TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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IP equals PSPACE

Statement

$I P = P S P A C E,$

where $I P$ is the class of languages with a constant-error interactive proof (The class IP) and $P S P A C E$ is the class of languages decidable in polynomial space. Equivalently, a language has an interactive proof with a probabilistic polynomial-time verifier, polynomially bounded interaction and constant gap between completeness and soundness if and only if it is decidable by a deterministic polynomial-space machine.

Facts & Assumptions

Given: The classes IP and PSPACE.

[A1]

$P S P A C E \subseteq I P$ : every language in PSPACE has an interactive proof with a probabilistic polynomial-time verifier, polynomially bounded interaction, perfect completeness and soundness error at most $1 / 3$ (PSPACE is contained in IP).

[A2]

$I P \subseteq P S P A C E$ : every language in IP is decidable in polynomial space (IP is contained in PSPACE).

[A3]

Two classes of languages are equal exactly when each is contained in the other; IP here is the class defined by the constant-error convention with completeness at least $2 / 3$ and soundness at most $1 / 3$ (The class IP).

Proof

technique · direct

1.1

The reverse containment $P S P A C E \subseteq I P$ is [A1]: for every $L \in P S P A C E$ there is an interactive proof whose verifier is probabilistic polynomial time and whose completeness and soundness satisfy the defining constants of IP, so $L \in I P$ .

A1A3given

1.2

The forward containment $I P \subseteq P S P A C E$ is [A2]: every language with such an interactive proof is decidable by a polynomial-space machine.

A2given

2.1

Since each of the two classes is contained in the other, they are equal by [A3]. In particular the equality does not require any strengthening of the space bounds used in either inclusion, and the constant-error convention used on both sides is the one fixed in the definitions.

step 1.1step 1.2A3given∎

Depends on

Used by

IP is closed under complement Corollary
IP admits perfect completeness Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Arora and Barak, Computational Complexity: A Modern Approach, Theorem 8.17, author-hosted draft (standard reference, not scraped)
A. Shen, IP = PSPACE: Simplified Proof, JACM 39(4) 1992, pp. 878–880 (standard reference, not scraped)