From $P\subseteq \oplus P \subseteq PSPACE$ and $P\subseteq PP \subseteq PSPACE$ we infer $\oplus P\neq PP$ gives that $$P\neq PSPACE.$$

Are there any other consequences?

  • $\begingroup$ Clearly that $\oplus\neq P$ and $P\neq 0$. *rimshot* $\endgroup$ – David Richerby May 2 at 14:55

Turns out, you are immediately able to separate $\mathrm{P}$ from $\mathrm{PP}$.

Assuming that $\oplus\mathrm{P\neq PP}$ but also assume to the contradiction that $\mathrm{P=PP}$, then $\mathrm{P^{PP}=P}$. But $\oplus\mathrm{P\subseteq P^{\#P[1]}=P^{PP}=P}$. So, $\oplus\mathrm{P=P=PP}$ violating our assumption and concluding our proof by contradiction.


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