*Chapter 11 :***Voter’s Paradox**

Voter’s Paradox

Trilemma deduction

Syllogisms by Trilemma

*11A. Voter’s Paradox*
Consider
these three orders on the three forms:

6 < 1
< 0

0 < 6
< 1

1 < 0
< 6

The
minima on these orders cyclically distribute; so do the maxima. Taken together
they form a “troika”, a.k.a. a “voter’s paradox”, which is at the heart of
Kenneth Arrow’s Impossibility Theorem. These logic knots have a habit of
bollixing political systems. These tiny tangles give politics its notorious
perversity.

To
simplify presentation of the Troika, I introduce three fictional characters;
none other than the Three Stooges.

General
Moe rules the Scissors Party with an iron hand. His politics are fascistic; he
favors Lies over Truth over Imagination. He prefers monarchy, most preferably
if the monarch is himself.

Judge
Larry is senior theoretician for the Paper Party. His politics are legalistic:
he favors Truth over Imagination over Lies. He prefers to govern by consensus.

Mayor
Curly is lead singer for the Rock Party. His politics are populistic: he favors
Imagination over Lies over Truth. He prefers to govern by majority rule.

Moe: Imagination
< Truth < Lies

Larry: Lies
< Imagination < Truth

Curly:
Truth < Lies
< Imagination

Each
single Stooge has a consistent linear ranking of imagination, lies and truth;
but when you put them all together, something has got to go. Two-thirds of the
Stooges (namely, Moe and Larry) put truth above imagination; Larry and Curly
put imagination above lies; and Curly and Moe put lies above truth.

Moe Larry Curly

Lies
< Imagination? no yes yes

Truth
< Lies? yes no yes

Imagination
< Truth? yes yes no

This
gives us a Condorcet Election, or “Voter’s Paradox”:

Truth

< <

Imagination > Lies

- by
2/3 majority each; yet they

*all*agree that the ranking is linear!
There
are several partial resolutions to this.

If we
appoint a single voter as tyrant (Moe, say) then we can decide this consistently;
but this is not a fair system.

If we
attempt to decide by consensus (as Larry suggests) then that is fair and
consistent; but we decide nothing, and that is a weak system.

If we
have faith in majority rule (as Curly professes) then we accept the non-linear
order,

*and*the linearity of the order. This is fair and decisive, but it is inconsistent.
Finally,
we can accept the non-linear ranking, and accept it as non-linear; this goes
with every 2/3 majority, but reverses a consensus; and that is perverse.

This
political knot is an instance of

**Arrow’s Impossibility Theorem**, which says that no voting system has all four of these virtues:
it is
fair: it gives all voters equal power

it is
decisive: it decides all questions posed to it

it is
logical: it does not believe contradictions

it is
responsive: it never defies a voter consensus.

In other
words, any government is at least one of:

cruel ;
weak ; absurd
; perverse.

Moe
prefers cruelty and lies, Larry prefers weakness and truth, and Curly prefers absurdity
and imagination;

*none*of them want perversity and paradox, but of course that is what they always get!

*11B. Trilemma deduction*

A
trilemma is a triple of propositions, any two of which can be true but not all
three. Any trilemma implies this deductive principle; from any two, derive the
negation of the third. For instance, this trilemma, the “disimplication glitch”:

A;

A
implies B;

Not
B.

implies these deductive rules:

From:
A; A implies B; deduce: B

From:
A implies B; Not B; deduce: not A

From:
not B; A; deduce: A does not imply B

So
modus ponens, modus tollens, and “anti-implication” are encoded by one
trilemma. A trilemma is an encapsuled near-absurdity; and near-absurdity is the
engine of deduction. Every deduction is a narrow escape, one trilemma wide.

And
for each trilemma, there’s a troika that generates it as a voter’s paradox. In
it, each of the three voters affirm two-thirds of the trilemma, and reject the
third.

In
this case:

Moe
says: A; A implies B; B

Larry
says: A implies B; Not B; not A

Curly
says: not B; A; A does not imply B

Take
the “strong or glitch”:

Not
A;

Not
B;

A or B.

It
implies these deductive rules:

From:
not A; not B; deduce: not A nor B

From:
not B; A or B; deduce: A

From:
A or B; not A; deduce: B

Here’s
a troika that generates it as a voter’s paradox:

Moe
says: not A; not B; not A nor B

Larry
says: not B; A or B; A

Curly
says: A or B; not A; B

Here’s
the “weak and glitch”:

A;

B;

not(A and B)

It
implies these deductive rules:

From:
A; B; deduce: A and B

From:
B; not(A and B); deduce: not(A)

This
is the “Failed Reductio” glitch:

A
implies B;

A
implies not B:

A.

It
implies these deductive rules:

From:
A implies B; A implies not B; deduce: not A

From:
A implies not B; A; deduce: A does not imply B

From:
A; A implies B; deduce: A does not imply not B

This
is the “disequality glitch”:

x
= y;

y
= z;

z ≠ x.

It
implies these deductive rules:

From:
x=y; y=z; deduce: z=x

From:
y=z; z≠x; deduce: x≠y

Here
is “Barbarism”:

All
A are B;

All
B are C;

Some
A are not C.

It
implies these deductive rules:

From:
All A are B; All B are C; deduce: All A are C

From:
All B are C; Some A are not C; deduce: Some A are not B

From:
Some A are not C; All A are B; deduce: Some B are not C

This is the “Disinduction
Trilemma”:

1 has property P;

For any n, P(n) implies
P(n+1);

For some N, not P(N).

It implies these deductive rules:

From: 1 has property P; For
any n, P(n) implies P(n+1);

Deduce: For all N, P(N).

From: For any n, P(n) implies
P(n+1); For some N, not P(N);

Deduce: 1
does not have property P.

From: For some N, not P(N); 1
has property P;

Deduce: For
some n, P(n) does not imply P(n+1).

This is the “Disintermediation
Trilemma”:

f(x) is real and continuous
on the interval [a,b];

f(x) does not equal zero
anywhere on [a,b];

f(a) and f(b) have different
signs.

It implies these deductive rules:

From:

f(x) is real and continuous
on the interval [a,b];

f(x) does not equal zero
anywhere on [a,b];

Deduce:

f(a) and f(b) have the same
sign.

From:

f(a) and f(b) have different
signs;

f(x) is real and continuous
on the interval [a,b];

Deduce:

f(x) does equal zero
somewhere on [a,b].

From:

f(x) does not equal zero
anywhere on [a,b];

f(a) and f(b) have different
signs.

Deduce:

f(x) is not both real and
continuous on the interval [a,b].

Here is the “First Cause
Trilemma”:

There is a first cause;

Every cause has a cause;

There are no causal
loops.

It
implies these deduction rules:

From: there is a first cause;
every cause has a cause; deduce:

there are causal loops.

From: every cause has a
cause; there are no causal loops; deduce:

there is no first cause.

From: there are no causal
loops; there is a first cause; deduce:

not every cause has a
cause.

The
trilemma can be written as:

There exists A, such that
for every B, A causes B.

For every A, there exists
B, such that B causes A.

There do not exist A and
B such that A causes B and B causes A.

This generalizes to any predicate R:

There exists x, such that
for every y, xRy;

For every x, there exists
y, such that yRx;

There do not exist x and
y such that xRy and yRx.

Call this the
“Loop Trilemma”. It’s a variant of the weak-and glitch.

When the predicate
is “explains”, you get the “Munchhausen Trilemma”:

Explanation is simple; there is a full explanation.

Explanation is complete; every explanation is
explained.

Explanation is not circular; no two explanations
explain each other.

Here’s
a “Set Loop Trilemma”

There exists an x, such
that for every y, yϵx;

For every x, there exists
a y, such that xϵy;

There does not exist an x
and a y such that xϵy and yϵx.

There’s
a universal set; every set’s an element; there are no set loops.

*11C.*

*Syllogisms by Trilemma*

Consider
these

**modal identities**:

*Swap:*
All
A are B = All not-B are not-A

No
A are B = No B are A

Some
A are B = Some B are A

Some
A are not B = Some not-B are not not-A

*Negation:*
Not
(all A are B) = Some A are not-B

Not(no
A are B) = Some A are B

Not(some
A are B) = No A are B

Not(some
A are not-B) = All A are B

*Mode Switch:*
All
A are B = No A are not-B

No
A are B = All A are not-B

Some
A are B = Some A are not not-B

From
one side of an equation, deduce the other.

To these rules, add one more:

**Some-All-None Trilemma**

**:**From any two of;

Some
A are B

All
B are C

No
A are C

deduce
the negation of the third.

Modal
identities plus some-all-none trilemma yields the core of classical syllogism
theory. Adding existential import to ‘all’ yields the rest. For instance, modal identities, substitutions
and swap can transform the Some-All-None Trilemma to Barbarism in three steps
thus:

Some A are not not-B

All
not-C are not-B

All
A are not-C (by
modal identities)

Some
X are not Z

All Y are Z

All
X are Y (substitute
X=A, Y=not-C, Z = not-B)

All
X are Y

All Y are Z

Some
X are not Z (swap)

An
anti-syllogism is not a syllogism itself, but it’s always ready to explode into
three conflicting syllogisms. For instance; the Barbarism trilemma defies
classical logic; yet encodes three classical logic rules:

From: All X are Y, All Y are
Z, deduce all X are Z.

From: All Y are Z, some X are
not Z, deduce not all X are Y.

From: Some X are not Z, All X
are Y, deduce not all Y are Z.

The last two can be changed,
by substitutions and swaps, to:

From: Some X are Z, no Y are
Z, deduce some X are not Y.

From: Some X are Z, All X are
Y, deduce some Y are Z.

*Exercise for the student***:**

Derive deduction rules and
troikas from these 24 trilemmas:

*Some days are bliss;*

*All bliss is perfect;*

*No days are perfect.*

*All heroes are immortal;*

*Some men are heroes;*

*All men are mortal.*

*No philosophers are liars;*

*Some philosophers are Cretans;*

*All Cretans are liars.*

*Some flattery is stupid;*

*Stupidity is always boring;*

*Flattery is never boring.*

*Some good deeds are wise;*

*Only foolish deeds are punished;*

*No good deed goes unpunished.*

*Some angels read comic books;*

*Only aliens read comic books;*

*No aliens are angels.*

*Only you can prevent forest fires;*

*Smokey the Bear can prevent forest fires;*

*You are not Smokey the Bear.*

*All Scots are canny;*

*All ghosts are uncanny;*

*Some Scots are ghosts.*

*Dragons are wise;*

*Dragons are brutal;*

*Dragons are not both wise and brutal.*

*I am a bum;*

*All bums are crooks;*

*I am not a crook.*

*I love Alice;*

*I do not love Bob;*

*I love Alice and Bob equally.*

*Bob is not a genius;*

*Bob is not an idiot;*

*Bob is a genius or an idiot.*

*If Alice fell down a rabbit hole, then I’ll be a monkey’s uncle;*

*If Alice did not fall down a rabbit hole, then I’ll be a monkey’s uncle;*

*I will not be a monkey’s uncle.*

*There are more saints than sages;*

*There are more sages than heroes;*

*There are more heroes than saints.*

*A crook is better than a fool;*

*A fool is better than a wimp;*

*A wimp is better than a crook.*

*The food is fast;*

*The food is cheap;*

*The food is good.*

*The pundit is honest;*

*The pundit is intelligent;*

*The pundit is partisan.*

*An attorney should investigate zealously;*

*An attorney should keep client’s secrets;*

*An attorney should report perjury.*

*Superman can fly;*

*Clark Kent can’t fly;*

*Clark Kent is Superman.*

*Some frogs are princes;*

*Some frogs are not princes;*

*All frogs are equally princes.*

*All lions are fierce;*

*Only coffee-drinkers are fierce;*

*Not all lions drink coffee.*

*No ducks waltz;*

*All officers waltz;*

*Some officers are ducks.*

*Time is unbounded;*

*Time is linear;*

*Time is finite.*

*There is a final effect;*

*All effects have effects;*

*Effectuation does not loop.*

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