Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I prove it?

  1. $A \le_p B \Leftrightarrow \overline{A} \le_p \overline{B}$.

  2. $A \in \mathcal P \Leftrightarrow \overline{A} \in \mathcal P$.

$\overline{A} = \Sigma^* \setminus A$

share|cite|improve this question
Both properties are really easy once you understand the definitions. – sdcvvc Jan 18 '12 at 22:46
up vote 1 down vote accepted

These can easily be proven by expanding definitions.

  1. If $A \leq_p B$, then there exists a polynomial time computible function f such that: \begin{align*} x \in A \iff f(x) \in B \end{align*} However, this is equivalent with \begin{align*} x \not \in A \iff f(x) \not\in B \end{align*} and thus $\overline{A}\leq_p \overline{B}$. The other direction is analogous.

  2. Suppose $A \in \mathcal{P}$, then there exists a polynomial time Turing machine $M$ that decides $A$. We now construct a Turing machine $M'$ that outputs exactly the opposite of $M(x)$. Clearly $M'$ decides $\overline{A}$ and still runs in polynomial time, thus $\overline{A} \in \mathcal{P}$. The other direction is analogous.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.