Let $A,B$ be two languages, for which we know:

  • $A \in PSPACE$
  • $A\le_LB$

Can we conclude from the above that $B \in PSPACE$ ?

I think the answer is no, however I don't know how to prove it. I guess I have to write a reduction that proves otherwise, but how?

Thank you!

  • $\begingroup$ Hint: LOGSPACE$\subseteq $PSPACE, and a logspace reduction has enough power to decide membership for a language in LOGSPACE. $\endgroup$
    – Shaull
    Commented Jun 30, 2018 at 8:47
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    – Raphael
    Commented Jun 30, 2018 at 10:00
  • $\begingroup$ What have you tried? Where did you get stuck? We do not want to just hand you the solution; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for tips on asking questions about exercise problems. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? $\endgroup$
    – Raphael
    Commented Jun 30, 2018 at 10:01

1 Answer 1


No. Note that given languages $A,B$ and a reduction $A\le_L B$, you get that $B$ is at least as hard as $A$, but that's about it.

You can contradict the above statement with the following extreme:

Take $A = \{0\}$, then $A\in \text{PSPACE}$, and take $B= H_{TM}\cup\{0\}$.

Define $f$ to be the following reduction $A \le_L H_{TM}\cup\{0\}$ (we can assume that $1\notin H_{TM}$):

$$f(0) = 0$$ $$\forall x\in \Sigma^*\setminus\{0\} . f(x)=1$$.

Clearly $f$ works in logarithmic space, but $B$ is clearly not in $\text{PSPACE}$. In fact, it isn't even in $\text{R}$.


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