0 \otimes_R R' = \Ker(E^0 \otimes_R R' \to E^1 \otimes_R R')$ for all $R \to R'$. Then the short exact sequence $0 \to Q^{-1} \to Z^0 \to H^0(E^\bullet) \to 0$ shows that $$ H^0(E^\bullet \otimes_R R') = H^0(E^\bullet) \otimes_R R' = A \otimes_R R' = A' $$ as desired. This equality furthermore gives the final assertion of the lemma.