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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A two-quantifier field transcript

Example

Work over the prime field F=Z/251, and take the closed prenex quantified Boolean formula Φ=∃x ∀y (x∨¬y), whose matrix is ψ=x∨¬y and which is true, since x=1 makes the matrix 1 for both values of y. The matrix has L=4 syntax nodes, so with n=2 the protocol has D=4, T=5, and N=241; p=251 is the first prime greater than N. Its matrix arithmetization is b(X,Y)=1−Y+XY, the multilinearized operator sequence of Φ is b  →RX  RXb  →RY  RYRXb  →AY  AYRYRXb  →RX  RXAYRYRXb  →EX  1, and the reverse protocol of The Shamir interactive protocol for TQBF processes the nodes EX, then RX, then AY, RY, RX, with challenges drawn in the order 2,3,4,5,6. With those challenges the honest claim sequence is 1→2→3→9→11→26 and the terminal comparison is b(6,5)=26.

Facts & Assumptions

Given: The field Z/251, the formula Φ=∃x ∀y (x∨¬y) and the protocol of The Shamir interactive protocol for TQBF for it.

[A1]

Arithmetization replaces ¬u and u∨v by 1−u and u+v−uv, variable leaves by their variables, and it agrees with the Boolean value of the formula at Boolean inputs (Arithmetization of Boolean formulas).

[A2]

The ordered arithmetization applies AXP=(P∣X=0)(P∣X=1) and EXP=1−(1−P∣X=0)(1−P∣X=1) to the quantifiers, inner quantifiers first (Field arithmetization of QBF quantifiers).

[A3]

The multilinearized sequence inserts RXP=(1−X)(P∣X=0)+X(P∣X=1) before each quantifier operation, and its blocks for the prefix ∃x∀y are RX,RY,AY followed by RX,EX (Multilinearization in one variable).

[A4]

In the protocol the rounds process the operator list backwards; in the round for a node the verifier tests c=s(0)s(1) for A, c=s(0)+s(1)−s(0)s(1) for E, and c=(1−a)s(0)+a s(1) with the current value a of the reduced variable for R, then samples a fresh challenge r, sets the reduced variable to r and c:=s(r), and finally accepts exactly when c=b(σ) (The Shamir interactive protocol for TQBF).

[A5]

The classes of Z/251 form a field under addition and multiplication modulo 251 (For every prime p, the two operations on Z/p make it a field), and Φ is a closed prenex quantified Boolean formula with the truth semantics of Quantified Boolean formulas and the language TQBF.

Verification

1.1

By [A1] the matrix arithmetization is X+(1−Y)−X(1−Y), which simplifies to 1−Y+XY=:b(X,Y). At the four Boolean points it takes the values b(0,0)=1, b(0,1)=0, b(1,0)=1, b(1,1)=1, matching x∨¬y in each case.

A1algebra
2.1

The reductions do not change b, because b is multilinear: b∣X=0=1−Y and b∣X=1=1 give RXb=(1−X)(1−Y)+X=1−Y+XY, and b∣Y=0=1 and b∣Y=1=X give RYb=(1−Y)+YX=1−Y+XY.

A3step 1.1algebra
3.1

Applying the quantifier operations: AYb=(b∣Y=0)(b∣Y=1)=1⋅X=X, then RXX=X since X∣X=0=0 and X∣X=1=1, and finally EXX=1−(1−0)⋅(1−1)=1. The constant so obtained is the truth value of Φ by [A2] and [A5], and indeed Φ is true.

A2step 2.1A5algebra
3.2

Reading the stages backwards, the honest messages are the restrictions of G4=RXX=X, G3=X, G2=RYRXb=1−Y+XY at x=3, G1=1−Y+XY at x=3, and G0=b at y=5; that is, in the order of the rounds, s1(T)=T, s2(T)=T, s3(T)=1+2T, s4(T)=1+2T and s5(T)=5T−4.

A3A4step 2.1algebra
4.1

The reverse protocol of [A4] therefore processes EX first, then RX, then AY, RY and RX. Take the challenges in this order to be 2,3,4,5,6, so that σ(x) is set to 2 and then to 3 and finally to 6, and σ(y) is set to 4 and then to 5.

A4step 3.1
4.2

The verifier's tests pass with these messages: EX checks 1=0+1−0; then RX with a=2 checks 2=(1−2)⋅0+2⋅1; then AY checks 3=1⋅3, where s3(0)=1 and s3(1)=3; then RY with a=4 checks 9=(1−4)⋅1+4⋅3; and finally RX with a=3 checks 11=(1−3)⋅(−4)+3⋅1. So the claim sequence is 1→2→3→9→11→26, where each new claim is obtained by evaluating the sent message at the new challenge.

A4step 3.2algebra
5.1

The point left at the end is σ=(6,5), and the terminal comparison gives b(6,5)=1−5+30=26, equal to the final claim 26; the verifier therefore accepts this transcript. All products and sums above are computed in Z/251, where (1−4)⋅1=−3≡248 and 248+12=260≡9(mod251), so the intermediate values agree with the field arithmetic.

A4step 4.2A5algebra∎

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