[EM] Inferring a method from an MMC axiom

[Craig Carey]

Axioms

[A1]. Right number of winners
[A2]. Truncation resistance
[A3]. P2
[A4]. MMC (undefined if >=4 candidates and /=1 winner)
[A5]. Monotonicity that gives way to the rules above
[A6]. A Proportionality that gives way to the rules above,
    fills in the remaining unsolved parts.

Note: there ought not be ambiguity/arbitrariness, eg. in the
use of axiom A6.

Election V has 10 papers showing exactly 0 to 2 preferences.
The preferential voting method is named W.
Using axiom [A3], the number of winners is constrained to equal 1.

V contains these papers and weights:

A_ a0
AB ab
AC ac
B_ b0
BA ba
BC bc
C_ c0
CA ca
CB cb

[A in W(V)] = (not [B in W(V)])(not [C in W(V)]), by [A1]

Now use [A2]. Here Tr(V,S) means truncate every paper just after
every element of S. E.g. Tr(37(BCAD),{C,D}) = 37(BC). If no
element of S shows then the paper is not altered.

[A in W(V)] = (not [B in W(Tr(V,{A,B}))])(not [C in W(Tr(V,{A,C}))])

The papers of Tr(V,{A,B}) are these

A_ a0+ab+ac
B_ b0+ba+bc
C_ c0
CA ca
CB cb

Now P2 can be used. It is above the "gives way to" axioms.
The above has the same winners as this:

A_ a0+ab+ac
B_ b0+ba+bc
C_ c0+2.ca
CA 0
CB cb-ca

Define a=a0+ab+ac, b=b0+ba+bc, c=c0+ca+cb. So c0=c-ca-cb.
That has the same winners as this:

A_ a - (c+ca-cb)
B_ b - (c+ca-cb)
CB cb-ca

Now MMC and whatnot is used. When B wins, is found.

This is unfinished. Since Blake Cretney was about the only
person advocating MMC (weakly) and to demo the easiness of
QE logic algebra, maybe he would finish this.

-----------

To Mr Schulze: did you find out why Mr Dummett had such a
weak rule ?. That rule had a Floor(x) function on it, reducing
the number of candidates that it said had to win. It seemed
something that did not look desirable, and it seemed to be
too distant from imposing a constraint.



G. A. Craig Carey