Inevitability of Trilemmas
Given a trilemma A;B;C, it’s easy to devise a troika that supports it. Let Moe vote for B and C but not A; let Larry vote for C and A but not B; let Curly vote for A and B but not C; then 2/3 majorities support each of A, B and C; but their conjunction fails unanimously. Therefore there is a troika for every trilemma.
But is the reverse true? Is there a voter’s paradox for every election of three voters? Yes, inevitably! For if Moe, Larry and Curly truly have three different agendas, then those agendas must differ by at least two bits; for one bit can distinguish only between two.
The three voters evaluate differently two propositions. Call them A and B; the voters can evaluate them four different ways:
Moe votes for: A, B
Larry votes for: A, not-B
Curly votes for: not-A, B
Moe votes for: not-A, B
Larry votes for: A, not-B
Curly votes for: not-A, not-B
Moe votes for: A, B
Larry votes for: A, not-B
Curly votes for: not-A, not-B
Moe votes for: A, B
Larry votes for: not-A, B
Curly votes for: not-A, not-B
Let a = not A, and b = not B;
Then the first troika supports these trilemmas:
A passes; B passes; not(A and B) passes: Weak And Glitch
(not a) passes; (not b) passes; (a or b) passes: Strong Or Glitch
A passes; B passes; (A does not equal B) passes: Equivalence Glitch
The second troika supports these trilemmas:
not A passes; not B passes; (A or B) passes: Strong Or
a passes; b passes; not(a and b) passes: Weak And
a passes; b passes; (a does not equal b) passes: Equivalence
The third troika supports the trilemma:
A passes; not B passes; (A equals B) passes: Equivalence
A passes; b passes; not(A and b) passes: Weak And
(not a) passes; (not B) passes; (a or B) passes: Strong Or
The fourth troika supports the trilemma:
a passes; not b passes; (a equals b) passes: Equivalence
a passes; B passes; not(a and B) passes: Weak And
(not A) passes; (not b) passes; (A or b) passes: Strong Or
So any triple of voters yields a weak-and glitch, a strong-or glitch, and an equivalence glitch.
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