MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
arXiv:2606.25954v1 [econ.TH] 24 Jun 2026
LAWRENCE S. MOSS
†
AND ARTHUR PAUL PEDERSEN
‡∗
Abstract. This theoretical note studies the finite axiomatizability of strict majority reasoning in finite social decision frames. Moss and Pedersen (2026) introduce a coherence criterion that characterizes exactly when qualitative majority judgments are representable by a finitely additive measure. The question addressed here is whether that coherence criterion can be replaced, in the finite setting, by any bounded finite fragment. We prove that it cannot. For every k ≥ 1, we construct a maximal standard frame whose shortest coherence violation has length exactly 2k + 2. Hence there is no uniform finite bound on the incoherence index of social decision frames, resolving Conjecture 5.7 from Moss and Pedersen (2026). The construction is geometric, in the sense that it proceeds via orthogonality and dimension in rational vector spaces, and self-contained: it isolates a symmetric family of half-sized voting blocs and extends it to a maximal frame in which every shorter balanced obstruction is excluded. Along the explicit infinite sequence of universe sizes obtained in the construction, this also establishes the middle-layer family predicted by Conjecture B.25 from Moss and Pedersen (2026). Together with the soundness and completeness theorem for the Moss-Pedersen minimal logic for strict majorities, this establishes that measurable social decision frames are not finitely axiomatizable in that language.
1. Introduction In the study of strict majority reasoning within finite electorates, qualitative majority judgments cannot always be represented by a finitely additive probability measure. When such a representation fails to exist, the corresponding social decision frame is incoherent. The minimal complexity of this incoherence — the length of the shortest sequence of voting blocs required to expose a structural contradiction — is measured by its index. The study of structural bounds on qualitative probability traces back to the Kraft-Pratt-Seidenberg cancellation conditions (Kraft et al., 1959). Within that framework, Fishburn (1996) has investigated the function f (n), which measures the minimal length of cancellation conditions required to guarantee representability for an n-element state space, proving that n − 1 ≤ f (n) ≤ n + 1 for n ≥ 5. The incoherence index established in this paper operates as the majoritarian analogue to Fishburn’s f (n), extending the analysis of representation bounds from full comparative probability to strict majorities. This places the problem in the broader tradition of representational measurement theory. From that standpoint, the central question is not merely whether qualitative judgments can be assigned numbers, but which structural conditions make such numerical representation legitimate. Classical measurement theory studies this question by formulating axioms on qualitative structures and proving representation theorems that connect those structures to numerical scales (Krantz et al., 1971; Luce et al., 1990). Scott’s linear-inequality approach, the Kraft-Pratt-Seidenberg cancellation conditions, and later work on qualitative probability all show that representability can depend on (†) Dept. of Mathematics, Indiana University, Bloomington. (‡) Dept. of Computer Science & the Intel Investigations Lab, the City College of New York; the Graduate Center & Remote Sensing Earth Systems Institute, the City University of New York. E-mail addresses: (†) [email protected], (‡) [email protected]. 1
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L.S. MOSS AND A.P. PEDERSEN
finite configurations whose complexity is not visible from the surface grammar of the judgments (Scott, 1964; Kraft et al., 1959; Fishburn, 1996; Narens, 1980). The present note identifies the corresponding phenomenon for strict majorities: representability is determined by a coherence scheme whose instances are finite but whose full force is not finitely exhaustible. To establish that this index is unbounded (Conjectures 5.7 and B.25 in Moss and Pedersen (2026)), one must construct families of subsets where the shortest logical contradiction requires an arbitrarily large sequence of sets. A combinatorial resolution to this conjecture was recently provided by Blanco (2026). That proof operates by mapping the winning and losing coalitions of trade-robust simple games into a self-dual selector, leveraging a theorem by Taylor and Zwicker (1995) on strongly rigid magic squares. Through a padding argument, Blanco establishes the existence of a frame with an index of exactly 2k + 2 for all sufficiently large integers n ≥ k(k + 1), achieving a highly efficient quadratic bound on the necessary size of the electorate. Our main contribution in this note is the development of an alternative, purely geometric proof of the unboundedness of the incoherence index. Rather than relying on block designs or external theorems from cooperative game theory, we map the properties of subset selection directly into the geometry of rational vector spaces. This approach allows us to reframe the search for incoherent sequences as an evaluation of linear dependencies within the Boolean hypercube. While the combinatorial proof of Blanco (2026) achieves a quadratic scaling of the electorate size and captures every sufficiently large n, it relies essentially on Taylor and Zwicker’s results on simple games. In contrast, our geometric proof is derived entirely from first principles. We define a highly symmetric base of subsets over a universe sized by central binomial coefficients, explicitly compute its linear span, and use a generic separating hyperplane to construct a maximal frame. Although this geometric construction yields an electorate that scales exponentially with respect to the index, it provides a transparent, self-contained mechanism governing why short balanced sequences are excluded outside a controlled core. In brief, our results show that representable qualitative majorities admit no finite structural axiomatization. We interpret finite axiomatizability in the formal language introduced by Moss and Pedersen (2026). In that language, terms are built by Boolean operations from atomic predicates, while the atomic sentences are of the forms A t and M t, expressing respectively that t is true of the whole universe of discourse and that t is true of a majority. The proof system for that language contains an infinite coherence scheme, indexed by finite sequences of terms. The construction in the present note shows that the infinitude of this scheme is inescapable: no finite set of sentences in the Moss-Pedersen language for strict majorities axiomatizes exactly the measurable social decision frames. 2. Frames, Coherence, and the Incoherence Index We study strict majority reasoning using social decision frames. For the remainder of this note, let k ≥ 1 be a fixed integer. All mathematical constructs are parameterized by k to explicitly track their dependencies. Let Wk be a finite universe of voters. We assume that the cardinality |Wk | is an even number, and write |Wk | = 2nk . A social decision frame is a pair Mk = (Wk , Mk ), where the designated family Mk ⊆ P(Wk ) contains distinguished subsets interpreted as voting blocs that form a strict majority. Define: o n Hk := A ∈ 2Wk : A ∈ / Mk and Wk \ A ∈ / Mk . The family Hk corresponds to exact ties.
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
3
Definition 2.1 (Maximal Standard Frames). A frame Mk = (Wk , Mk ) is said to be standard if n o A ∈ 2Wk : |A| > nk ⊆ Mk and A ∈ Mk =⇒ |A| ≥ nk . A standard frame is said to be maximal if for every A ∈ 2Wk with |A| = nk , exactly one of A and Wk \ A belongs to Mk . ♠ Thus, by resolving every possible complementary pair, a maximal frame ensures that Hk = ∅. Definition 2.2 (Perfectly Balanced Sequences). A finite sequence of subsets A1 , . . . , Am ⊆ Wk is perfectly balanced if it uniformly covers the universe of voters exactly m/2 times. Algebraically, this is expressed using the standard binary indicator function 1A as m X i=1
1A i =
m 1W k , 2
where 1Wk is the vector of all ones over the universe.
♠
The central property under investigation is coherence, a structural condition which ensures that a frame behaves consistently with an underlying finitely additive measure. Definition 2.3 (Coherence and the Incoherence Index). A frame Mk = (Wk , Mk ) is coherent if: (c) For every positive integer m and sequence of sets A1 , . . . , Am ⊆ Wk : m
If
(c1) Ai ∈ Mk ∪ Hk
for each i = 1, . . . , m,
and
(c2)
X m 1W k ≥ 1A i , 2 i=1
m
then
(c3) Ai ∈ Hk
for each i = 1, . . . , m,
and
(c4)
X m 1Wk = 1A i . 2 i=1
The frame Mk is said to be incoherent if it fails to be coherent.
♠
Here and below, inequalities between vectors in QWk are understood pointwise. Thus, condition (c2) dictates that no voter is covered by more than half of the sets in the sequence. If this condition holds, coherence requires that the sequence must be perfectly balanced (c4) and consist entirely of exact ties (c3). In a maximal frame, there are no exact ties since Hk = ∅. Therefore, for any non-empty sequence satisfying (c1), condition (c3) cannot hold. Consequently, a maximal frame is coherent if and only if no non-empty sequence of sets drawn from Mk satisfies the coverage bound (c2). If such a sequence exists, the frame is incoherent. The minimal length m of such a sequence violating coherence is defined to be incoherence index of the frame. 3. Bipolar Indicator Vectors and Zero-Sum Conditions To analyze sequences of subsets algebraically, we map each subset A ⊆ Wk to a bipolar indicator vector xA ∈ {−1, 1}Wk . We define xA (v) = 1 if v ∈ A, and xA (v) = −1 if v ∈ / A. This translates the standard binary indicator function 1A via the affine transformation (1)
xA = 21A − 1Wk .
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L.S. MOSS AND A.P. PEDERSEN
Two vectors x, y ∈ QWk form an antipodal pair if y = −x. In this geometry, the affine transformation (1) ensures that complementary subsets A and Wk \ A correspond precisely to an antipodal pair of vectors xA and −xA . A sequence of sets A1 , . . . , Am ⊆ Wk is perfectly balanced if and only if the corresponding bipolar indicator vectors sum to 0 in QWk . Summing the affine transformation (1) over the sequence yields ! m m m X X X xA i = 21Ai − 1Wk = 2 1Ai − m1Wk . i=1
i=1
i=1
Setting the right-hand side to 0 is algebraically equivalent to the condition express this zero-sum condition as m X x Ai = 0 .
Pm
m i=1 1Ai = 2 1Wk . We
i=1
This zero-sum condition is the point of contact with cooperative game theory. In that literature, a sequence satisfying this exact balance condition for a length of 2j is formally known as a jtrade, or a j-balanced sequence over a self-dual selector (Blanco, 2026). Where combinatorial approaches leverage magic-square games to construct these trades, our framework evaluates their linear dependencies by mapping them directly into zero-sum bipolar vectors. Let Xk ⊂ {−1, 1}Wk be the set of all bipolar indicator vectors whose components sum to zero: ( ) X Wk Xk = x ∈ {−1, 1} : x(v) = 0 . v∈Wk
Thus Xk is the bipolar encoding of the middle layer n o A ∈ 2Wk : |A| = nk . Equivalently, vectors in Xk correspond precisely to subsets of Wk of size exactly nk = |Wk |/2. 4. The Core Construction We explicitly construct the components of a frame whose incoherence index requires a sequence of length 2k + 2. Let the base set of elements be Vk = {1, . . . , 2k + 2}. Define the universe of voters Wk to be the set of all (k + 1)-element subsets of Vk : n o Wk = v ⊆ Vk : |v| = k + 1 . The total number of voters is |Wk | = 2k+2 k+1 . This central binomial coefficient is even for k ≥ 1: the complement map v 7−→ Vk \ v is a fixed-point-free involution on the set of (k + 1)-element subsets of Vk . Hence |Wk | is even. We denote 1 2k + 2 nk = . 2 k+1 Definition 4.1 (Dictator Blocs). For each base element j ∈ Vk , we define a dictator bloc Mk,j ⊆ Wk consisting of all voters who possess the element j: n o Mk,j = v ∈ Wk : j ∈ v .
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
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Indeed, fixing j ∈ Vk , a voter v ∈ Wk belongs to Mk,j exactly when v = {j} ∪ s, where s ⊆ Vk \ {j} and |s| = k. Hence: 2k + 1 1 2k + 2 |Mk,j | = = = nk . k 2 k+1 Thus, each Mk,j contains exactly half of the voters.
♠
Every voter v ∈ Wk is a subset of Vk containing exactly k + 1 elements. Therefore, every voter belongs to exactly k + 1 of the sets in the sequence Mk,1 , . . . , Mk,2k+2 . The sequence has length m = 2k + 2, meaning the coverage for every voter is exactly m/2. Consequently, this specific sequence of subsets is perfectly balanced. Definition 4.2 (The Core Set). Let yk,j = xMk,j ∈ Xk be the bipolar indicator vector corresponding to Mk,j . Because the sequence of blocs is perfectly balanced, the corresponding vectors satisfy the zero-sum condition 2k+2 X yk,j = 0 . j=1
We define the core set Ck as the set of these 2k + 2 balanced vectors: n o Ck = yk,1 , . . . , yk,2k+2 . ♠
5. Algebraic Properties of the Core Sequence Lemma 5.1 (Minimality). Any non-empty perfectly balanced sequence formed by drawing sets exclusively from Mk,1 , . . . , Mk,2k+2 must have a length that is a multiple of 2k + 2. ■ Proof. Let cj ≥ 0 denote the integer number of times Mk,j appears in the sequence. By hypothesis, the sequence is perfectly balanced, meaning the corresponding bipolar vectors sum to zero in QWk : 2k+2 X
cj yk,j = 0 .
j=1
P2k+2
Let m = j=1 cj be the total length of the sequence. Evaluating the sum at an arbitrary voter v ∈ Wk yields 2k+2 X cj yk,j (v) = 0 . j=1
Substituting yk,j (v) = 21Mk,j (v) − 1 using (1), we obtain 2k+2 X
cj 21Mk,j (v) − 1
= 0.
j=1
Distributing the summation yields 2
2k+2 X j=1
! cj 1Mk,j (v)
−
2k+2 X j=1
cj = 0 .
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L.S. MOSS AND A.P. PEDERSEN
Since a voter v belongs to Mk,j if and only if j ∈ v, the indicator evaluates to 1Mk,j (v) = 1 strictly P when j ∈ v. Substituting the sequence length 2k+2 j=1 cj = m, we find ! X X m 2 cj − m = 0 =⇒ cj = . 2 j∈v
j∈v
This demonstrates that the sum of the coefficients corresponding to the elements inside any voter v must equal the constant m/2. Let a and b be two distinct elements in Vk . We construct two adjacent voters va and vb in Wk . Choose a subset S0 ⊂ Vk \ {a, b} of size exactly k. This is possible because |Vk \ {a, b}| = 2k ≥ k. Define va = S0 ∪ {a} and vb = S0 ∪ {b}. Both va and vb have cardinality k + 1 and are therefore valid voters in Wk . The balanced sum condition requires X X m m cj = cj = and . 2 2 j∈va
j∈vb
Subtracting the two equations eliminates the shared elements in S0 : ! ! X X ca + cj − cb + cj = 0 =⇒ j∈S0
ca = cb .
j∈S0
Since the indices a and b are arbitrary, all coefficients cj must equal a single uniform constant c ≥ 0. The total length of the sequence is therefore given by m =
2k+2 X
cj =
j=1
2k+2 X
c = c(2k + 2) .
j=1
Since the sequence is non-empty, we must have c ≥ 1. Thus every non-empty perfectly balanced sequence drawn from the core consists of exactly c copies of each dictator bloc Mk,j . In particular, its length is a multiple of 2k + 2, with the minimal non-zero length being exactly 2k + 2. □ Alternatively, Lemma 5.1 may be reformulated as follows. Lemma 5.2. Let S1 , . . . , Sr be a non-empty sequence of sets, and suppose that for each i there is a number n(i) such that Si = Mn(i) . Assume that S1 , . . . , Sk is perfectly balanced. Then r ≥ m. Proof. First, fix a voter v ∈ W . We begin by determining the size of the set Z, where Zv = {(i, j) : 1 ≤ i ≤ r, j ∈ v, and n(i) = j} in two ways. On the one hand, |Zv | =
X
|{i ≤ r : n(i) = j}|.
j∈v
On the other, Zv = {i : n(i) ∈ v} = {i : v ∈ Sn(i) }. As a result, we see that X m (2) |Z| = |{i ≤ r : n(i) = j}| = . 2 j∈v
This holds for all v ∈ W . As a result, |Zv | is independent ofP v. For each j ≤ m, let cj = |{i ≤ r : n(i) = j}|. Then from our work above, we see that for all v, j∈v cj = m 2.
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
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Now let a, b ∈ [m]; we show that ca = cb . Let v0 ⊆ [m] be any set such that v0 ∪ {a} and v0 ∪ {b} belong to W . (Such a set v0 exists since m ≥ 4.) Then X X m cj = ca + c(j) = 2 j∈v j∈v0 ∪{a} 0 P m Similarly, 2 = cb + j∈v0 cj . It follows that c(a) = c(b), as claimed. P So the function a 7→ ca is constant on [m]. The value ofPthis function cannot be 0, since □ a∈[m] ca = r > 0. So each number ca is at least 1. Hence r = a∈[m] ca > |[m]| = m. Lemma 5.3 (Linear Intersection Lemma). Let Lk = spanQ (Ck ) be the linear span of Ck over the rational numbers. The only bipolar indicator vectors representing sets of size nk that lie in Lk are precisely the core vectors and their antipodes. We write this as: n o Lk ∩ Xk = ±yk,1 , . . . , ±yk,2k+2 . ■
P2k+2
Proof. Let x = j=1 αj yk,j ∈ Lk ∩ Xk . For any voter v ∈ Wk , we apply the affine transformation to express the component x(v): ! 2k+2 2k+2 2k+2 X X X X αj − αj . x(v) = αj yk,j (v) = αj 21Mk,j (v) − 1 = 2 j=1
j=1
j∈v
j=1
P2k+2
We define βj = 2αj and let the total sum be S = j=1 αj . This yields X x(v) = βj − S . j∈v
Because x ∈ Xk , its components must satisfy x(v) ∈ {−1, 1}. We require X X βj − S ∈ −1, 1 =⇒ βj ∈ S − 1, S + 1 for all v ∈ Wk . j∈v
j∈v
P
We denote this sum by S(v) = j∈v βj . We utilize the adjacent voters va = S0 ∪ {a} and vb = S0 ∪ {b} constructed in Lemma 5.1. The difference S(va ) − S(vb ) evaluates to ! ! X X S(va ) − S(vb ) = βa + βj − βb + βj = βa − βb . j∈S0
j∈S0
Since both S(va ) and S(vb ) belong to the set {S − 1, S + 1}, their absolute difference is at most 2. Thus, we must have βa − βb ∈ {−2, 0, 2} . It follows that the values βj can take at most two distinct numerical values over the index set Vk : Indeed, if three distinct values occurred, then the largest and smallest would differ by at least 4, contradicting the fact that every pairwise difference belongs to {−2, 0, 2}. If all βj were identically equal, then S(v) would be constant for all v ∈ Wk , forcing x to be a constant vector. However, since x ∈ Xk , its components sum to zero. As Wk is non-empty ( 2k+2 6), a constant vector summing to zero must be the zero vector 0. This contradicts ≥ k+1 x(v) ∈ {−1, 1}. Therefore, the elements βj must take exactly two distinct numerical values, and these values must differ by exactly 2.
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L.S. MOSS AND A.P. PEDERSEN
Let these two values be λ and λ − 2. Let p be the integer number of components equal to λ. The remaining 2k + 2 − p components equal λ − 2. We sort the sequence of coefficients in descending order, writing β(1) ≥ · · · ≥ β(2k+2) . The first p elements of this sorted sequence equal λ, and the remaining elements equal λ − 2. Because Wk consists of all subsets of Vk of size k + 1, there exists a voter vmax ∈ Wk containing the elements corresponding to the k + 1 largest values of βj , and another voter vmin ∈ Wk containing the k + 1 smallest values. Thus, the maximum possible value of S(v) over Wk is the sum of the k + 1 largest values, and the minimum is the sum of the k + 1 smallest values. Since S(v) must fall into {S − 1, S + 1} and is not constant, the maximum sum must attain S + 1 and the minimum sum must attain S − 1. We strictly require the difference between the maximum sum and the minimum sum to be exactly 2: S(vmax ) − S(vmin ) =
k+1 X i=1
β(i) −
2k+2 X
β(i) =
i=k+2
k+1 X
β(i) − β(i+k+1)
= 2.
i=1
The difference term β(i) − β(i+k+1) must evaluate to either 0 or 2. Exactly one term in the summation equals 2, while all other terms equal 0. This requires exactly one index i∗ where β(i∗ ) = λ and β(i∗ +k+1) = λ − 2. Due to the sorted descending order, β(i∗ ) = λ implies i∗ ≤ p. Simultaneously, β(i∗ +k+1) = λ − 2 implies i∗ + k + 1 > p. We combine this condition to p − k ≤ i∗ ≤ p . The index i∗ is constrained by the bounds of the summation, meaning 1 ≤ i∗ ≤ k + 1. Therefore, the index i∗ must satisfy max(1, p − k) ≤ i∗ ≤ min(p, k + 1) . For i∗ to be uniquely determined (as required by the exact sum of 2), the number of valid integers in this interval must exactly equal 1. We analyze the length of this discrete interval: min(p, k + 1) − max(1, p − k) + 1 = 1 . We evaluate this equation over the full domain 1 ≤ p ≤ 2k + 1: Case 1 (1 ≤ p ≤ k): The length simplifies to p − 1 + 1 = p. Setting this to 1 yields p = 1. Case 2 (p = k + 1): The length simplifies to (k + 1) − 1 + 1 = k + 1. Since k ≥ 1, we have k + 1 ≥ 2 ̸= 1. No solution exists in this range. Case 3 (k + 2 ≤ p ≤ 2k + 1): The length simplifies to (k + 1) − (p − k) + 1 = 2k + 2 − p. Setting this to 1 yields p = 2k + 1. Thus, there are exactly two integer solutions: p = 1 and p = 2k + 1. Suppose first that p = 1, and let j0 be the unique index with βj0 = λ. Then S=
2k+2 X j=1
2k+2 1 X 1 αj = βj = λ + (2k + 1)(λ − 2) = (k + 1)λ − (2k + 1). 2 2 j=1
If j0 ∈ v, then X
βj = λ + k(λ − 2) = (k + 1)λ − 2k,
j∈v
and hence x(v) = 1. If j0 ∈ / v, then X βj = (k + 1)(λ − 2) = (k + 1)λ − 2k − 2, j∈v
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
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and hence x(v) = −1. Therefore x(v) = 21Mk,j0 (v) − 1 = yk,j0 (v) for every v ∈ Wk , so x = yk,j0 . Suppose next that p = 2k + 1, and let j0 be the unique index with βj0 = λ − 2. Then S=
2k+2 X j=1
2k+2 1 X 1 αj = βj = (2k + 1)λ + (λ − 2) = (k + 1)λ − 1. 2 2 j=1
If j0 ∈ v, then X
βj = (λ − 2) + kλ = (k + 1)λ − 2,
j∈v
and hence x(v) = −1. If j0 ∈ / v, then X
βj = (k + 1)λ,
j∈v
and hence x(v) = 1. Therefore x(v) = 1 − 21Mk,j0 (v) = −yk,j0 (v) for every v ∈ Wk , so x = −yk,j0 . The intersection Lk ∩ Xk contains no other vectors.
□
6. A Generic Vector Avoiding Finitely Many Hyperplanes We recall a classic result. Lemma 6.1 (Finite-Union Lemma, Bialynicki-Birula et al. (1959)). Let K be an infinite field, and let V be a finite-dimensional vector space over K. If U1 , . . . , Ur are proper linear subspaces of V , then r [ V ̸= Uℓ . ℓ=1 ■
Proof. We argue by induction on r. The case r = 1 is immediate. Suppose V =
r [
Uℓ
ℓ=1
with each Uℓ proper, and assume r is minimal. Then none of the Uℓ is contained in the union of the others. Choose r [ a ∈ U1 \ Uℓ and b ∈ V \ U1 . ℓ=2
For each t ∈ K, set vt = a + tb. Since b ∈ / U1 , the affine line {vt : t ∈ K} meets U1 only at t = 0. For each ℓ ≥ 2, the set of t ∈ K such that vt ∈ Uℓ has at most one element: if a + tb and a + sb both belong to Uℓ with t = ̸ s, then b ∈ Uℓ , and hence a ∈ Uℓ , contradicting the choice of a. Thus Sr the line contains infinitely many points but meets the finite union ℓ=1 Uℓ in only finitely many points, a contradiction. □ We use to establish the following lemma.
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L.S. MOSS AND A.P. PEDERSEN
Lemma 6.2 (Generic Vector Avoiding Finitely Many Hyperplanes). There exists a rational vector u∗k ∈ QWk possessing the following three properties: X (1) u∗k (v) = 0. v∈Wk
(2) u∗k · yk,j = 0 for all j ∈ {1, . . . , 2k + 2}. (3) u∗k · x ̸= 0 for all x ∈ Xk \ Lk . ■
Proof. Let Zk be the subspace of vectors in QWk whose components sum to zero: ( ) X Wk Zk = u∈Q : u(v) = 0 . v∈Wk
Since Wk has size 2nk =
2k+2 k+1
, the dimension of Zk is 2k + 2 dim(Zk ) = − 1. k+1
Each vector in Ck belongs to Xk , and hence to Zk . Moreover, the core relation 2k+2 X
yk,j = 0
j=1
gives one non-trivial linear dependence among the 2k + 2 generators. Therefore dim(Lk ) ≤ (2k + 2) − 1 = 2k + 1. Let Uk = Zk ∩ L⊥ k. Since Lk ⊆ Zk and the standard dot product restricts non-degenerately to Zk , this is the orthogonal complement of Lk inside Zk , and dim(Uk ) = dim(Zk ) − dim(Lk ). Consequently, 2k + 2 2k + 2 dim(Uk ) ≥ − 1 − (2k + 1) = − 2k − 2. k+1 k+1 For k = 1, this lower bound is 6 − 4 = 2. For k ≥ 2, it is at least 20 − 6 = 14, and hence is also at least 2. Thus dim(Uk ) ≥ 2 for all k ≥ 1. The set Xk \ Lk is finite. Each vector x in this set defines an orthogonal constraint u · x = 0. For each x ∈ Xk \ Lk , the functional u 7→ u · x is not identically zero on Uk . Indeed, if u · x = 0 for every u ∈ Uk , then x ∈ Uk⊥ ∩ Zk = Lk , since Zk = Lk ⊕ Uk orthogonally inside Zk , a contradiction. Therefore n o Ux = u ∈ Uk : u · x = 0 is a proper hyperplane of Uk . By Lemma 6, the rational vector space Uk is not the union of the finitely many proper hyperplanes Ux : x ∈ X k \ L k .
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
11
Choose [
u∗k ∈ Uk \
Ux .
x∈Xk \Lk
Since u∗k ∈ Uk ⊂ Zk , its components sum to zero (Property 1). Since u∗k ∈ L⊥ k , it is orthogonal to all core vectors yk,j (Property 2). Since u∗k avoids the hyperplanes defined by x ∈ Xk \ Lk , the dot product u∗k · x is non-zero for all such vectors (Property 3). □ 7. Resolution of the Conjecture Theorem 7.1. For any integer k ≥ 1, there exists a maximal standard frame Mk = (Wk , Mk ) whose incoherence index is exactly 2k + 2. Moreover, the length-2k + 2 witness is the core sequence Mk,1 , . . . , Mk,2k+2 . ■
Proof. We use the hyperplane-avoiding vector u∗k from Lemma 6.2 to define a tie-breaking family of bipolar indicator vectors Fk : n o Fk = x ∈ Xk \ Lk : u∗k · x > 0 ∪ Ck . Here Fk is a family of bipolar vectors. The corresponding family of middle-layer subsets is n o Fkset = A ∈ 2Wk : |A| = nk and xA ∈ Fk . For any x ∈ Xk \ Lk , exactly one of the vectors ±x yields a strictly positive dot product with u∗k . Exactly one is included in Fk . For vectors inside Lk ∩ Xk , Lemma 5.3 proved these are exactly the antipodal pairs ±yk,j . If a = ̸ b, choose v ∈ Wk with a, b ∈ v, which is possible since k + 1 ≥ 2. Then yk,a (v) = 1
and
yk,b (v) = 1,
so yk,a (v) ̸= −yk,b (v). Hence yk,a ̸= −yk,b whenever a = ̸ b. We manually include yk,j ∈ Ck and exclude −yk,j . Thus, the set Fk precisely contains exactly one vector from every antipodal pair in Xk . We define the frame Mk = (Wk , Mk ) by setting: n o n o Mk := A ∈ 2Wk : |A| > nk ∪ A ∈ 2Wk : |A| = nk and xA ∈ Fk . o n = A ∈ 2Wk : |A| > nk ∪ Fkset . Thus every member of Mk has size at least nk , and every subset of Wk of size strictly greater than nk belongs to Mk . On the middle layer, membership is determined by Fk . Because Fk contains exactly one vector from every antipodal pair in Xk , exactly one of A and Wk \ A belongs to Mk for every A ⊆ Wk with |A| = nk . Hence Mk is a maximal standard frame, and Hk = ∅. Since Hk = ∅, any non-empty sequence A1 , . . . , Am with Ai ∈ Mk for all i violates coherence exactly when it satisfies the coverage bound (c2). Indeed, (c1) then holds automatically, while (c3) is impossible. We write the coverage bound as m X i=1
1Ai (v) ≤
m 2
for all v ∈ Wk .
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L.S. MOSS AND A.P. PEDERSEN
We sum the sizes of these sets, evaluating over all voters: m X X
1Ai (v) ≤
v∈Wk i=1
X m v∈Wk
2
m m = 2nk = mnk . 2 2
= |Wk |
By reversing the order of summation, we double-count the total sizes of the sets: m X X
m X
1Ai (v) =
i=1 v∈Wk
|Ai | .
i=1
Pm This yields i=1 |Ai | ≤ mnk . Since every member of Mk has size at least nk , we also have Pm i=1 |Ai | ≥ mnk . Hence equality holds in both estimates. It follows that |Ai | = nk for every i, and that the pointwise inequalities in (c2) are all equalities: m X
m 2
1Ai (v) =
i=1
for all v ∈ Wk .
Any sequence violating coherence must therefore be a perfectly balanced sequence of size-nk sets. We translate this perfectly balanced sequence to bipolar indicators using equation (1), xAi = 21Ai − 1Wk : m X i=1
xA i =
m X
21Ai − 1Wk
i=1
= 2
m X
! − m1Wk
1Ai
i=1
= 2
m 1W k 2
! − m1Wk
= 0. Each xAi corresponds to a set of size nk in Mk , so xAi ∈ Fk . We take the dot product of this zero-sum condition with u∗k : m X u∗k · xAi = 0 . i=1 ∗ By construction of Fk , we have uk · x > 0 for every x ∈ Fk \ Ck , while u∗k · x = 0 for every x ∈ Ck . Thus u∗k · x ≥ 0 for all x ∈ Fk . Since a sum of non-negative rational numbers is zero only when
every term is zero, we must have u∗k · xAi = 0
for every i.
By Lemma 6.2, this implies xAi ∈ Lk for every i. 1 By Lemma 5.3, the only vectors in Fk ∩ Lk are the core vectors in Ck . Therefore, any perfectly balanced sequence in Fk is formed exclusively by copies of the original dictator blocs. Conversely, the core sequence Mk,1 , . . . , Mk,2k+2 1I feel that there’s a missing line here, when we go back from the xs to the ys. That is, this ending is too fast,
especially considering the meticulous detail in the rest of the proof.
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
13
is contained in Mk and is perfectly balanced. Since Hk = ∅, it witnesses incoherence at length 2k + 2. By Lemma 5.1, no shorter witnessing sequence exists. Hence the incoherence index of Mk is exactly 2k + 2. □ Corollary 7.2 (Resolution of Conjecture 5.7 and an Explicit B.25-Type Middle-Layer Construction). Conjecture 5.7 from Moss and Pedersen (2026) is true: there is no uniform finite bound on the incoherence index of social decision frames. Moreover, for each k ≥ 1, the construction gives a complement-free family in the middle layer with the balancedness properties required in Conjecture B.25 for the explicit universe size 2k + 2 2nk = . k+1 ■
Proof. We address both conjectures explicitly. Proof of Conjecture 5.7: We must show that there is no uniform finite bound on the incoherence index of social decision frames. Suppose, for a reductio ad absurdum, that there exists a uniform finite upper bound M ∈ N on the incoherence index across all frames. By definition, this asserts that every incoherent finite social decision frame must exhibit at least one sequence of subsets of length m ≤ M that structurally violates the coherence conditions. By the Archimedean property, we may choose an integer k ≥ 1 sufficiently large such that 2k + 2 > M . By Theorem 7.1, there exists a maximal standard frame Mk = (Wk , Mk ) whose incoherence index is exactly 2k + 2. Because its incoherence index is 2k + 2, the frame Mk is incoherent. However, the incoherence index represents the minimal length of any sequence in Mk that violates coherence. Thus, any sequence of subsets in Mk of length m < 2k + 2 must perfectly satisfy the coherence conditions. Since M < 2k + 2, it follows that there is no sequence of length m ≤ M that violates coherence in Mk . This directly contradicts the assumption that every incoherent frame possesses a coherence violation of length ≤ M . Therefore, no uniform finite bound M can exist. Explicit B.25-type middle-layer construction: Let 1 2k + 2 , n = nk = 2 k+1 so that |Wk | = 2nk . After identifying Wk with [2nk ], the middle layer is precisely the set of all subsets of Wk of size nk . Let n o Fkset = A ⊆ Wk : |A| = nk and xA ∈ Fk . We verify the required properties of this family of nk -subsets: (1) Fkset has no balanced subfamilies of sizes 2, 4, . . . , 2k: Any perfectly balanced sequence of subsets drawn from Fkset corresponds algebraically to a zero-sum sequence of vectors in Fk . The proof of Theorem 7.1, together with Lemma 5.1, shows that any such non-empty sequence has length at least 2k + 2. Therefore, no balanced sequence of length at most 2k exists in Fkset . (2) Fkset contains a balanced subfamily of size 2k + 2: The core vectors Ck ⊆ Fk correspond precisely to the 2k + 2 dictator blocs Mk,j . As established in Section 4, this sequence of 2k + 2 subsets covers every voter exactly k + 1 times, and hence is perfectly balanced. (3) Fkset is as large as possible among complement-free subfamilies of the middle layer, hence 2nk k of size 12 2n : The set X consists of all k nk nk bipolar indicator vectors corresponding to subsets of Wk of size nk . The vector family Fk contains exactly one vector from every
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L.S. MOSS AND A.P. PEDERSEN
antipodal pair {x, −x} in Xk . Therefore Fkset contains no complementary pair of subsets, and 1 2nk set |Fk | = . 2 nk This establishes the B.25-type middle-layer properties for the explicit infinite sequence of universe sizes 2k + 2 2nk = . k+1 The form of Conjecture B.25 that applies to “all sufficiently large n” requires an additional padding argument. □ The preceding corollary rules out any finite truncation of the coherence scheme. The next corollary strengthens this to finite axiomatizability in the full Moss-Pedersen term language: no finite set of sentences, whether or not drawn from the coherence scheme, defines exactly the measurable frames. Corollary 7.3 (Measurable majorities are not finitely axiomatizable). The class of measurable social decision frames is not finitely axiomatizable in the term language of Moss and Pedersen (2026). ■ Proof. Suppose, for a contradiction, that there is a finite set Γ of sentences in the Moss-Pedersen language whose finite-frame models are exactly the measurable social decision frames. Since every sentence in Γ is valid on measurable frames, the completeness theorem of Moss and Pedersen (2026) gives, for each γ ∈ Γ, a proof of γ from the proof system with its infinite coherence scheme. Each proof is finite, and Γ itself is finite. Hence only finitely many instances of the coherence scheme occur across all of these proofs. Let M be the largest sequence length appearing in any of those instances. We use the following bounded-soundness observation. Any sentence derivable using only coherence instances of length at most M is valid in every finite frame satisfying all coherence instances of length at most M , since the remaining axioms and inference rules of the Moss-Pedersen proof system are sound on arbitrary finite frames. Choose k ≥ 1 with 2k + 2 > M . By Theorem 7.1, there is a maximal standard frame Mk whose incoherence index is exactly 2k + 2. Thus Mk satisfies every coherence instance of length at most M , but fails coherence at length 2k + 2. In particular, Mk is not measurable. By bounded soundness, however, Mk validates every sentence in Γ. This contradicts the assumption that Γ axiomatizes exactly the measurable finite frames. □ 8. Conclusion and Future Directions This paper has established that there is no uniform finite bound on the incoherence index of social decision frames, thereby resolving Conjecture 5.7 from Moss and Pedersen (2026). This paper also gives a direct geometric construction of the middle-layer families predicted by Conjecture B.25 from Moss and Pedersen (2026). It does so along the explicit infinite sequence of universe sizes 2k + 2 2nk = . k+1 Together with the completeness theorem from Moss and Pedersen (2026), it follows from the unboundedness of this index that measurable social decision frames admit no finite structural axiomatization in the Moss-Pedersen language for strict majorities. The proof-theoretic reason is simple. By completeness, any finite axiomatization in the MossPedersen language would be derivable using only finitely many instances of the coherence scheme,
MEASURABLE MAJORITIES ARE NOT FINITELY AXIOMATIZABLE
15
and hence only coherence instances up to some finite sequence length M . Theorem 7.1 supplies an incoherent frame Mk whose shortest coherence violation has length 2k + 2 > M . That frame satisfies every coherence instance of length at most M , while nevertheless failing representability. Thus, much like the classical demonstration by Kraft et al. (1959) for comparative probability, measurability for strict majorities requires an infinite coherence scheme. The significance is not that strict majority reasoning resists numerical representation. On the contrary, Moss and Pedersen (2026) show that representability is exactly characterized by coherence. The present result shows instead that the boundary between representable and non-representable majority frames has unbounded finite complexity. Every obstruction is finite, but there is no finite ceiling on the length of the obstruction required. In this sense, the measurement-theoretic content of strict majority reasoning is not exhausted by any finite stock of Moss-Pedersen language conditions. This result has immediate consequences for the formal logic of majority reasoning. In Moss and Pedersen (2026), a natural term logic was introduced to reason about propositions of the form “most of everything is an X.” That logic achieves soundness and completeness by means of an infinite coherence axiom scheme. The geometric construction presented here shows that this infinitude is not eliminable: no finite set of sentences in the Moss-Pedersen language defines exactly the measurable frames. 1. Extremal Bounds on Electorate Size. There is a substantial efficiency gap between the known constructions of highly incoherent frames. The geometric proof presented here uses a symmetric dictator core, yielding an electorate that scales exponentially with the index √ 2k + 2 |Wk | = ∼ O(4k / k). k+1 In contrast, the Taylor–Zwicker magic-square construction used by Blanco (2026), together with a padding argument, achieves quadratic scaling. An open extremal problem is to determine the least electorate size 2n required to support a maximal standard frame with incoherence index m. 2. Generalization to Fractional Thresholds. The geometric framework mapping subsets to the rational vector space QWk naturally generalizes beyond strict majority rule, corresponding to the 1/2 threshold. One might consider super-majoritarian frames, such as 2/3- or 3/4threshold frames. The finite-hyperplane avoidance technique developed here provides a general algebraic tool for analyzing the linear dependencies of these fractional structures, raising the question of whether similar unboundedness theorems hold for arbitrary quota rules. Ultimately, the inherent complexity of strict majority reasoning is not merely an artifact of measure theory, but a deep combinatorial reality embedded in the geometry of finite vector spaces.
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