## Friday, June 28, 2013

### On Troikas: Disintermediation, Still Growth and the Tree

Disintermediation, Still Growth and the Tree

Consider this Disintermediation Trilemma:

* The function f(x) is real and continuous on the interval [a,b];
* f(x) equals zero nowhere on interval [a,b];
* f(a) and f(b) are of opposite sign.

Two of the trilemma imply the negation of the third, by the Intermediate Value Theorem:

If f is continuous and never zero, then it does not change sign;
If f is never zero but changes sign, then it is discontinuous;
If f changes sign and is continuous, then it is somewhere zero.

Let each Stooge affirm all terms of each of these implications, to make a troika supporting the trilemma.

A real continuous nonzero sign-changing function is a curious beast; it’s a kind of continuous Heap, changing sides without crossing the boundary. Therefore let us call such a function a “Smuggler”, and this the Smuggler Trilemma.
Consider the tree. First it is less than ten million microns tall, later it is more than ten million microns tall; it grows continuously; yet you cannot find a microsecond in which it is exactly ten million microns tall! So let us call this the Paradox of the Tree.

Now consider this Unmean Values Trilemma:
* f(x) is differentiable on the interval [a,b];
* f(a) < f(b) ;
* df/dx is positive nowhere on [a,b].
You can also call this the Still Growth Trilemma. Any two of the trilemma implies the negation of the third, by the Mean Values Theorem. As ever, this unpacks into three deduction rules, and expands into a troika.
Consider the tree. It increases in size, at a continuous growth rate of zero. That is the Second Paradox of the Tree.

## Thursday, June 27, 2013

### On Troikas: Disinduction and the Heap

Disinduction and the Heap

Consider mathematical induction:
Given:             1 has property P;
And:                For any positive integer n, if n has property P then so does n+1;
Deduce:           Any positive integer N has property P.
The middle term says that P is ‘inductive’; which means an infinite conjunction of implications:   (P(1) implies P(2)) and (P(2) implies P(3)) and (P(3) implies P(4)) and …   Therefore mathematical induction is a sorites of infinite length.

Denying the conclusion creates a trilemma:
1 has property P;
For any positive integer n, if n has property P then so does n+1;
Some positive integer N does not have property P.

This is the Disinduction Trilemma. It is also known as the Paradox of the Heap: for let P = ‘does not make a heap of sand grains’. Surely 1 grain of sand is not a heap; and surely adding one single grain of sand to a non-heap will not make it a heap, so non-heapness is inductive; yet surely there is some number N, denumerating a sand heap!

Disinduction can be expanded into this troika:
Moe: 1 has property P, P is inductive, and every N has property P.
Larry: P is inductive, some N does not have property P, and neither does 1.
Curly: 1 has property P, some N does not, and P is not inductive.

It also can be unpacked into these three deduction rules:
If 1 has property P, and P is inductive, then every N has property P.
If  P is inductive, and some N does not have property P, then neither does 1.
If  1 has property P, and some N does not, then P is not inductive.

## Wednesday, June 26, 2013

### On Troikas: Anti-Sorites

Anti-Sorites

An anti-sorites is a multilemma; that is, a set of statements, not all of which can be true. Therefore all but one true implies the last is false; a form of sorites reasoning. Any anti-sorites encodes several sorites at once.
You make an anti-sorites by appending the negation of a sorite’s classical conclusion. This you can then ‘untangle’, by row swaps and modal identities. For instance, take this sorites:

Some lunches are nutritious;
Anything nutritious is good for you;
Only valuable things are good for you;
Anything valuable is paid for;
free means paid for.

The logical conclusion to this sorites is
Some lunch is free.
- being nutritious, good for you, valuable, paid for, and hence free. After all, you paid for it.
To make an anti-sorites, replace that conclusion with “no lunch is free”,  and get:

Some lunches are nutritious;
Anything nutritious is good for you;
Only valuable things are good for you;
Anything valuable is paid for;
Free means paid for;
No lunch is free.

This anti-sorites, untangled by row swaps and modal identities, becomes;

Some lunches are nutritious;
Anything nutritious is good for you;
Anything good for you is valuable;
Anything valuable is paid for;
Anything paid for is free;
No lunch is free.

This is a SAAAAN anti-sorites. The inner “all” sequence collapses to “Anything nutritious is free”, resulting in a Some-All-None trilemma. Either the initial “some” statement is false, or the final “none” statement is false, or one of the chain of “all”s fails.
Here’s an anti-sorites, row-swapped and then remodulated:

Butterflies are free;
Not all lunches are bland;
Only butterflies are beautiful;
All unbeautiful things are bland;
There’s no such thing as a free lunch.

Not all lunches are bland;
All unbeautiful things are bland;
Only butterflies are beautiful;
Butterflies are free;
There’s no such thing as a free lunch.

Some lunches are not bland;
All not-bland things are beautiful;
All beautiful things are butterflies;
All butterflies are free;
No lunch is free.

Here are some anti-sorites derived from sorites by Lewis Carroll, then untangled:

No interesting poems are unpopular among people of real taste;
No modern poetry is free of affectation;
All of your poems are on the subject of soap-bubbles;
No affected poetry is popular among people of real taste;
No ancient poem is on the subject of soap-bubbles;
Some of your poems are interesting.

Some interesting poems are yours;
All of your poems are on the subject of soap-bubbles;
All poems on the subject of soap-bubbles are modern;
All modern poetry is affected;
All affected poetry is unpopular among people of real taste;
No interesting poems are unpopular among people of real taste.

No kitten that loves fish is unteachable;
No kitten without a tail will play with a gorilla;
Kittens with whiskers always love fish;
No teachable kitten has green eyes;
No kittens have tails unless they have whiskers;
Some kitten with green eyes will play with a gorilla.

Some kitten with green eyes will play with a gorilla;
Any kitten that will play with a gorilla has a tail;
All kittens with tails have whiskers;
All kittens with whiskers love fish;
All kittens who love fish are teachable;
No kittens with green eyes are teachable.

Things sold on the street are of no great value;
Nothing but rubbish can be had for a song;
Eggs of the Great Auk are very valuable;
It is only what is sold on the street that is really rubbish;
Eggs of the Great Auk can be had for a song.

Eggs of the Great Auk can be had for a song;
Anything that can be had for a song is really rubbish;
Anything that is really rubbish is sold on the street;
Anything sold on the street is not very valuable;
Eggs of the Great Auk are very valuable.

This multilemma has a single object at beginning and end, with a property inverting during the A sequence; so an “xAAA~x” anti-sorites.
We know that one of the statements in the Great Auk Anti-Sorites is false. This is a lot more diffuse than a trilemma. An anti-sorites is like the game of telephone, with an error happening somewhere in the deductive chain; then a ‘some’ vanishes into a ‘none’, or the Great Auk’s eggs turn upside down.
All of these can be supported by troikas. Each Stooge need merely deny one of three different terms of the anti-sorites. Perhaps Moe could deny the Some statement, Larry could deny one of the Alls, and Curly deny the None.

## Tuesday, June 25, 2013

### On Troikas: Finite Boundless Lines

Finite Boundless Lines

Consider this Time Troika:
Moe says that time is linear, bounded, and finite; an interval.
Larry says that time is linear, unbounded, and infinite; a line.
Curly says that time is circular, unbounded, and finite; a loop.
2/3 (KM) say: time is finite.
2/3 (LK) say: time is unbounded.
2/3 (ML) say:  time is linear.
But all agree that time can’t be all three.

If Moe gives time a beginning but no end, then the trilemma is: time is finite; past time is unbounded; past time is linear. If Moe gives time an end but no beginning, then the trilemma is: time is finite; future time is unbounded; future time is linear.
These are both what I call a Line/Ray/Loop Troika. The line, ray and loop needn’t be in time; they can be any ordered sequence, continuous or discrete. In a line/ray/loop troika:
Moe sees a Ray: the sequence has an endpoint.
Larry sees a Line: the sequence extends to infinity.
Curly sees a Loop: the sequence cycles.
Majorities affirm this “Finite Boundless Line” trilemma:
The sequence is finite;
The sequence is unbounded;
The sequence is linear;
-        but all agree that the sequence is not all three!

Now consider this “Agrippa Troika”:
Moe: There is an unexplained first explanation.
Larry: Explanations regress to infinity.
Curly: Explanation is circular.
In this troika, dogmatic Moe has an explanatory Ray, infinite-regressing Larry has an explanatory Line, and circular-reasoning Curly has an explanatory Loop. Majorities affirm:
Explanation is finite;
Explanation is complete;
Explanation is noncircular.
But not all three! That is the “Munchausen Trilemma”:
Any explanation can be at most two of:
Finite; that is, some explanation explains all others;
Complete; that is, every explanation is explained;
Noncircular, that is, there are no explanatory loops.
The Munchausen Trilemma describes a finite boundless line of explanations. Finite, complete and noncircular is called normality; an illusion created by paradox.

The line/ray/loop troika applies to the Paradox of the First Cause. Define the First Cause as the cause of all causes; then the Paradox of the First Cause is; what caused the first cause? If there be any such cause, then let us call it a zeroth cause. Is there a zeroth cause?
There are three possibilities:
In the Line, there is no first cause, but instead an endless backwards sequence of causation.  The Line is a theory of infinitely deep causation.
In the Ray, there is a first cause but no zeroth cause; so the first cause is uncaused. The Ray is a theory of chaotic causation.
And in the Loop, there is a first cause and a zeroth cause; and these cause each other. The Loop is a theory of circular causation.

The Line corresponds to chaos theory, with infinite fractal complexity. The Ray corresponds to quantum theory, based upon intrinsic random chance. The Loop corresponds to Goedelian metamath, with self-referential paradox. Chaos, quantum and metamath are the 20th century’s three-fold retort to determinism. Of the three:
In Ray and Loop, causation has finite depth;
In Loop and Line, causation is recursive;
In Line and Ray, causation does not loop.

So by 2/3 majorities: causation is finite; causation is recursive; causation is linear; but not all three! That’s the Causation Trilemma, a finite boundless line. Cause has a beginning, every cause has a cause, cause is one-way; deny one!

Dual to the First Cause is the Final Effect; the effect of all effects. Does the Final Effect have any effect? Either there is no Final Effect, or the Final Effect  is ineffectual, or there is a Postfinal Effect.
This translates, as before, into Line, Ray and Loop. In the Line, effect is linear, recursive and infinite; there is no final effect. In the Ray, effect is linear, nonrecursive, and finite; the final effect has no effect. In the Loop, effect is circular, recursive and finite; postfinal and final are effects of each other.
Of  Line, Ray and Loop, by 2/3 majority each: effect is linear; effect is finite; effect is recursive: but never all three. That’s the Effectuation Trilemma, another finite boundless line.  Effect has an end, every effect has an effect, effect is one-way; deny one!