Monday, June 24, 2013

On Troikas: Inevitability of Trilemmas



          Once again I am overtaken by writing, and must blog more chapters of “On Troikas and Trilemmas”. Over the next four days I shall blog:
          Inevitability of Trilemmas
          Finite Boundless Lines
          Anti-Sorites
          Disinduction and the Heap

               *******

        Inevitability of Trilemmas


            Given a trilemma A;B;C, it’s easy to devise a troika that will yield 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? Does every election of three voters produce a voter’s paradox? 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 at least two propositions differently.  Call those propositions 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 the trilemma:
                        A passes; B passes; not(A and B) passes:  the “Weak And glitch”.
                        (not a) passes; (not b) passes; (a or b) passes:  the “Strong Or glitch”.
                        A passes; B passes; (A does not equal B) passes:  “Disequivalence 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:  Disequivalence
                                               
            The third troika supports the trilemma:
                        A passes; not B passes; (A equals B) passes:   Disequivalence
                        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:   Disequivalence
                        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, a strong-or, and a disequivalence glitch. 

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