Abstract
This corpus already carries several impossibilities (Papers 25, 163, 178, 179). All of them have the form “this object cannot be built”. This paper adds another form — a set of reasonable-looking requirements that cannot hold at the same time. NO NEW MATHEMATICAL THEOREM IS CLAIMED.
First, with two options there is no tension at all. Majority rule satisfied anonymity, neutrality and monotonicity across every profile for two through five voters (Section 2).
Second, with three options majority rule already breaks. Cyclic profiles are $1/18$ of all profiles at three voters. Yet at four voters there are none at all, and at six they return — and a counting argument says why (Section 3).
Third, and this is the core. Independence collapses the space of rules to a finite one, so the smallest case can be swept by machine. Of $4096$ rules, exactly two satisfy transitivity and unanimity together, and both are dictatorships (Section 4).
Fourth, there are four escapes and each surrenders something. Give up independence and the Borda count becomes available, but in a four-voter example the social verdict reversed while no voter changed how they ranked $a$ against $b$ (Section 5).
Fifth, the fence is the number of options. With two everything holds; with three nothing survives but a dictator (Section 6).
Scope (what is not claimed)
NO NEW MATHEMATICAL THEOREM IS CLAIMED — the results of Condorcet, Arrow, May and Black are all standard. No political or institutional claim of any kind is made — what is treated here is only the mathematics of whether a set of conditions is compatible. On Paper 25: that paper split “unsolvable” three ways with the separator being which tools are allowed. The separator here is which requirement is surrendered — not a matter of adding tools but of removing demands. What is added is verifying Arrow's theorem in the smallest case by exhaustive search rather than quoting it, showing by a counting argument that cycles cannot occur with four voters, searching out a Borda violation of independence and finding that a strict reversal exists at four voters and not at three, and setting the four escapes side by side by what each gives up.
1. Notation and position
The options are $a$, $b$ and $c$, and each voter ranks them. A collection of everyone's rankings is a profile. Of a rule returning a social ranking we ask: unanimity (if everyone puts $x$ above $y$ then so does society), independence (the social ranking of $x$ against $y$ depends only on how each voter ranks $x$ against $y$), transitivity (the social ranking has no cycle) and non-dictatorship (no single voter's ranking is always the social one). Every count below is exhaustive.
Position: Paper 179 split “cannot be constructed” three ways, and its impossibilities were about objects — not writable, not searchable, no algorithm. The impossibility here is about a set of demands — objects abound, and none of them meets every condition.
2. With two options there is no tension
Verification 2.1: Take two options, apply majority rule, and check the conditions on every profile.
| voters | profiles | anonymity | neutrality | monotonicity |
|---|---|---|---|---|
| $2$ | $4$ | holds | holds | holds |
| $3$ | $8$ | holds | holds | holds |
| $4$ | $16$ | holds | holds | holds |
| $5$ | $32$ | holds | holds | holds |
(QED)
Observation 2.2 All three hold together at every size. While there are only two options nothing has to be given up — majority rule simply works.
3. With three options majority rule already breaks
Verification 3.1: With three options, count the profiles whose pairwise majorities form a cycle.
| voters | profiles | cyclic | share |
|---|---|---|---|
| $3$ | $216$ | $12$ | $1/18$ |
| $4$ | $1296$ | $\mathbf{0}$ | $\mathbf{0}$ |
| $5$ | $7776$ | $540$ | $5/72$ |
| $6$ | $46656$ | $180$ | $5/1296$ |
| $7$ | $279936$ | $21000$ | $875/11664$ |
(QED)
Observation 3.2 Not one at four voters. This is not “rare because small”: it cannot happen. The reason is a count — whichever of the six rankings a voter holds, their contributions to the three margins ($a$ against $b$, $b$ against $c$, $c$ against $a$) sum to either $+1$ or $-1$. So over $n$ voters the three margins sum to at most $n$ in size. A cycle needs all three margins to share a sign, and when $n$ is even every margin is even and so at least $2$ in size, forcing the sum to be at least $6$. At $n=4$ that cannot be met.
Verification 3.3: The contributions of each ranking.
| ranking | three contributions | sum |
|---|---|---|
| $abc$ | $(+1,+1,-1)$ | $+1$ |
| $acb$ | $(+1,-1,-1)$ | $-1$ |
| $bac$ | $(-1,+1,-1)$ | $-1$ |
| $bca$ | $(-1,+1,+1)$ | $+1$ |
| $cab$ | $(+1,-1,+1)$ | $+1$ |
| $cba$ | $(-1,-1,+1)$ | $-1$ |
Every sum is $\pm1$. (QED)
Observation 3.4 At six voters they return ($180$ profiles). The sum can reach $6$ only if every voter contributes $+1$ and all three margins are exactly $2$ — it passes by the narrowest margin available.
4. The core: sweeping the smallest case by machine
Observation 4.1 Independence collapses the space. If the social ranking of $x$ against $y$ depends only on each voter's ranking of $x$ against $y$, then a rule is nothing but one function per pair. With two voters each such function has $2^{2^2}=16$ possibilities, so there are $16^3=4096$ rules in all.
Verification 4.2: Sweep those $4096$ and count those satisfying transitivity on every profile together with unanimity.
| count | |
|---|---|
| rules swept | $4096$ |
| transitive and unanimous | $\mathbf{2}$ |
| of those, dictatorships | $\mathbf{2}$ |
| non-dictatorial survivors | $\mathbf{0}$ |
(QED)
Observation 4.3 Every survivor is a dictatorship. This is not Arrow's theorem quoted but the smallest case confirmed by machine. “The four conditions are incompatible” is here the result of counting to $4096$.
Observation 4.4 An exhaustive sweep is not a proof of the theorem — what was checked is the case of two voters and three options. The general statement is Arrow's ($1951$).
5. There are four escapes, and each surrenders something
Verification 5.1: Give up independence — the Borda count. Search exhaustively over profiles in which every voter ranks $a$ against $b$ the same way, and see how far the social verdict can move.
| voters | verdict can change | strict reversal |
|---|---|---|
| $2$ | yes | yes |
| $3$ | yes | no |
| $4$ | yes | yes |
(QED)
Verification 5.2: The reversal at four voters.
| profile | scores | verdict |
|---|---|---|
| $abc\;|\;abc\;|\;bac\;|\;bca$ | $a=5,\ b=6,\ c=1$ | $b$ wins |
| $abc\;|\;acb\;|\;bac\;|\;bac$ | $a=6,\ b=5,\ c=1$ | $\mathbf{a}$ wins |
No voter changed how they rank $a$ against $b$ — only where $c$ sits moved. (QED)
Observation 5.3 Nobody changed anything about $a$ and $b$, and the social verdict on them swapped. That is what independence forbids and what the Borda count costs. And at three voters no strict reversal occurs — a win becomes a tie and no further. As with the cycles, the parity of the number of voters is doing work here.
Observation 5.4 Give up transitivity — keep majority rule and live with the cycles (Section 3).
Verification 5.5: Give up the universal domain — place the options on a line $a<b<c$ and admit only rankings with a single peak. The admissible rankings are $abc$, $bac$, $bca$ and $cba$.
| voters | single-peaked profiles | cyclic |
|---|---|---|
| $3$ | $64$ | $\mathbf{0}$ |
| $5$ | $1024$ | $\mathbf{0}$ |
(QED)
Observation 5.6 The cycles vanish. The price is restricting in advance what anyone is allowed to think — the two rankings $acb$ and $cab$ are refused at the door.
Verification 5.7: Give up unanimity — the constant rule that always returns $a>b>c$. It is independent (it never looks at the voters), transitive, and has no dictator. Profiles where everyone agrees and the rule contradicts them: $27$. (QED)
Observation 5.8 The fourth escape genuinely exists — and nobody takes it. The list of conditions is long precisely in order to shut escapes of this kind.
6. The census
Observation 6.1
| what is given up | what is gained | the price |
|---|---|---|
| independence | the Borda count | where an irrelevant third sits flips the verdict |
| transitivity | majority rule | the social ranking cycles |
| universal domain | single peaks, median voter | some opinions are refused in advance |
| unanimity | a constant rule | nobody's opinion counts |
Observation 6.2 The fence is the number of options.
| options | what happens |
|---|---|
| $2$ | every condition holds (Ver. 2.1) |
| $3$ | nothing survives but a dictator (Ver. 4.2) |
Observation 6.3 It joins the corpus's row of fences — linearity in Paper 171, integrality in Paper 173, compactness in Paper 177, and here there being only two options. Each stands at a place where adding one more breaks everything.
Observation 6.4 A word on the difference from Paper 25. Its separator was which tools are allowed, and allowing more tools made things solvable. The separator here is which condition is surrendered, and there is nothing to add — only something to remove.
7. Conclusion
- Two options bring no tension (Ver. 2.1): all three conditions at every size.
- Three options bring cycles (Ver. 3.1): $1/18$ at three voters.
- Cycles cannot occur at four voters (Obs. 3.2): because the contributions sum to $\pm1$.
- Two survivors out of $4096$, both dictators (Ver. 4.2).
- Borda breaks independence (Ver. 5.2): moving $c$ alone reverses $a$ against $b$.
- Single peaks kill the cycles (Ver. 5.5): zero.
- The fence is the number of options (Obs. 6.2).
References
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Reference
Yamagishi, Y. (2026). Incompatibility Begins with the Third Option [Paper 184]. YOI: https://www.yamagishiyuuki.com/yoi/184
DOI: 10.5281/zenodo.22105459
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 184. This paper is a preprint and has not been peer reviewed.