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✓ 17 results · all verified · 11 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The IP = PSPACE Theorem

1 · Prerequisites

2 · Summary

This page proves IP=PSPACE by arithmetizing quantified Boolean formulas. A closed prenex formula is first read as a polynomial over a prime field, with the quantifier operators A and E reproducing the Boolean semantics on Boolean assignments, and with the multilinearization reductions R inserted before every quantifier operation so that each individual degree stays at most D=max⁡{L,2} through an operator list of length T=n(n+3)/2. A deterministic trial-division search supplies a prime field of polynomial bit length with p>12TD, which is what keeps the root bound 2TD/p below 1/6, and hence the soundness error below 1/3.

The Shamir protocol then reverses that operator list: the prover sends the univariate restriction at each node, the verifier checks the node identity at 0 and 1, samples a fresh challenge, and finally tests the matrix value at the accumulated point. The page proves the honest-prover invariant, the polynomial communication and verifier-time bounds, perfect completeness, and the round-by-round soundness estimate that a false claim becomes true with probability at most 2D/p; a union bound over the T nodes turns this into soundness error at most 2TD/p<1/3. With the published inclusion IP⊆PSPACE and the reduction route through TQBF completeness, the equality follows, together with closure under complement and the perfect-completeness form of the theorem. False statements on this page record what the proof uses: neither the degree reduction nor the verifier's terminal evaluation of the matrix can be dropped.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Field arithmetization of QBF quantifiers

Definition

Let F be a field (Field) and let m≥1. For a polynomial P∈F[X1,…,Xm] and an index 1≤i≤m, write P∣Xi=0,P∣Xi=1 for the two polynomials in F[X1,…,Xi−1,Xi+1,…,Xm] obtained by substituting the field elements 0 and 1 for Xi. Define the field quantifier operators AXiP:=(P∣Xi=0) (P∣Xi=1),EXiP:=1−(1−P∣Xi=0)(1−P∣Xi=1), both of which again lie in F[X1,…,Xi−1,Xi+1,…,Xm].

Let Φ=Q1x1 Q2x2⋯Qnxn ψ be a closed prenex quantified Boolean formula (Quantified Boolean formulas and the language TQBF) with quantifier-free matrix ψ, and let b:=Pψ∈F[X1,…,Xn] be the arithmetization of the matrix (Arithmetization of Boolean formulas). The ordered arithmetization of Φ over F is the constant obtained from b by processing the quantifiers from the innermost to the outermost: put P(n):=b and, for j=n,n−1,…,1, put P(j−1):=AXjP(j) when Qj=∀ and P(j−1):=EXjP(j) when Qj=∃. The resulting P(0) is a polynomial in no variables, hence a field element.

Remarks

  • The operators act on the polynomial and the index, not on a formula tree: each specialization is a well-defined polynomial in the remaining variables, and the two displayed expressions are computed in that polynomial ring. No enumeration, decomposition, isomorphism or choice is made in the definition, and no expansion into monomials is required.
  • The convention "inner quantifiers first" is used throughout this page: the prefix is read left to right and Qn, the innermost quantifier, is processed first, so Q1 is processed last. The operator list of Multilinearization in one variable refines this ordered convention.
  • The field elements 0 and 1 are the two Boolean values, and they are distinct in a field. The gate operations reproduce the truth tables only at Boolean inputs; that agreement is the subject of Quantifier polynomials agree with QBF semantics on Boolean assignments, and no such agreement is asserted here.
  • The definition itself does not assert a truth-value theorem. For the multilinearized refinement of this sequence in Multilinearization in one variable, agreement with QBF truth is proved in The ordered arithmetization evaluates to the quantified Boolean truth value.
  • For n=0 the matrix is a formula with no variable occurrences, b∈F is a constant, and P(0)=b.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Quantifier polynomials agree with QBF semantics on Boolean assignments

Statement

Let F be a field, let φ(x,y1,…,ym) be a quantifier-free Boolean formula on the variables x,y1,…,ym with m≥0, and let P:=Pφ∈F[X,Y1,…,Ym] be its arithmetization. Then for every Boolean assignment a∈{0,1}m to y1,…,ym the two operators of Field arithmetization of QBF quantifiers satisfy (AXP)(a)=the truth value of ∀x φ(x,a),(EXP)(a)=the truth value of ∃x φ(x,a), where Boolean truth values are embedded in F as 0 and 1.

Facts & Assumptions

Given: A field F, a quantifier-free Boolean formula φ(x,y1,…,ym) with m≥0, its arithmetization P=Pφ, and a Boolean assignment a to y1,…,ym.

[A1]

Arithmetization replaces ¬ by 1−Pψ, ∧ by PψPθ, and ∨ by Pψ+Pθ−PψPθ, and the Boolean value of a formula is obtained from the usual truth tables, with false and true identified with the field elements 0 and 1 (Arithmetization of Boolean formulas).

[L1]

For every Boolean assignment z to x,y1,…,ym, the arithmetization Pφ(z) equals the Boolean value of φ at z, embedded as 0 or 1 in F (Arithmetization preserves Boolean values).

[L2]

The operators are AXP=(P∣X=0)(P∣X=1) and EXP=1−(1−P∣X=0)(1−P∣X=1), evaluated at the remaining variables (Field arithmetization of QBF quantifiers).

Proof

technique · direct
1.1

On the four bit pairs (u,v)=(0,0),(0,1),(1,0),(1,1) the product uv takes the values 0,0,0,1, and the expression u+v−uv takes the values 0,1,1,1. These are exactly the truth tables of the conjunction and the disjunction of two bits, and the computations are polynomial identities valid in every field, including fields of characteristic two.

A1algebra
1.2

Let b denote 0 or 1. By [L1] the value P∣X=b(a)=Pφ(b,a) is the Boolean value of the formula φ(b,a) at the assignment a, hence an element of {0,1}⊂F. This includes the case m=0, where a is the empty assignment and φ has no free variable besides x, and the case in which φ is a constant formula.

L1given
2.1

Put u:=P∣X=0(a) and v:=P∣X=1(a). By step 1.2 both are bits. Hence by [L2] (AXP)(a)=uv,(EXP)(a)=u+v−uv. By step 1.1 the first value is 1 exactly when both φ(0,a) and φ(1,a) are true, which is the truth value of ∀x φ(x,a), and the second is 1 exactly when at least one of the two is true, which is the truth value of ∃x φ(x,a). Both operator values are therefore themselves bits.

step 1.1step 1.2L2algebra
3.1

Consequently AX and EX, evaluated at any Boolean assignment to the remaining variables, return the truth values of the corresponding quantifications of the arithmetized subformula.

step 2.1∎
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Multilinearization in one variable

Definition

Let F be a field and let P∈F[X1,…,Xm] with m≥1. For 1≤i≤m define the multilinearization of P in Xi by RXiP:=(1−Xi) (P∣Xi=0)+Xi (P∣Xi=1). This is a polynomial in F[X1,…,Xm] of degree at most one in Xi; in every other variable its degree is at most the corresponding degree of P.

Let Φ=Q1x1⋯Qnxnψ be a closed prenex quantified Boolean formula (Quantified Boolean formulas and the language TQBF) with quantifier-free matrix ψ and matrix arithmetization b=Pψ (Field arithmetization of QBF quantifiers). The multilinearized ordered arithmetization of Φ is obtained from b by applying, for j=n,n−1,…,1, first the j reductions RX1,RX2,…,RXj in this order and then the quantifier operator Oj, where Oj:=AXj when Qj=∀ and Oj:=EXj when Qj=∃. The operator list of Φ is the resulting sequence of operations read in application order, RX1,…,RXn, On,RX1,…,RXn−1, On−1,…,RX1, O1, and its length is denoted T. Then T=n(n+1)/2+n=n(n+3)/2, and for n=0 the list is empty.

Write P(n):=b. For j=n,…,1, write Mj for the polynomial obtained from P(j) by the j reductions of the j-th block, and P(j−1):=Oj(Mj). The final P(0) is a constant.

Remarks

  • Substituting Xi=0 into RXiP gives (1−0)P∣Xi=0+0=P∣Xi=0, and substituting Xi=1 gives P∣Xi=1; so RXiP agrees with P where the substituted variable is Boolean, and its coefficient of Xi is P∣Xi=1−P∣Xi=0.
  • Reductions in distinct variables commute with one another, because RXi alters only the Xi-dependence. The displayed order is the one the interactive protocol of The Shamir interactive protocol for TQBF reverses; it is recorded here so that the protocol has a definite node list.
  • The operator list is not the same as the ordered arithmetization of Field arithmetization of QBF quantifiers: the reductions are inserted before each quantifier operation, and they are what keep the individual degrees of the intermediate polynomials bounded. That bound, together with the agreement of the sequence with the Boolean semantics, is proved in Multilinearization preserves Boolean values and bounds individual degree and The ordered arithmetization evaluates to the quantified Boolean truth value.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Multilinearization preserves Boolean values and bounds individual degree

Statement

Let F be a field and RXi the reduction of Multilinearization in one variable.

  1. For every P∈F[X1,…,Xm] and b∈{0,1}, (RXiP)∣Xi=b=P∣Xi=b. Consequently, if P agrees with a function on the Boolean cube at every Boolean point, then so does RXiP.
  2. RXiP has degree at most one in Xi, and degree in every other variable at most that of P there.
  3. Let Φ=Q1x1⋯Qnxnψ be a closed prenex quantified Boolean formula whose quantifier-free matrix ψ has L syntax nodes, let b=Pψ and D:=max⁡(L,2), and let P(j) and Mj be the polynomials of the multilinearized ordered arithmetization. Then each P(j) with j≤n−1 has individual degree at most 2 in each of x1,…,xj, each Mj is multilinear in x1,…,xj, and every node polynomial of the operator list, after substituting arbitrary field elements for all variables other than its active variable, has degree at most D in that active variable. The operator list has T=n(n+3)/2=O(n2) entries.

Facts & Assumptions

Given: A field F, a polynomial P over it, and a closed prenex quantified Boolean formula Φ with quantifier-free matrix ψ of L syntax nodes.

[A1]

Multilinearization is RXiP=(1−Xi)P∣Xi=0+XiP∣Xi=1; the operator list of Φ runs, for j=n down to 1 through the blocks RX1,…,RXj,Oj, and has T=n(n+1)/2+n entries (Multilinearization in one variable).

[A2]

The matrix arithmetization b=Pψ is built by the gate rules of Arithmetization of Boolean formulas, and Oj is AXj or EXj according to Qj (Field arithmetization of QBF quantifiers).

[L1]

If φ is a formula with s≥1 syntax nodes in which ti leaves are labelled Xi, then Pφ has individual degree at most ti in Xi, hence at most s; and Pφ can be evaluated at a supplied point in O(s) field operations (Formula degree and point-evaluation cost).

[L2]

Write pb=p∣Xi=b. If deg⁡Xjp≤dj for j≠i, then p0+p1 has bound dj in Xj, while p0p1 and p0+p1−p0p1 have bound 2dj (Degree under Boolean elimination operators).

Proof

technique · direct
1.1

Substituting Xi=0 in the right side of [A1] leaves P∣Xi=0 and substituting Xi=1 leaves P∣Xi=1. The coefficient of Xi is P∣Xi=1−P∣Xi=0, so the Xi-degree is at most one; specialization, multiplication by 1−Xi or by Xi, and addition do not raise the degrees in the other variables. This proves claims (1) and (2), including for constant and zero polynomials.

A1algebra
2.1

If P agrees with a function g at every Boolean point, then at each Boolean point the value of RXiP is the value of P at the same point by step 1.1 and hence equals g there. So RXiP also agrees with g on the Boolean cube, whichever bits g takes.

step 1.1given
2.2

By [L1] the matrix arithmetization b=Pψ has individual degree at most L in every variable. Let 1≤j≤n and suppose P(j) has individual degree at most 2 in x1,…,xj when j≤n−1. By step 1.1 the reductions of the block j do not increase degrees in xj+1,…,xn and make each of x1,…,xj have degree at most one, so Mj is multilinear in x1,…,xj and its individual degrees in xj+1,…,xn are at most those of P(j).

L1A1A2step 1.1
3.1

Applying Oj specializes Mj at Xj=0 and Xj=1 and multiplies or combines the two results, so [L2] gives P(j−1) individual degree at most twice that of Mj in each remaining variable. At j=n the polynomial P(n−1) has individual degree at most 2 in x1,…,xn−1, because Mn is multilinear; inductively the same bound holds at every level j≤n−1.

L2step 2.2algebra
4.1

For the degree of a node polynomial in its active variable, consider the blocks in turn. The reductions of the first block act on b and on its successive reductions; repeated use of claim (2) of step 1.1 shows that the input polynomial of the reduction RXi in that block has Xi-degree at most that of b, namely at most L. Every later reduction acts on some P(j) with j≤n−1, whose individual degrees are at most 2 by step 3.1, and the quantifier node Oj acts on the multilinear Mj, of Xj-degree at most one. Substituting arbitrary field elements for the other variables cannot raise any of these degrees, so every node polynomial specializes to a univariate of degree at most D=max⁡(L,2).

step 1.1step 2.2step 3.1L1algebra
5.1

The operator list consists of ∑j=1nj=n(n+1)/2 reductions and n quantifier operators, so T=n(n+3)/2, which is O(n2); for n=0 the list is empty and all degree assertions are vacuous.

A1step 4.1algebra∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

The ordered arithmetization evaluates to the quantified Boolean truth value

Statement

Let F be a field, let Φ=Q1x1⋯Qnxnψ be a closed prenex quantified Boolean formula with quantifier-free matrix ψ (Quantified Boolean formulas and the language TQBF), let b=Pψ be its matrix arithmetization (Arithmetization of Boolean formulas), and let P(n)=b,P(n−1),…,P(0) be the multilinearized ordered arithmetization of Φ (Multilinearization in one variable). For 0≤j≤n put Ψj:=Qj+1xj+1⋯Qnxn ψ, so that Ψn=ψ and Ψ0=Φ, and Ψj has free variables x1,…,xj. Then for every j and every Boolean assignment a∈{0,1}j, P(j)(a)=the Boolean value of Ψj at a, the truth values being embedded in F as 0 and 1. In particular P(0), a constant, is the truth value of Φ.

Facts & Assumptions

Given: A field F and a closed prenex quantified Boolean formula Φ=Q1x1⋯Qnxnψ.

[A1]

The polynomials P(j) and Mj are the stages of the multilinearized ordered arithmetization: P(n)=b, Mj is obtained from P(j) by the reductions RX1,…,RXj, and P(j−1)=Oj(Mj) (Multilinearization in one variable).

[A2]

The subformulas Ψj are given by Ψn=ψ and Ψj−1=Qjxj Ψj; their Boolean values are defined by the usual recursive semantics (Quantified Boolean formulas and the language TQBF).

[L1]

RXiP agrees with P at Xi=0 and Xi=1, and leaves the other variables' degrees no larger; in particular, if P agrees with a function at all Boolean points of the cube, then so does RXiP (Multilinearization preserves Boolean values and bounds individual degree).

[L2]

The matrix arithmetization agrees with the Boolean value of ψ at every Boolean assignment (Arithmetization preserves Boolean values).

[L3]

For every polynomial P, the operators are AXP=(P∣X=0)(P∣X=1) and EXP=1−(1−P∣X=0)(1−P∣X=1) (Field arithmetization of QBF quantifiers).

Proof

technique · induction
1.1

Base case j=n: by [A1] and [A2], P(n)=b=Pψ and Ψn=ψ; by [L2] the two agree at every Boolean assignment to x1,…,xn, including the assignment-free case n=0.

L2A1A2givenbase
1.2

Induction hypothesis: suppose 1≤j≤n and P(j) agrees with Ψj at every point of the cube {0,1}j.

ih
2.1

Under the hypothesis of step 1.2, Mj is obtained from P(j) by the reductions RX1,…,RXj, each of which preserves agreement on Boolean points by [L1]; hence Mj also agrees with Ψj on {0,1}j. In particular, for every a∈{0,1}j−1 and each b∈{0,1}, the specialized value Mj∣Xj=b(a) is the Boolean value of Ψj(a,b), and the two values are the two bits whose universal and existential quantification define the truth values of Ψj−1(a)=QjxjΨj.

step 1.2L1A1A2
3.1

Fix a Boolean assignment a to the remaining variables and put u=Mj∣Xj=0(a) and v=Mj∣Xj=1(a). Both are bits by step 2.1. By [L3], the universal operator returns uv and the existential operator returns 1−(1−u)(1−v). On the four pairs (u,v)=(0,0),(0,1),(1,0),(1,1) these expressions give, respectively, (0,0,0,1) and (0,1,1,1) in every field. Thus they compute conjunction and disjunction, precisely the semantics of Qj in [A2]. Since P(j−1)=Oj(Mj) by [A1], it agrees with Ψj−1 at every Boolean a.

step 2.1L3A1A2algebra
4.1

Step 1.1 supplies the base case and step 3.1 the induction step, so agreement holds at every index j=n,n−1,…,0. For j=0 the cube {0,1}0 has the single empty assignment and Ψ0=Φ, so the constant P(0) is the truth value of Φ; the case n=0 is the base case itself.

step 1.1step 1.2step 3.1discharge-induction∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

A polynomial-size prime field meets the soundness budget

Statement

Let T≥1 and D≥1 be integers and put N:=max⁡{2, 12TD+1}. Then:

  1. (Existence and size.) There is a prime p with N<p<2N, and every such prime satisfies p>12TD and p<2N≤max⁡{4, 24TD+2}. Hence p=O(TD) and ⌈log⁡2p⌉=O(log⁡(TD)).
  2. (Deterministic search.) Call an integer m with N<m<2N admissible when no integer d with 2≤d≤N divides m. The first admissible m in the order N+1,N+2,…,2N−1 is prime; a search that tests every candidate against every such d performs at most N2 trial divisions and O(N2log⁡2N) bit operations, using no randomness and no primality test.
  3. (Residue arithmetic.) In F:=Z/p addition and multiplication make F a field in which each class has exactly one representative in {0,1,…,p−1}; representatives are stored in ⌈log⁡2p⌉ bits, adding or subtracting two representatives costs O(log⁡p) bit operations and multiplying them costs O(log⁡2p) bit operations. Consequently Horner evaluation of a polynomial of degree at most D at a point of F costs O(D) operations in F, hence O(Dlog⁡2p) bit operations.
  4. (Sampler with point mass at most 2/p.) Let k:=2⌈log⁡2p⌉, let U be uniform on {0,1}k and let r be the integer U∈{0,…,2k−1} reduced modulo p. Then Pr⁡[r=a]≤2/p for each a∈{0,…,p−1}, and Pr⁡[r∈S]≤2∣S∣/p for every set S of residues.

Facts & Assumptions

Given: Integers T≥1 and D≥1, N=max⁡{2,12TD+1}, and a uniform string U∈{0,1}k.

[A1]

For every integer n>1 there is a prime p with n<p<2n (Bertrand's postulate).

[A2]

An integer p is prime when p>1 and every positive divisor d of p satisfies d=1 or d=p; an integer n>1 that is not prime is composite, and then n has a positive divisor d with d≠1 and d≠n (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[A3]

For every prime p the operations of addition and multiplication on Z/p make it a field (For every prime p, the two operations on Z/p make it a field).

[A4]

For every positive integer n, each class in Z/n contains exactly one integer r with 0≤r<n, and ∣Z/n∣=n (For n≥1, every class in Z/n has one representative r with 0≤r<n, so ∣Z/n∣=n; while Z/0 is in bijection with Z).

Proof

technique · direct
1.1

Since T,D≥1 we have N≥max⁡{2,13}=13>1, so [A1] with n:=N yields a prime p with N<p<2N. Writing N=max⁡{2,12TD+1} gives p>N≥12TD+1>12TD in both cases of the maximum, and p<2N≤max⁡{4,24TD+2}; since T,D≥1, also 24TD+2≤26TD, so p=O(TD) and ⌈log⁡2p⌉≤log⁡2p+1=O(log⁡(TD)). This is claim (1).

A1givenalgebra
1.2

Let m be an integer with 1<m<2N that is not prime, so m is composite and has a positive divisor d0 with d0≠1 and d0≠m by [A2]. Write m=d0m1; the divisor d:=d0 can be replaced by the smaller of the pair {d0,m1}, so we may take 2≤d≤m. From m<2N≤N2, valid because N≥2, we get m<N, hence d≤N. Therefore every non-prime m with 1<m<2N is inadmissible: admissibility forces primality.

A2givenalgebra
1.3

By [A3] the operations of Z/p make it a field, and p>N≥2 is a positive integer, so by [A4] applied to n:=p every class in Z/p has exactly one representative in {0,1,…,p−1} and the field has exactly p elements. Representatives are integers below 2N, hence below 21+⌈log⁡2N⌉: each occupies at most 1+⌈log⁡2N⌉=O(log⁡N) bits, and schoolbook add/subtract and multiply on such integers cost O(log⁡N) and O(log⁡2N) bit operations.

A3A4algebra
1.4

Let a∈{0,…,p−1} and count the strings of {0,1}k whose integer value U satisfies U≡a(modp); these are exactly the integers in the arithmetic progression a,a+p,a+2p,… below 2k, so there are at most ⌈2k/p⌉≤2k/p+1 of them. Because k=2⌈log⁡2p⌉ gives 2k≥p2, division by 2k yields Pr⁡[r=a]≤1/p+1/p2≤2/p, the last step using p≥2.

A4givenalgebra
2.1

Run the search of claim (2). By step 1.1 the prime p of the interval (N,2N) is one of the candidates N+1,…,2N−1, and it is admissible: a divisor d of p with 2≤d≤N would satisfy d≠p because d≤N<p, and d≠1, contradicting primality. So the scan halts, and by step 1.2 its first admissible candidate is prime. There are at most N−1 candidates and at most N−1 divisors tested for each, so at most N2 trial divisions are performed, each on integers below 2N with O(log⁡N) bits, for O(N2log⁡2N) bit operations in total. This is claim (2).

step 1.1step 1.2algebra
2.2

By step 1.3 each class of Z/p is represented by a unique integer in {0,…,p−1} of O(log⁡N)=O(log⁡(TD)) bits, so one field operation costs O(log⁡2p) bit operations: O(log⁡p) for addition or subtraction of representatives and O(log⁡2p) for a schoolbook product followed by reduction modulo p. Horner's rule evaluates a polynomial of degree at most D with D multiplications and D additions of representatives, that is O(D) field operations or O(Dlog⁡2p) bit operations. Since p<2N≤max⁡{4,24TD+2} by step 1.1, this is O(Dlog⁡2(TD)) bit operations. This is claim (3).

step 1.1step 1.3algebra
2.3

For a set S of residues, Pr⁡[r∈S]=∑a∈SPr⁡[r=a]≤2∣S∣/p by step 1.4, the residues outside {0,…,p−1} being represented by their unique representative in that range. With ∣S∣=1 this reduces to step 1.4, and with S=∅ the bound reads 0≤0. This is claim (4).

step 1.4algebra
3.1

Claims (1)–(4) are steps 1.1, 2.1, 2.2 and 2.3, so for all integers T,D≥1 the interval (N,2N) contains a prime p above the budget 12TD, findable deterministically in O(N2log⁡2N) bit operations, whose residue field admits O(log⁡2p)-cost arithmetic and a 2/p-point-mass sampler. Every bound is polynomial in T and D, which is the content of the statement. ∎

step 1.1step 2.1step 2.2step 2.3
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

The Shamir interactive protocol for TQBF

Definition

Let Φ=Q1x1 Q2x2⋯Qnxn ψ be a closed prenex quantified Boolean formula (Quantified Boolean formulas and the language TQBF) whose quantifier-free matrix ψ has L syntax nodes, let b:=Pψ be the arithmetization of the matrix (Arithmetization of Boolean formulas), and put D:=max⁡{L,2},T:=n(n+3)/2. Let N:=max⁡{2,12TD+1}. If T=0, put p:=3; if T≥1, let p be the first admissible integer of the search of A polynomial-size prime field meets the soundness budget. In either case p is prime and 12TD<p<2N; put F:=Z/p, a field with the residue operations (For every prime p, the two operations on Z/p make it a field), and k:=2⌈log⁡2p⌉. Let N1,N2,…,NT be the operator list of the multilinearized ordered arithmetization of Φ in application order, with Nt acting in its active variable, and let G0:=b,Gt:=Nt(Gt−1)(t=1,…,T), so that GT is the arithmetized value of Φ (Multilinearization in one variable, Field arithmetization of QBF quantifiers). Recall that each node is either a reduction RXi or the quantifier operator Oj of a block j, and that the operator list ends with RX1,O1. The operators make the individual degree of the node polynomials bounded by D in the following sense: for every t and every substitution of arbitrary field elements for all variables of Gt−1 other than its active variable, the resulting univariate polynomial over F has degree at most D (Multilinearization preserves Boolean values and bounds individual degree).

The Shamir protocol for Φ. The protocol runs for T rounds. The verifier V holds two pieces of state: a claim c∈F and a current point σ, a partial function from {1,…,n} to F; initially c:=1andσ:=∅, the empty assignment. On an input that is not the encoding of a closed prenex quantified Boolean formula, V rejects immediately; this convention makes V total.

Round t. The rounds run for t=T,T−1,…,1, so that the nodes are processed in the reverse of the application order; let M:=Nt and let xi be the active variable of M.

  1. Prover message. The prover sends a list (a0,a1,…,aD)∈FD+1 of field elements, read as the coefficient vector of the univariate polynomial s(X):=a0+a1X+⋯+aDXD. A message that is not such a list, or a list whose entries are not elements of F, is rejected.
  2. Verifier check. If M is the quantifier node Oj of block j, then V tests c=s(0) s(1)when Qj=∀,c=s(0)+s(1)−s(0)s(1)when Qj=∃. If M is a reduction RXi, then V tests c=(1−a) s(0)+a s(1)where a:=σ(i), and if σ(i) is undefined, V rejects. If the applicable test fails, V rejects and the exchange stops.
  3. Challenge and update. If the test passes, V reads the next k bits of its random tape, forming the block Ut∈{0,1}k, sets rt:=(the integer denoted by Ut) mod p,σ(i):=rt,c:=s(rt), and proceeds to the next round. Assigning σ(i):=rt replaces any value assigned to xi earlier in the exchange.

Terminal step. After round t=1, V evaluates the matrix arithmetization b at the point σ, using the arithmetic circuit of the formula ψ and O(L) operations in F (Formula degree and point-evaluation cost), and V accepts  ⟺  c=b(σ). This terminal evaluation is the only place in the protocol where the formula ψ and the polynomial b are used; every earlier test involves only the claim, the current point and the prover's last message.

The honest prover. In the round with node M=Nt and active variable xi, the honest prover Ph sends the coefficient vector of qi(X):=Gt−1∣xj:=σ(j) for j≠i, xi:=X, the restriction of the current node polynomial Gt−1 to the current point, in the variable X ranging over F. By the degree bound recalled above, qi has degree at most D, so this is a legal message; Ph's choice depends only on the messages exchanged so far, which determine σ and c.

Remarks

  • A run has at most T prover messages and at most T verifier challenges rT,rT−1,…,r1; a failed check stops before its challenge is drawn, while a run that passes every check has all T challenges and then the terminal step. The residue challenges are sent to the prover, so its strategy uses the public residue transcript. This residue protocol does not reveal the full random blocks Ut; the public-coin variant of TQBF has a polynomial-round interactive proof sends each whole block.
  • The order of the rounds is the reverse of the operator list, which is what makes the checks local: each check relates the current claim to the received polynomial at the two Boolean points 0,1, and, for a reduction, to the current value a of the reduced variable. Processing Oj before the reductions of block j is what lets the reductions of a block see a value for xj; the reductions of block j then overwrite xj,xj−1,…,x1, so at the end every variable carries the last challenge drawn for it.
  • The verifier never evaluates the matrix before the terminal step, and the prover is never required to be efficient: only the number of rounds, the message lengths and the verifier's work are bounded, as required by Completeness and soundness of an interactive protocol.
  • Both displayed forms of the existential test agree, since s(0)+s(1)−s(0)s(1)=1−(1−s(0))(1−s(1)) in every field; the second is the form used in Field arithmetization of QBF quantifiers for the operator EXj, and the first is the form used by Shen.
  • For n=0 the operator list is empty, T=0, there are no messages, σ is the empty assignment and the protocol reduces to the test 1=b, where b is the constant value of the arithmetization of a quantifier-free closed formula. For n=1, T=2 and the two rounds process first the quantifier node O1 and then the reduction RX1; both cases are covered by the statements below.
  • The prime p is a function of the input alone: D and T depend only on the formula and the search of A polynomial-size prime field meets the soundness budget is deterministic, so the verifier makes no choice here. The only randomness in the protocol is at most Tk bits of the random tape, read in blocks of k bits only after passed checks; by the point-mass bound of that lemma every residue is sampled with probability at most 2/p, and the challenge is drawn after the prover's message has been fixed.
  • The degree bound D is the cap the verifier enforces on the received polynomial; the honest prover always meets it, and a prover that sends a longer or malformed list is rejected before any test is made. The factorization of the verifier's work into O(TD) field operations for the rounds and O(L) for the terminal evaluation, and the completeness and soundness of the protocol, are established in Honest prover maintains the field-value claim, Explicit communication, round, and evaluation bounds, Shamir protocol has perfect completeness, A false field claim becomes true in one round with bounded probability, Total TQBF soundness by the first repaired claim and Shamir verifier runs in polynomial time.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Honest prover maintains the field-value claim

Statement

Let Φ be a true closed prenex quantified Boolean formula on n variables, let F, D, T, the operator list N1,…,NT, the stage polynomials G0,…,GT and the Shamir protocol with its honest prover Ph be as in The Shamir interactive protocol for TQBF. Run the protocol on Φ against Ph with an arbitrary fixed random tape, and for t=T,T−1,…,1 let σt be the verifier's point at the start of round t, so that σT is the empty assignment, and let ct be the claim at the start of round t, so that cT=1. Then:

  1. In every round the message of Ph is a legal coefficient list, of degree at most D.
  2. For every t=T,T−1,…,1, the point σt assigns a field element to every variable that occurs in Gt, and ct=Gt(σt).
  3. Every verifier test in the run passes and the terminal test accepts.

Facts & Assumptions

Given: A true closed prenex quantified Boolean formula Φ, its Shamir protocol, its honest prover Ph, and an arbitrary fixed random tape of that protocol.

[A1]

The rounds run for t=T,T−1,…,1 and process the nodes NT,NT−1,…,N1; in the round with node M=Nt and active variable xi the prover sends a coefficient list of degree at most D, the verifier tests c=s(0)s(1) for a universal node, c=s(0)+s(1)−s(0)s(1) for an existential node, and c=(1−a)s(0)+a s(1) with a=σ(i) for a reduction RXi, then draws its next challenge rt, sets σ(i):=rt and c:=s(rt), and finally accepts exactly when c=b(σ) (The Shamir interactive protocol for TQBF).

[A2]

The honest message of the round with node M=Nt is the coefficient list of qt(X):=Gt−1 restricted to the current point in the active variable X, and qt has degree at most D (The Shamir interactive protocol for TQBF, Multilinearization preserves Boolean values and bounds individual degree).

[A3]

Every node Nt is either a quantifier node Oj of a block j, whose input Gt−1 is Mj, or a reduction RXi of a block j; Mj and every intermediate polynomial in block j involve only x1,…,xj, and the reverse protocol schedule assigns every variable in this set except the active one before the round, then assigns the active variable at the challenge update (Multilinearization in one variable, The Shamir interactive protocol for TQBF).

[A4]

The operators are AXiP=(P∣Xi=0)(P∣Xi=1), EXiP=1−(1−P∣Xi=0)(1−P∣Xi=1) and RXiP=(1−Xi)(P∣Xi=0)+Xi(P∣Xi=1) (Field arithmetization of QBF quantifiers, Multilinearization in one variable).

[L1]

The constant GT=P(0) is the truth value of Φ, so GT=1 because Φ is true, and the values of the stage polynomials at Boolean points follow the quantified semantics (The ordered arithmetization evaluates to the quantified Boolean truth value, Quantifier polynomials agree with QBF semantics on Boolean assignments).

Proof

technique · induction
1.1

Base case t=T: if n≥1 then NT=O1, so by [A3] the constant GT=O1(GT−1) has no variables at all and the empty point σT assigns all of its variables vacuously; by [L1] and the hypothesis that Φ is true, cT=1=GT(σT). If n=0 there are no rounds, T=0, σ is empty and the terminal test reads 1=b, which holds by [L1] because b=G0=GT=1.

A1A3L1givenbase
1.2

Induction hypothesis: for some t with 1≤t≤T, the point σt assigns a value to every variable occurring in Gt and ct=Gt(σt); let M=Nt be the round-t node with active variable xi.

ih
1.3

Write q(X) for the restriction of Gt−1 to the current point in the active variable X, so that q is the honest message of round t by [A2]. By [A3], Gt−1 involves only the variables in its block prefix, and the reverse protocol schedule has assigned every one except the active variable, whether or not some of them disappeared from Gt through polynomial cancellation. Thus the restriction is a well-defined univariate polynomial with field values at 0,1. By [A2] its degree is at most D, so the message is legal.

A2A3given
2.1

If M=Oj, then Gt−1=Mj, and [A4] gives Gt(σt)=(Gt−1∣xj=0)(σt)⋅(Gt−1∣xj=1)(σt) when Qj=∀ and Gt(σt)=1−(1−(Gt−1∣xj=0)(σt))(1−(Gt−1∣xj=1)(σt)) when Qj=∃, which are exactly q(0)q(1) and q(0)+q(1)−q(0)q(1). If M=RXi, then [A4] gives Gt(σt)=(1−a)(Gt−1∣xi=0)(σt)+a (Gt−1∣xi=1)(σt) with a=σt(i), that is Gt(σt)=(1−a)q(0)+a q(1). In both cases the value the verifier computes from the message equals Gt(σt)=ct by step 1.2, so the round-t test passes.

step 1.2A1A4algebra
3.1

Let r=rt be the challenge of round t and let σt−1 and ct−1 be the point and claim after the round. By [A1] the verifier sets the active variable to r, leaving every other value of σt unchanged, and ct−1=q(r); by the definition of q, this is Gt−1 evaluated at σt−1. The schedule in [A3] ensures that after this update every variable of Gt−1 is assigned, including when variables were absent from Gt because they cancelled in a preceding operator. Therefore the induction hypothesis holds at t−1, which completes the induction step and ensures the next honest message is legal by step 1.3.

step 1.3step 2.1A1A3ih
4.1

Descending induction from the base case of step 1.1 gives ct=Gt(σt) for every t=T,…,1. At the end of round 1 the point σ0 assigns every variable occurring in G0=b by the statement of claim 2, so the terminal comparison is c0=b(σ0) and it passes; all earlier tests passed by step 2.1. This proves claims 1, 2 and 3, including the case n=0 handled in step 1.1.

step 1.1step 2.1step 3.1A1discharge-induction∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Explicit communication, round, and evaluation bounds

Statement

Let Φ be a closed prenex quantified Boolean formula on n variables whose quantifier-free matrix ψ has L syntax nodes, let T=n(n+3)/2 and D=max⁡{L,2}, and let p be the prime chosen by the verifier of the Shamir protocol of The Shamir interactive protocol for TQBF, so that p<2N for N=max⁡{2,12TD+1}. Then:

  1. The protocol has at most T=O(n2) prover messages, one per reached round (exactly T if all round checks pass), and no other prover messages.
  2. Each prover message is a coefficient list with at most D+1 entries, each represented by an integer in {0,…,p−1}, hence by ⌈log⁡2p⌉=O(log⁡(TD+2)) bits. The prover sends at most T(D+1)⌈log⁡2p⌉ bits; including the residue challenges, total communication is at most T(D+2)⌈log⁡2p⌉, which is polynomial in the input length.
  3. The verifier draws at most 2T⌈log⁡2p⌉ random bits, in blocks of 2⌈log⁡2p⌉ bits, and evaluates each legal received polynomial at 0,1 and, if the check passes, at the challenge with O(D) field operations per round.
  4. If reached, the terminal evaluation of the matrix arithmetization costs O(L) field operations, and every field operation used costs O(log⁡2p) bit operations. The rounds and the terminal step together cost O ⁣((TD+L)log⁡2p) bit operations. Including the deterministic prime search, the verifier's total cost is O ⁣(N2log⁡2N+(TD+L)log⁡2p) bit operations, still polynomial in the input length; when T=0, setting p=3 costs only constant time.

Facts & Assumptions

Given: A closed prenex quantified Boolean formula Φ with n variables and matrix ψ of L syntax nodes, and the protocol of The Shamir interactive protocol for TQBF.

[A1]

The protocol runs for at most T rounds, stopping at a failed check, one prover message per round, each message a list of D+1 coefficients of a univariate of degree at most D; the challenge is read as one block of k=2⌈log⁡2p⌉ random bits and reduced modulo p; the terminal evaluation of b at the current point uses the arithmetic circuit of ψ (The Shamir interactive protocol for TQBF).

[A2]

The operator list has T=n(n+3)/2 entries and every node polynomial, after substituting arbitrary field elements for the variables other than its active variable, has degree at most D=max⁡{L,2} in that active variable (Multilinearization preserves Boolean values and bounds individual degree).

[A3]

The prime p satisfies 12TD<p<2N≤max⁡{4,24TD+2}, hence for T≥1 one has p=O(TD) and ⌈log⁡2p⌉=O(log⁡(TD+2)); for T=0 the protocol sets N=2 and p=3, so the same logarithmic bound holds; each residue occupies ⌈log⁡2p⌉ bits, addition and subtraction of residues cost O(log⁡p) bit operations and multiplication costs O(log⁡2p) bit operations, so Horner evaluation of a degree-D polynomial costs O(D) field operations and O(Dlog⁡2p) bit operations (A polynomial-size prime field meets the soundness budget).

[L1]

The arithmetization b=Pψ of the matrix, with L syntax nodes, is evaluated at any supplied point with O(L) field operations (Formula degree and point-evaluation cost).

Proof

technique · direct calculation
1.1

The protocol has one prover message in each reached round of its T-round schedule and no other prover message, because the initial claim 1 is the verifier's own constant and the terminal step uses the verifier's evaluation of b; the operator list has T=n(n+3)/2=n2/2+3n/2 entries by [A2], so at most T=O(n2) prover messages are sent, with equality if all checks pass, and claim 1 holds. This includes n=0, where T=0 and there is no message at all.

A1A2algebra
1.2

Each message is a list of D+1 coefficients by [A1], and a coefficient is a residue represented by an integer in {0,…,p−1}; by [A3] each such integer needs ⌈log⁡2p⌉=O(log⁡(TD+2)) bits, so a message costs (D+1)⌈log⁡2p⌉ bits and the prover communication is at most T(D+1)⌈log⁡2p⌉ bits. At most T residue challenges add at most T⌈log⁡2p⌉ bits, giving the stated two-way bound. Since n and L are bounded by the input length and D=max⁡{L,2}, this is polynomial in the input length, which is claim 2.

A1A3algebra
1.3

Each round whose check passes reads exactly one block of k=2⌈log⁡2p⌉ bits, so the run reads at most 2T⌈log⁡2p⌉ random bits in total. In a round with a legal message the verifier evaluates it at the two Boolean points and, if the check passes, at the challenge, which is at most three Horner evaluations of a degree-D polynomial: by [A3] this costs O(D) field operations, or O(Dlog⁡2p) bit operations, per round. This is claim 3.

A1A3algebra
2.1

If the terminal step is reached, the verifier evaluates b at one point of Fn using the arithmetic circuit of ψ; by [L1] this costs O(L) field operations, hence O(Llog⁡2p) bit operations by [A3]. The rounds contribute O(TDlog⁡2p) bit operations by step 1.3, including the constant-size identity checks, fixed-width message checks and reduction of each challenge block modulo p. Thus the rounds and terminal step together cost O((TD+L)log⁡2p) bit operations.

step 1.3A1A3L1algebra
3.1

For T≥1, the prime-search clause of A polynomial-size prime field meets the soundness budget supplies the additional O(N2log⁡2N) bit-operation cost: there are at most N2 trial divisions, each on O(log⁡N)-bit integers. For T=0, [A1] sets p=3 in constant time, also within that bound. Adding this setup cost to step 2.1 gives claim 4. Since N=max⁡{2,12TD+1} and n,L are bounded by the input length, the total bound is polynomial. All bounds hold for every random tape and every prover, with malformed messages rejected at the fixed message-length cap, so they are worst-case bounds.

step 2.1A1A3algebra∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Shamir protocol has perfect completeness

Statement

Let Φ be a true closed prenex quantified Boolean formula and let the Shamir protocol of The Shamir interactive protocol for TQBF run on Φ with its honest prover Ph. Then Ph is a legal prover strategy and the verifier accepts on every random tape, so the protocol has completeness 1 on Φ. Consequently every true TQBF input has an honest prover that is accepted with probability one.

Facts & Assumptions

Given: A true closed prenex quantified Boolean formula Φ and the Shamir protocol for Φ with its honest prover Ph.

[A1]

For every fixed random tape, every verifier test of the run passes and the terminal test accepts (Honest prover maintains the field-value claim).

[A2]

The honest prover's message in a round is determined by the node index and the public current point and claim, both of which are produced by the verifier from the messages exchanged so far; illegal or malformed prover messages are rejected, and an input that is not a well-formed closed prenex formula is rejected immediately (The Shamir interactive protocol for TQBF).

[A3]

c-completeness of a protocol means that on every input of the language an honest prover strategy makes the verifier accept with probability at least c, the probability being over the verifier's coins; the class IP requires some constant-gap protocol with completeness at least 2/3 (The class IP).

Proof

technique · direct
1.1

Fix a random tape. By [A1] every test of the run passes and the terminal test accepts, so this tape leads to acceptance; since the tape was arbitrary, the honest prover is accepted for every one of the 2p(∣x∣) random tapes of the verifier, hence with probability one on Φ.

A1given
1.2

The honest prover is a legal strategy: in each round its message is a coefficient list of degree at most D as stated in The Shamir interactive protocol for TQBF, it is determined by the public information of the current point and claim, and it never waits on private verifier history; the verifier's own rejection rules for malformed messages therefore never fire.

A2given
2.1

Combining the two steps, on the true input Φ the honest prover makes the verifier accept with probability 1, so Φ has completeness 1; in particular 1≥2/3, and since Φ was an arbitrary true TQBF input the same holds for every true input, which is exactly the statement.

step 1.1step 1.2A3given∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

A false field claim becomes true in one round with bounded probability

Statement

Let Φ be a closed prenex quantified Boolean formula, with n variables and the parameters D and p of its Shamir protocol, and let an arbitrary prover run against the verifier of The Shamir interactive protocol for TQBF. Fix a round t∈{1,…,T} and condition on a reached prefix of positive probability whose current point σt assigns a value to every variable of the stage polynomial Gt and whose current claim satisfies ct≠Gt(σt). Suppose the round-t message s is a legal list of degree at most D that passes the round's check. Then the probability, over the fresh challenge rt that the verifier draws after the message is fixed, that the updated claim equals the true predecessor value is at most 2D/p; here the true predecessor value at the challenge r is q∗(r), where q∗ is the restriction of Gt−1 to σt in the active variable of the round.

Facts & Assumptions

Given: A closed prenex quantified Boolean formula Φ, its Shamir protocol, an arbitrary prover, a round index t, a reached prefix with ct≠Gt(σt), and a legal message s passing the round's check.

[A1]

In the round with node M=Nt and active variable xi the verifier tests c=s(0)s(1) for a universal node, c=s(0)+s(1)−s(0)s(1) for an existential node, and c=(1−a)s(0)+a s(1) with a=σt(i) for a reduction; then it reads a fresh challenge block, sets σt−1(i)=rt, σt−1(k)=σt(k) for k≠i and ct−1=s(rt), and the challenge is drawn only after the message is fixed. The honest (true) round polynomial is the restriction of Gt−1 to the current point in the active variable (The Shamir interactive protocol for TQBF).

[A2]

Every node polynomial of the operator list, after substituting arbitrary field elements for all variables other than its active variable, has degree at most D=max⁡{L,2} in that active variable; in particular q∗ has degree at most D (Multilinearization preserves Boolean values and bounds individual degree).

[A3]

The verifier's challenge is the integer value of a block of 2⌈log⁡2p⌉ uniform bits reduced modulo p, and this sampler gives every residue of F probability at most 2/p (A polynomial-size prime field meets the soundness budget).

[A4]

The operators of the arithmetization are AXiP=(P∣Xi=0)(P∣Xi=1), EXiP=1−(1−P∣Xi=0)(1−P∣Xi=1) (Field arithmetization of QBF quantifiers) and RXiP=(1−Xi)(P∣Xi=0)+Xi(P∣Xi=1) (Multilinearization in one variable).

[A5]

At a reached round in block j, the reverse schedule has already sampled every variable in the input polynomial's block prefix except the active variable: before Oj these are x1,…,xj−1, and before a reduction RXi the preceding quantifier and reduction rounds have assigned the other variables among x1,…,xj (Multilinearization in one variable, The Shamir interactive protocol for TQBF).

[L1]

A nonzero polynomial of degree at most D over a field has at most D distinct roots in that field (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

Proof

technique · direct calculation
1.1

By [A5], the restriction q∗ of Gt−1 to the current point in the active variable is well defined. Let M=Nt be the round's node. By [A4] the true value of the current stage at σt is q∗(0)q∗(1) when M is a universal node and 1−(1−q∗(0))(1−q∗(1)), equivalently q∗(0)+q∗(1)−q∗(0)q∗(1), when M is an existential node. If M is the reduction RXi, put a=σt(i); this value is defined in that case because the protocol rejects a reduction round with undefined σt(i), and the stage value is (1−a)q∗(0)+a q∗(1). In each case the round's check is the same expression evaluated at s(0),s(1), using a only in the reduction case.

A1A4A5givenalgebra
1.2

The polynomial q∗ has degree at most D by [A2], and the received message s has degree at most D by hypothesis; both are univariate polynomials over F.

A2given
2.1

If s=q∗ as polynomials, then the value the check computed, namely ct, would equal the expression of step 1.1 with q∗ in place of s, which is Gt(σt); this contradicts the hypothesis ct≠Gt(σt). Hence s≠q∗.

step 1.1given
2.2

By step 1.2 the difference s−q∗ is a nonzero polynomial of degree at most D; by [L1] it has at most D distinct roots in F.

step 1.2L1
2.3

The updated claim is s(rt) and the true value of the updated stage is q∗(rt), because Gt−1 agrees with q∗ in the active variable and σt−1 differs from σt only there. So the updated claim equals the true value exactly when rt is a root of s−q∗.

A1step 1.1algebra
3.1

The challenge rt is drawn after the message is fixed and puts probability at most 2/p on each residue by [A3], so the probability that it lies in the root set of s−q∗, a set of at most D residues, is at most 2D/p. Averaging over any randomization of the prover's message preserves the bound, since the verifier's fresh bits are independent of the message; when 2D/p≥1 the bound is trivially true.

step 2.2step 2.3A3algebra∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Total TQBF soundness by the first repaired claim

Statement

Let Φ be a false closed prenex quantified Boolean formula, with n variables, matrix length L, and the parameters T=n(n+3)/2, D=max⁡{L,2} and prime p of its Shamir protocol. Then for every prover strategy the probability that the verifier of The Shamir interactive protocol for TQBF accepts is at most 2TD/p, which is less than 1/6 and hence at most 1/3. A randomized prover is allowed, its coins being independent of the verifier's future challenges.

Facts & Assumptions

Given: A false closed prenex quantified Boolean formula Φ, its Shamir protocol, and an arbitrary prover strategy.

[A1]

The rounds run for t=T,T−1,…,1; in each round the verifier assigns its freshly drawn challenge to the active variable of the round's node, so the sequence of points depends on the verifier's random tape alone and not on the prover's messages; after the last round the verifier accepts exactly when the current claim equals the matrix value b at the current point (The Shamir interactive protocol for TQBF).

[A2]

For each round t, conditional on a reached prefix with ct≠Gt(σt) and on a fixed message passing the round's check, the probability that the updated claim equals the true predecessor value is at most 2D/p (A false field claim becomes true in one round with bounded probability).

[A3]

If T≥1, the prime satisfies p>12TD by A polynomial-size prime field meets the soundness budget; if T=0, the protocol sets p=3>12TD=0 (The Shamir interactive protocol for TQBF).

[A4]

The final stage constant GT equals the truth value of Φ, embedded in F as 0 or 1, and the initial stage polynomial is G0=b; for the false Φ this gives GT=0≠1 (The ordered arithmetization evaluates to the quantified Boolean truth value).

Proof technique: direct calculation.

Proof

1.1

Because the point sequence follows the fixed round schedule from the verifier's random tape alone by [A1], the points are well defined for every prover; the stage polynomials Gt depend only on variables that the schedule has already assigned when round t begins, so the true values vt:=Gt(σt) are well-defined random variables on the verifier's tape, and vT=GT=0.

A1A4given
2.1

The run begins with the claim cT=1; by step 1.1 the initial claim is false, that is cT≠vT, and this holds pointwise for every random tape and every prover.

step 1.1A1given
2.2

For t=1,…,T let Bt be the event that the run reaches round t with ct≠vt, that the round's message passes its check, and that ct−1=vt−1. By [A2] the conditional probability of the transition, given any such reached prefix and any fixed message passing the check, is at most 2D/p; averaging over the reached prefixes and over the prover's randomization gives Pr⁡[Bt]≤2D/p.

step 1.1A2algebra
3.1

If the verifier accepts, every check passed and the terminal test c0=b(σ0) holds; since G0=b by [A4], this says c0=v0. Together with step 2.1 this forces the claim to pass from false to true at some round, so the acceptance event is contained in ⋃t=1TBt. Pointwise the indicator of this union is at most the sum of the indicators, and averaging gives Pr⁡[accept]≤∑tPr⁡[Bt]≤T⋅2D/p=2TD/p.

step 2.1step 2.2A1A4algebra
4.1

By [A3] we have 2TD/p<2/12=1/6, and 1/6<1/3, so the acceptance probability of every prover on the false input Φ is at most 1/3. If n=0 then T=0, there are no rounds, the terminal test compares 1 with G0=GT=0 and fails, and the acceptance probability 0 equals the bound 2TD/p; if a prover randomizes, its coins are fixed before the challenge that decides Bt and the averaging of step 2.2 already covers it.

step 3.1A3A4algebra∎
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Shamir verifier runs in polynomial time

Statement

Let an input Φ be a closed prenex quantified Boolean formula on n variables whose matrix has L syntax nodes, let ℓ be the input length, and let T=n(n+3)/2, D=max⁡{L,2}, N=max⁡{2,12TD+1} and p<2N be the parameters of the Shamir protocol of The Shamir interactive protocol for TQBF. Then the verifier of that protocol:

  1. uses p=3 when T=0 and otherwise performs the deterministic prime search of A polynomial-size prime field meets the soundness budget in O(N2log⁡2N) bit operations; it rejects a malformed input immediately;
  2. runs at most T rounds, in each of which it performs O(D) field operations on the received message and, only if that round passes, reads one block of 2⌈log⁡2p⌉ fresh random bits; if all round checks pass, it then evaluates the matrix arithmetization at one point in O(L) field operations;
  3. consequently runs in worst-case time O ⁣(N2log⁡2N+(TD+L)log⁡2p) bit operations, which is polynomial in ℓ, and uses at most 2T⌈log⁡2p⌉ random bits.

The verifier is therefore a probabilistic polynomial-time machine in the sense of Interactive-proof transcripts, rounds, and prover strategies: a fixed deterministic procedure whose running time on every input and every random tape is bounded by a polynomial in the input length.

Facts & Assumptions

Given: A closed prenex quantified Boolean formula Φ with n variables and a matrix of L syntax nodes, and the protocol of The Shamir interactive protocol for TQBF.

[A1]

On input that is not a well-formed closed prenex formula the verifier rejects immediately; otherwise it runs at most T rounds, checks the format and the identity of each received list, draws one challenge block of k=2⌈log⁡2p⌉ bits per passed round, and, if no round rejects, finishes with the evaluation of b at the current point; the prime p depends on the input alone, being 3 for T=0 and otherwise found by the deterministic search of A polynomial-size prime field meets the soundness budget (The Shamir interactive protocol for TQBF).

[A2]

The protocol has at most T messages, each with at most D+1 entries; the verifier reads at most 2T⌈log⁡2p⌉ random bits; per round it performs O(D) field operations, and the terminal evaluation costs O(L) field operations (Explicit communication, round, and evaluation bounds).

[A3]

For T≥1, the deterministic search for p performs at most N2 trial divisions and O(N2log⁡2N) bit operations, and p=O(TD) with ⌈log⁡2p⌉=O(log⁡(TD)) (A polynomial-size prime field meets the soundness budget). For T=0 the protocol sets N=2 and p=3 directly. In both cases ⌈log⁡2p⌉=O(log⁡(TD+2)), and every field operation on residues of Z/p costs O(log⁡2p) bit operations.

[L1]

A probabilistic polynomial-time verifier is a deterministic machine with a read-only random tape whose computation time is bounded by a polynomial in the input length on every input and every random tape (Interactive-proof transcripts, rounds, and prover strategies).

Proof technique: direct calculation.

Proof

1.1

The verifier first parses the input and, if the input is malformed, rejects; otherwise it uses p=3 when T=0, and when T≥1 it runs the deterministic search for p, which by [A3] uses at most N2 trial divisions and O(N2log⁡2N) bit operations. The direct T=0 choice also fits this bound.

A1A3given
1.2

Fix a round and a received message. Format checking inspects at most D+1 coefficients; evaluating the message at 0 and at 1 costs O(D) field operations, and the identity test then costs a constant number of field operations. If the check passes, drawing the challenge block costs one read of k=2⌈log⁡2p⌉ bits and one reduction modulo p, and the update evaluates s at the challenge, again O(D) field operations. So each round costs O(D) field operations and at most k random bits.

A1A2A3algebra
2.1

There are at most T rounds, so the rounds cost O(TD) field operations and at most Tk=2T⌈log⁡2p⌉ random bits in total; no step of the verifier loops unboundedly, since the round count and the message-length caps are fixed functions of the input.

step 1.2A1A2algebra
3.1

If the terminal step is reached, the verifier evaluates b at one point of Fn, which costs O(L) field operations by [A2]; together with step 2.1 the field-operation count of the whole run is O(TD+L), which by [A3] is O((TD+L)log⁡2p) bit operations.

step 2.1A2A3algebra
4.1

Adding the search cost of step 1.1 gives the bound O(N2log⁡2N+(TD+L)log⁡2p) claimed in item 3; here T=n(n+3)/2, n≤ℓ, L≤ℓ, D=max⁡{L,2} and p<2N. When T≥1, N=12TD+1 and hence 2N=24TD+2; when T=0, N=2 and the protocol directly sets p=3<4=2N. Thus N and p are polynomially bounded in ℓ, so the displayed bit bound is polynomial in ℓ. The verifier is a deterministic machine with an explicit polynomial bound on its steps for every input and every random tape, so by [L1] it is a probabilistic polynomial-time machine, as claimed.

step 1.1step 3.1A1A3L1algebra∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

TQBF has a polynomial-round interactive proof

Statement

TQBF∈IP: the language of true closed prenex quantified Boolean formulas has a public-coin interactive proof with perfect completeness and soundness error at most 1/3, in which the verifier is a probabilistic polynomial-time machine and the number of rounds and the communicated bits are bounded by polynomials in the input length. Concretely, on input Φ the verifier runs the Shamir protocol of The Shamir interactive protocol for TQBF for Φ with the following public-coin message convention: after a passed round check it sends the full fresh block Ut∈{0,1}2⌈log⁡2p⌉ instead of just its residue rt, and both parties compute rt=Ut mod p; the honest prover is accepted with probability 1 when Φ∈TQBF, and no prover is accepted with probability more than 1/3 when Φ∉TQBF, including malformed inputs.

Facts & Assumptions

Given: The language TQBF of true closed prenex quantified Boolean formulas under a fixed effective encoding.

[A1]

The class IP consists of the languages having an interactive protocol with a probabilistic polynomial-time verifier, polynomially bounded rounds and communication, completeness at least 2/3 and soundness at most 1/3 (The class IP).

[A2]

On every true input the honest prover of the Shamir protocol is accepted with probability one (Shamir protocol has perfect completeness).

[A3]

On every false input the acceptance probability of every prover is at most 2TD/p<1/6<1/3 (Total TQBF soundness by the first repaired claim).

[A4]

The verifier is a probabilistic polynomial-time machine; it rejects malformed inputs immediately, parses it otherwise, and its running time, randomness consumption and communication are polynomial in the input length (Shamir verifier runs in polynomial time).

[A5]

The protocol has at most T=O(n2) rounds with one message per round, each message of at most D+1 field elements of O(log⁡(TD+2)) bits, and its verifier state updates and tests use each sampled block only through its residue (Explicit communication, round, and evaluation bounds, The Shamir interactive protocol for TQBF).

Proof technique: direct.

Proof

1.1

By [A4] the verifier runs in polynomial time and, by [A5], the interaction has polynomially many rounds and messages of polynomial total length; in the variant specified in the statement each verifier message is its full fresh random block, and these blocks are all its coins. Thus this variant is public-coin. It sends at most 2T⌈log⁡2p⌉ verifier bits and at most T(D+1)⌈log⁡2p⌉ prover bits; computing the same residues leaves the verifier time polynomial.

A4A5given
1.2

If Φ∈TQBF then Φ is true and the honest prover computes each residue from the revealed block and otherwise uses the original strategy. On every random tape the messages and verifier state coincide with the original run, so [A2] gives acceptance with probability one, so completeness is 1≥2/3.

A2given
1.3

For any prover in the public-coin variant, construct a randomized prover in the residue-message protocol as follows. Upon receiving rt, independently sample a uniformly distributed block from the nonempty finite set {U∈{0,1}k:U mod p=rt}, and give this block to the simulated prover. Conditional on rt, this is exactly the distribution of the verifier's real block; the verifier uses no other information from that block. Induction over reached rounds therefore gives the same joint law of prover messages, residues and acceptance in the two protocols. The simulation uses no future verifier coins, so it is an allowed randomized strategy in [A3]. Thus on a well-formed false input [A3] gives acceptance probability at most 2TD/p<1/3 also in the public-coin variant. Malformed inputs reject immediately by [A4].

A3A4A5given
2.1

Steps 1.1, 1.2 and 1.3 verify every clause of the definition of IP in [A1] for the language TQBF: a probabilistic polynomial-time verifier, polynomially bounded interaction, completeness at least 2/3 and soundness at most 1/3; hence TQBF∈IP, which is the statement.

step 1.1step 1.2step 1.3A1∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

PSPACE is contained in IP

Statement

PSPACE⊆IP: every language in PSPACE has an interactive proof with perfect completeness and soundness error at most 1/3, whose verifier is a probabilistic polynomial-time machine with polynomially many rounds and polynomially bounded communication. Explicitly, for L∈PSPACE there is a polynomial-time computable reduction f with x∈L if and only if f(x)∈TQBF, and the protocol that runs the verifier of TQBF has a polynomial-round interactive proof on f(x) is such a proof for L.

Facts & Assumptions

Given: A language L in PSPACE.

[A1]

TQBF is PSPACE-complete: TQBF lies in PSPACE and every language in PSPACE polynomial-time many-one reduces to TQBF, a reduction being a polynomial-time computable map that preserves yes and no instances exactly (TQBF is PSPACE-complete).

[A2]

TQBF has an interactive proof with a probabilistic polynomial-time verifier, perfect completeness and soundness error at most 1/3; the verifier's running time, round count and communication are polynomial in its input length (TQBF has a polynomial-round interactive proof).

[A3]

A language lies in IP when it has an interactive protocol with a probabilistic polynomial-time verifier, polynomially bounded interaction, completeness at least 2/3 and soundness at most 1/3 (The class IP).

Proof technique: direct.

Proof

1.1

By [A1] there is a polynomial-time computable map f with x∈L if and only if f(x)∈TQBF; since f runs in polynomial time, there is a polynomial q with ∣f(x)∣≤q(∣x∣) for every input x.

A1given
2.1

Define the verifier VL for L as follows: on input x, compute Φx:=f(x) deterministically and then run the TQBF verifier V of [A2] on input Φx, forwarding the prover's messages to V and V's messages to the prover. Since f is computable in polynomial time and V runs in time polynomial in ∣Φx∣≤q(∣x∣), the machine VL is a probabilistic polynomial-time machine, and the protocol has polynomially many rounds and polynomially bounded communication because those bounds for V are polynomial in ∣Φx∣.

step 1.1A2construct
2.2

If x∈L then Φx∈TQBF by step 1.1, so by [A2] the honest prover for V on Φx, used as the prover for VL, is accepted with probability one, in particular at least 2/3.

step 1.1A2
2.3

If x∉L then Φx∉TQBF by step 1.1, so by [A2] every prover for V on input Φx is accepted with probability at most 1/3; a prover for VL on x induces such a prover for V on Φx, the reduction being deterministic, so no prover for VL exceeds 1/3.

step 1.1A2given
3.1

Steps 1.1 and 2.1 exhibit for L a probabilistic polynomial-time verifier with polynomially bounded interaction, and steps 2.2 and 2.3 give completeness 1≥2/3 and soundness at most 1/3; hence L∈IP by [A3]. Since L was an arbitrary language in PSPACE, PSPACE⊆IP, with perfect completeness for every language in the class.

step 2.1step 2.2step 2.3A3given∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

IP equals PSPACE

Statement

IP=PSPACE,

where IP is the class of languages with a constant-error interactive proof (The class IP) and PSPACE 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]

PSPACE⊆IP: 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]

IP⊆PSPACE: 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 PSPACE⊆IP is [A1]: for every L∈PSPACE there is an interactive proof whose verifier is probabilistic polynomial time and whose completeness and soundness satisfy the defining constants of IP, so L∈IP.

A1A3given
1.2

The forward containment IP⊆PSPACE 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∎
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

IP is closed under complement

Statement

If L∈IP then its complement L‾ also lies in IP. Consequently IP is closed under complement.

Facts & Assumptions

Given: A language L∈IP.

[A1]

IP=PSPACE: every language in IP lies in PSPACE and every language in PSPACE lies in IP (IP equals PSPACE).

[A2]

If L∈PSPACE then its complement lies in PSPACE; this is closure of deterministic polynomial space under complement (PSPACE equals NPSPACE and is closed under complement).

Proof technique: direct.

Proof

1.1

Since L∈IP, the containment IP⊆PSPACE of [A1] gives L∈PSPACE.

A1given
2.1

In particular there is a deterministic polynomial-space machine M deciding L: it halts on every input with the correct yes or no answer and uses at most p(n) cells for some polynomial p.

step 1.1given
3.1

By [A2] the complement of L lies in PSPACE; concretely, flipping the accept and reject states of the machine M of step 2.1 yields a deterministic polynomial-space machine deciding L‾, since M is total on all inputs.

step 2.1A2construct
4.1

Applying the containment PSPACE⊆IP of [A1] to the language L‾ gives L‾∈IP, which is the claim. This argument uses complementation of a deterministic space-bounded machine and does not claim that complementing an arbitrary interactive protocol preserves completeness or soundness.

step 3.1A1given∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

IP admits perfect completeness

Statement

For every language L∈IP there is an interactive proof for L with perfect completeness and soundness error at most 1/3: an honest prover is accepted with probability one on every input of L, every prover is accepted with probability at most 1/3 on every input outside L, and the verifier is a probabilistic polynomial-time machine with polynomially many rounds and polynomially bounded communication. The protocol need not be the given protocol for L.

Facts & Assumptions

Given: A language L∈IP.

[A1]

Every language in IP lies in PSPACE (IP is contained in PSPACE, IP equals PSPACE).

[A2]

Every language in PSPACE has an interactive proof with a probabilistic polynomial-time verifier, polynomially bounded interaction, completeness one and soundness error at most 1/3; the proof is obtained by reducing the input to TQBF and running the arithmetized sum-check-style protocol there (PSPACE is contained in IP).

Proof technique: direct.

Proof

1.1

By [A1] the language L lies in PSPACE.

A1given
2.1

By [A2] the language L, being in PSPACE, has an interactive proof with perfect completeness and soundness error at most 1/3, with a probabilistic polynomial-time verifier and polynomially bounded interaction.

step 1.1A2given
3.1

That protocol is an interactive proof for L with perfect completeness and constant soundness error, as required; it is constructed from the deterministic polynomial-time reduction of L to TQBF and therefore may differ from any particular protocol for L that was used to witness L∈IP.

step 2.1A2given∎
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

False: IP = PSPACE needs no degree reduction in this arithmetization

Statement

False assertion. If the multilinearization operators Rx are omitted from the quantified arithmetization and the resulting protocol uses only the quantifier operators Ax and Ex in inner-to-outer order, then the prover messages still have degree polynomial in the input length: for every closed prenex quantified Boolean formula of size m, the polynomial carried at every operator node of the naive sequence has degree at most a polynomial in m, so no degree reduction is needed for this proof route.

Facts & Assumptions

Given: The naive inner-to-outer quantified arithmetization without multilinearization operators, and the family of formulas defined below.

[A1]

The ordered arithmetization replaces NOT, AND and OR by 1−a, ab and a+b−ab, applies AXP=(P∣X=0)(P∣X=1) to a universally quantified variable and EXP=1−(1−P∣X=0)(1−P∣X=1) to an existentially quantified variable, and processes inner quantifiers first (Field arithmetization of QBF quantifiers).

[A2]

The multilinearized ordered arithmetization inserts, before each quantifier operation, the reductions RXiP=(1−Xi)(P∣Xi=0)+Xi(P∣Xi=1); the variant considered here omits exactly these reductions and keeps the same quantifier operations and the same inner-to-outer order (Multilinearization in one variable).

[A3]

A closed prenex quantified Boolean formula is Q1x1⋯Qkxkψ with quantifier-free matrix ψ, and its truth value is defined by the usual Boolean semantics of the quantifiers over {0,1} (Quantified Boolean formulas and the language TQBF).

[A4]

Arithmetization maps a variable leaf xi to Xi and a matrix built from these leaves by the gate rules of [A1] (Arithmetization of Boolean formulas).

Refutation

1.1

For every k≥1 let Φk:=∃y ∀x1⋯∀xk (y); its matrix is the variable y, and its prefix has k+1 quantifiers. Fix the standard effective encoding with binary variable indices and fixed punctuation, so the encoded size of this family is mk=O(klog⁡(k+1)). By [A4] the arithmetization of the matrix is b=Y.

A3A4construct
1.2

In the naive inner-to-outer sequence the universal quantifiers are processed first, starting with the innermost xk and ending with x1, and the outer existential quantifier in y is processed last, by the ordering rule of [A1]; the omitted operations are precisely the reductions RXi of [A2].

A1A2
2.1

We claim that after the first j universal quantifiers the carried polynomial is Y2j, by induction on j. For j=0 this is b=Y; for the step, Y2j does not involve the next variable Xi, so both of its specializations at Xi=0 and Xi=1 equal Y2j, and AXi gives the product Y2j⋅Y2j=Y2j+1.

A1step 1.1algebra
3.1

Taking j=k in step 2.1, the polynomial carried at the input of the outer quantifier node Ey is Y2k, whose degree as a polynomial in the active variable y is 2k; the message that the protocol requires at that node is the restriction of this polynomial in the active variable, so it also has degree 2k.

step 1.2step 2.1algebra
4.1

Since mk=O(klog⁡(k+1)) for the encoding fixed in step 1.1, every polynomial q(mk) is bounded above by a polynomial in klog⁡(k+1) and is eventually smaller than 2k. Hence the asserted polynomial degree bound fails for the family Φk under this allowed encoding, which refutes the false assertion. The count of syntax nodes alone would not give this encoded-length bound under an arbitrary effective encoding.

step 1.1step 3.1algebra
5.1

The failure is one of degree and not of Boolean semantics: for every k the formula Φk is true, because ∀x1⋯∀xk (y) holds exactly when y=1 and then ∃y is satisfied; so the refuted degree bound is not rescued by any appeal to the truth of the instances.

A3step 3.1algebra∎
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

False: the verifier can trust the final field value

Statement

False assertion. The Shamir protocol remains sound if its terminal step is modified so that the verifier accepts whenever the current claim equals a field value v announced by the prover, instead of evaluating the arithmetized matrix b at the current point: on every false closed prenex quantified Boolean formula the modified verifier would still accept with probability at most 1/3 against every prover.

Facts & Assumptions

Given: The modified protocol, in which the round checks of the Shamir protocol are kept unchanged and the terminal comparison c=b(σ) is replaced by the comparison c=v for a value v announced by the prover.

[A1]

The rounds run for t=T,…,1, the initial claim is c=1, a round with node M checks c=s(0)s(1) for a universal node, c=s(0)+s(1)−s(0)s(1) for an existential node, and c=(1−a)s(0)+a s(1) with a=σ(i) for a reduction, and then sets c=s(rt) at the fresh challenge; the unmodified terminal step accepts exactly when c=b(σ), and this evaluation is the only place where the matrix arithmetization is used (The Shamir interactive protocol for TQBF).

[A2]

Soundness s of a protocol means that for every input outside the language and every prover strategy the acceptance probability is at most s, so soundness at most 1/3 fails as soon as one false instance is accepted with probability exceeding 1/3 (Completeness and soundness of an interactive protocol).

[A3]

A closed prenex quantified Boolean formula ∃x φ(x) is true exactly when φ holds for x=0 or for x=1 (Quantified Boolean formulas and the language TQBF).

[A4]

Arithmetization sends the leaf x to X and forms X(1−X) for x∧¬x, and it agrees with the Boolean value of the matrix at Boolean inputs (Arithmetization of Boolean formulas).

[L1]

The unmodified protocol has soundness at most 1/3 on false inputs, achieved through the terminal evaluation against b (Total TQBF soundness by the first repaired claim).

Refutation

1.1

Let Φ:=∃x (x∧¬x). By [A3] the formula is false: φ(x)=x∧¬x holds neither at x=0 nor at x=1, so the existential quantification is false. Its matrix arithmetization is b=X(1−X).

A3A4construct
1.2

Consider the prover strategy that sends, in every round, the constant coefficient list s≡1, and announces v=1 at the modified terminal step; this is a legal message in every round, since the constant polynomial has degree 0≤D.

A1construct
2.1

We verify that every round test passes and that the claim remains 1. The claim starts at c=1. For a universal node the test is c=s(0)s(1)=1⋅1=1; for an existential node it is c=s(0)+s(1)−s(0)s(1)=1+1−1=1; for a reduction with current value a it is c=(1−a)s(0)+a s(1)=(1−a)+a=1. In every case the test passes, and the update c:=s(rt) leaves c=1 for every challenge rt.

A1step 1.2algebra
3.1

Consequently, after all rounds the claim is c=1; the modified verifier then compares it with the announced value v=1 and accepts. This holds for every random tape, because the strategy s≡1 and the announcement v=1 do not depend on the challenges.

step 2.1A1given
4.1

The modified verifier therefore accepts the false input Φ with probability 1, which is greater than 1/3; by [A2] the quality that fails here is exactly soundness, so the false assertion is refuted. This also shows why the terminal evaluation c=b(σ) cannot be delegated to the prover: it is what forces the final claim to agree with a value computed from the input.

step 3.1A2L1given∎

5 · Examples, counterexamples and false statements

None yet.

Sources