Proof. To prove this we choose a free resolution $F^\bullet \to R'$ of $R'$ as an $R$-module. Because $A$ and $R'$ are Tor independent over $R$ we see that $F^\bullet \otimes _ R A$ is a free $A$-module resolution of $A'$ over $A$. By our general construction of the derived tensor product above we see that
as desired. $\square$
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