Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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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.
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Rosser incompleteness from consistency

Statement

Every consistent effective theory extending Q is incomplete. The same holds for a consistent effective theory with an effective interpretation of Q, using proof predicates for translated arithmetic sentences.

Facts & Assumptions

[F1]

The syntactic diagonal lemma: For every formula ψ(v) with no other free variables in an effective signature extending arithmetic, there is a sentence θ such that Q in that signature proves θ↔ψ(⌜θ⌝). The construction is effective and requires neither consistency nor soundness.

[F2]

The standard certified provability predicate: For a fixed effective theory T, let Prf⁡T(p,a) be the chosen numeralwise arithmetic representation of certified proof checking, with proof code first. Use lem-primitive-recursive-syntax-and-proof-checking and the strengthened representation constructed in thm-primitive-recursive-numeralwise-representability. Retain also the finite PA proof of equivalence to its syntactic Σ1 computation form. Thus “Sigma1” for this chosen predicate may mean PA-Sigma1; it does not assert Q equivalence.

Put Prov⁡T(a):=∃pPrf⁡T(p,a) and Con⁡(T):=¬Prov⁡T(⌜⊥⌝), where ⊥=∃v0¬(v0=v0) is in the appropriate signature and corner brackets denote the numeral of a code. External consistency means that there is no actual finite T-refutation; the displayed Con is an arithmetic formula.

For theories extending Q, 0=1 may replace the fixed contradiction: Q proves 0≠S0, so from 0=S0 explosion gives ⊥; conversely reflexivity refutes ⊥ and explosion gives 0=S0. Appending these fixed finite proof blocks gives primitive-recursive transformations between refutation certificates, verified in PA. We use the fixed ⊥ throughout. Correctness only on standard numerals is insufficient to replace this predicate in a derivability or second-incompleteness theorem.

[F3]

Q calculates numerals and finite bounded cases: Q decides every closed arithmetic atomic formula and every closed bounded formula. For every external n, it proves x≤nˉ→⋁i=0nx=iˉ and x≤nˉ∨n+1‾≤x. These are metatheoretic schemes; induction on n here is not an induction axiom in Q.

[F4]

Interpretation transports derivations and inconsistency: An interpretation as defined above sends every S-derivation of ϕ to a T-derivation of GFV(ϕ)→ϕI. In particular a source contradiction gives a target contradiction, so external Con(T) implies Con(S). Effective certificate data gives an effective translation. A formal Con implication additionally follows in any base B that verifies a total map from S-contradiction certificates to T-contradiction certificates.

Proof

Given: Consistent effective T and either a Q extension or the specified effective interpretation of Q.

1.1F1F2F3given

In the extension case, apply F1 to obtain R↔∀p(Prf⁡T(p,⌜R⌝)→∃q≤pPrf⁡T(q,⌜¬R⌝)). The syntax and proof checks in F2 are numeralwise expressible in Q. If m proves R, consistency implies that no q proves not-R. In particular Q refutes the finitely many checks for q≤m; F3 combines them into a bounded universal. Q also proves the positive check at m. Therefore Q refutes the right side of the fixed point, giving not-R in T, a contradiction.

2.1F2F3step 1.1

If n proves not-R, consistency implies that Q refutes every proof-of-R check for p at most n. F3 gives their bounded universal and splits arbitrary p into p≤nˉ or n+1‾≤p. The first case has false antecedent. In the second, the numeral order calculation in F3 gives nˉ≤p; the positive check at n witnesses the consequent. Q thus proves the right side for every p and hence proves R, again a contradiction. Neither R nor its negation is provable.

3.1F1F2F3F4step 1.1step 2.1∎

For an interpretation I, define arithmetic predicates P(p,e) and N(p,e) to check proofs of the translated sentence with code e and its translated negation. The formula translation itself is primitive recursive from its finite syntactic data, so these are primitive-recursive checks; the supplied effective axiom-proof certificates are used only when translating Q theorems. Diagonalize the Rosser formula in Q using P,N. An actual proof of RI gives by the finite argument of step 1.1 a Q proof of not-R; F4 translates it to a T proof of ¬RI. An actual proof of ¬RI is a proof of (¬R)I, since translation commutes with negation; step 2.1 gives a Q proof of R, which F4 translates. Thus both alternatives contradict consistency in this case as well.

Depends on

Used by

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Sources