Box Algebra , Boundary Mathematics,

Logic and Laws of Form

By Lou Kauffman

I. Section One

We will work with a formal system based on one symbol, a rectangle.

An *expression* in this system is a finite collection of non-overlapping
rectangles in the plane. For example:

In an expression, one can say about any two rectangles whether one is inside or outside of the other.

We allow two rules of transformation on expressions.

**1.
Cancellation. ** Two nested rectangles with the inner rectangle empty and the
space between the two rectangles empty, can be replaced by the absence of the two
rectangles. That is they can be erased.

**2. Condensation.** Two empty adjacent rectangles can be replaced by one
of them.

**Proposition. ** Any finite expression in rectangles can be reduced by a
series of applications of cancellation and condensation to either a single
rectangle or to an empty plane.

Example.

**The Problem:** Write a proof of this proposition. Your proof should be
clearly worded and it should explain why the hypothesis of finiteness for
expressions is needed.

**Hints:** It is useful to use a concept of depth in an expression. I have
labeled the spaces in the expression below by their depth. The depth of a space
is the number of inward crossings needed to reach it.

There must be rectangles enclosing spaces of maximal depth in the expression. Such rectangles are empty (else they would not be deepest). Call such a rectangle a deepest rectangle. Now you should be able to show that a deepest rectangle either has an empty rectangle next to it (so that condensation is possible), or it is surrounded by a rectangle forming a nest of two rectangles (so cancellation is possible), or the deepest rectangle is the only rectangle in the expression. These remarks will provide an algorithm for reducing any given expression to either a rectangle or to an empty plane.

**Further Work:** If you are interested you can carry this theory further
in the following ways.

**1.** Allow the opposite of condensation and cancellation in expressions.
That is, allow *creation* and *expansion *where expansion allows you
to put a new empty rectangle next to an empty rectangle, and creation allows you
to create a nest of two rectangles in any empty space.

(a) Creation.

(b) Expansion.

**For example:**

Note that in this example we used only creation and expansion.

Also, it should be apparent by now that we only care about the relationships of the rectangles not their sizes.

**Another example.** We can use combinations of creation, cancellation,
expansion and condensation together.

We shall call two expressions equivalent if one can be obtained from another by a finite sequence of these moves (creation, cancellation, expansion and condensation).

We have shown that any expression is equivalent to either a rectangle or to the empty plane.

**Problem. **Show that the rectangle and the empty plane are not equivalent
to one another.

Once you have shown this, it follows that the simplification of an expression
to rectangle or empty plane is unique. *We will call the *

*single
rectangle the marked state and the empty plane the unmarked state.* Thus every
expression is equivalent to either the marked state or the unmarked state.

**Hint:** Find an independent way to evaluate an expression as marked or as
unmarked, and then show that your method of evaluation does not change under the
elementary equivalences of creation, cancellation, expansion and
condensation.

**2.** One can then make an algebra in relation to this arithmetic just as
we make algebra in relation to ordinary arithmetic. Given any expressions A and
B, we let AB denote their juxtaposition in the plane. We let denote the result of putting a box around
the expression A. Since each expression A or B represents either a marked or an
unmarked value, the algebra that results will be an analog of Boolean algebra, or
in other words an algebra of logic.

In fact we can interpret this box algebra for logic as follows:

(a) A box stands for the value TRUE. An empty space or a doubled box stands for the value FALSE.

(b) stands for NOT A.

(c) AB stands for A OR B.

We leave it as an exercise for you to see that

stands for A AND B and that

stands for A IMPLIES B.

With this dictionary one can learn to do logic using the box algebra. It has a number of interesting features. If you want to explore this topic further take a look at the book "Laws of Form" by G. Spencer-Brown where a version of this arithmetic and algebra first appeared.

A very similar algebra for logic was invented by the late nineteenth and early twentieth century logician and philosopher Charles Sanders Peirce.

II. Section Two -- A Dialogue

This section is a dialogue between Lou, the author of this piece, and George, an hypothetical reader.

**George. ** I have been working with your box algebra for a while now and
I have some questions. My first question is this: You said that we can cancel
two empty boxes if they are adjacent to one another.

After awhile I realized that you probably meant that

are adjacent and that we should be allowed to condense them to a single box. But I also would like to condense the two large empty boxes in the following expression. Are they adjacent?

**Lou. ** Those two boxes are not directly next to one another. I would not
call them adjacent if they were tables in a restaurant.

However it is true that we can condense them and allow the move:

So we need a technical definition of "adjacent" that will work for our

purposes. We will say that two boxes are **in the same
space **if it is possible to draw a path from the outside of one to the outside
of the other that does not cross any boundaries in the expression. This path is
not necessarily a straight line. So in the situation we are discussing we see

with the connection between these two boxes showing that they are in the same
space. Now we shall simply define two boxes to be **adjacent** if they are in
the same space and let it go at that. This notion of adjacency is not quite as
local as the English use of the term, but it will work for our purposes.

**George.** Ok. I am satisfied with that, but I notice that another way we
could handle adjacency would be to only allow the boxes to occur in rows. Thus I
would take an expression like

and rewrite it (without using rules expect my ability to rearrange the boxes) to the form below.

Then I can take adjacency to mean that one box is directly to the right or the left of the other, and that neither box contains the other.

**Lou. **That is a good point! You are observing that as far as valuation
of an expression is concerned, the only relationship we need to know about two
boxes is whether one box is inside the other or not. Thus we can rearrange the
boxes so that they line up horizontally, and proceed to do cancellation and
condensation on them using direct adjacency. Notice that with respect to this
ordering from left to right, you are assuming the commutative law. That is you
are assuming that AB is the same as BA when AB denotes the result of juxtaposing
two expressions in direct adjacency. In particular we have

This is of course perfectly compatible with our evaluations.

**George. ** In fact, if we linearly order the expressions from left to
right, then they are the same in structure as parentheses. I mean I can write [
[] [[ ]] ] instead of the left expression in boxes above.

**Lou.** That is certainly true and it is a good way to abbreviate these
boxes and a good way to input these expressions into a computer. We can also use
the parentheses to translate box algebra into more traditional forms of algebra.
I will come back to this point in later sections of the discussion.

**George.** But I would like to know how you will prove that there is no
way to transform the empty box to an empty space.

**Lou**. Did you try to prove it?

**George.** No.

**Lou.** Well try to do it, and then we will talk about that.

III. The Dialogue Continued

**George.** I have been thinking about
that problem, and I am confused. It seems quite obvious to me that there is no
way to transform a box to an empty space, but how do you prove it?

Lou. Well first of all, it helps to see that if we had defined thinks differently then the catastrophe could happen! Consider a parenthesis based system with the rules

A. >< =

B. <<>> =

Then we have < >< > = < > = < > which is just our condensation/expansion rule. But look at this:

< > = << >> < > = << > >< >

= << > > = << >> = .

So in this system it is possible for the box (in parenthesis form) to disappear!

**George.** That was not fair! You divided the box into the left and
right pieces and then let the pieces act separately!!

**Lou.** It is true. I did that. But it was an example of a formal system
in which creation/cancellation, expansion/condensation occur and nevertheless the
"box" can disappear. This example shows that we should make a proof for our box
arithmetic.

**George. ** All right. I will work on it.

**Lou.** Here is a hint. Think of a box expression as a signal processor.
Suppose that there is a signal in each empty box, that every time the signal
"goes out" into boxes of smaller depth they are inverted when they cross a box
boundary.

Here I have illustrated this idea using two signal values. One value is n (for
not marked). The other value is m (for marked). In the expression above, we give
the deepest space the value n and watch the signal bubble upward. It crosses one
boundary and becomes m. Then it crosses another boundary and becomes n, and
finally a third boundary and becomes m in the space of depth zero. Incidentally,
I will call the space of depth zero the **shallowest space.**

**George.** I see! And the final value is in fact the value of the
expression consisting of the nest of three boxes. But what will you do with this
one I wonder?

Here I have put unmarked signals in the deep empty spaces of the expression, and I have bubbled them up according to your rules as far as I could. But I end up with an m and an n in the same space at depth 1. What do I do now?

**Lou.** Well, if m means marked and n means unmarked, what would m next to
n mean?

**George.** Oh! m next to n means marked next to unmarked but that would
still be marked. I see. We can use the following rules:

mm = m

mn = nm = m

nn = n

When it is marked then it is marked. The only way to be unmarked is to be unmarked. Is this philosophy? Well look, I will continue to calculate:

So the calculation says that this expression is unmarked, and indeed if I transform it will go away!

**Lou. ** Right. This is a method for calculating in a completely standard
way a value of n or m for any expression. Now you can do the exercise to show
that if you change the expression according to one of our rules, this value does
not change. And clearly the single empty box has value m while the empty space
has value n. Since m and n are distinct, this proves our result.

**George.** How do you know that m and n are distinct?

**Lou. ** Careful! You are crossing a boundary with that question.

It is given that m and n are distinct symbols that we are calculating with. We see that any calculation with m's and n's is just a string of them like mmnmnnn. And the value of such a string is m if there is an m in the string. The value of the string is n is there is no m in the string. The use of m and n depends upon our ability to distinguish at all. If we could not make two distinct symbols to work with, then there would be no hope of doing any mathematics at all. In this sense the possibility of doing mathematics is the same as the possibility that there is a distinction.

**George.** Well. Where do we go from here?

**Lou.** Algebra and onward! But lets take a rest right now.

IV. The Dialogue Continues

**George. ** Wait. I don't want to quit
just yet! I have another question.

**Lou.** Yes?

**George.** What about infinite expressions? I know that we cannot reduce
them all to marked or unmarked states, but can't we study them anyway?

**Lou.** Certainly! Lets slow down a bit. Now the expression J that I am
drawing here is perhaps the simplest example of an infinite expression. It is an
infinitely descending nest of boxes.

There is no way to evaluate J using our rules, since J has no deepest space, and hence no instance of condensation or cancellation. Notice that I cannot actually write J. What I write contains those famous three dots in a deepest space. The three dots mean that you have to

go on in the same pattern "forever". This expression J is a mathematical idealization. It does not exist as an entity that can be fully written out. However, J does have a very nice description in terms of itself. We can write J equal to a box around itself!

In other words, this infinite expression J is sitting inside itself.

We say that *J reenters its own indicational space.*

From this reentering property you can see how J has no hope to be equal to either a marked state or an unmarked state. For suppose that J is marked. Then J would be equal to a box around a box, but this is unmarked! And if J is unmarked, then the equation will tell us that J is marked.

We already knew that J could not be reduced to a marked state or to an
unmarked state, but now we know that it would be inconsistent to assign it such a
value. If J is to be a value in logic, then it is a value that heads on out
beyond true and false. J is analogous to numbers like the square root of minus
one, numbers that cannot be part of the real line. such numbers are called
complex or imaginary. Similarly, we shall call J an *imaginary (logical)
value!*

**George.** I see what happened to you. You fell into the
thinking that mathematicians always fall into. You think that the infinite nest
of marks in J exists "all at once" in some timeless domain. And then you look
inside the mark around the outside of J and see that inifinity in the same
pattern. And so you get your equation.

That is fine for idealists, but I am going to take a different view about what you just told me. I am going to to understand that an equals sign in the form A = B really means "replace A by B". That is the way I use it when I am programming a computer. The mind is just another computer. You think things can be "equal". I deny it!

The equals sign just means that one "thing" can be replaced by another thing and in either order, unless one decides upon an order.

Now lets see what your equation says. Suppose J = , then the equation says " replace a box by a box around a box" or "replace a box around a box by a box around that." I will emphasize this by writing a double arrow instead of that equals sign.

But the equation does not say that you cannot do it again so I really should write as follows.

If you follow the process to the right, it starts building the infinite form.
Now I know it never gets there, but is this not a much better and more concrete
view of the situation? The clock ticks, and at each tick a new box is added to
surround the old nest of boxes.

At each tick of the clock the value of the
expression changes

from marked, to unmarked, to marked , to unmarked, and so on.

Your J is just an *oscillator. *It is no more paradoxical than
a doorbell or a buzzer. Furthermore J is either marked or unmarked at any given
time! You got confused when confronted with temporality. You want to collapse
the passage of time into an eternal and everlasting geometry of the present.

**Lou. ** There is certainly merit in what you say, but I do want to treat
the algebra in a way that does not depend upon time. If we take J to satisfy the
equation and use the usual rules of
substitution and replacement in algebra, then it is easy to arrive at a
contradiction. For example, lets suppose that J satisfies all the rules that
finite expressions satisfy at the algebraic level. Then since

for any P, we have .

But if and , then and this is a contradiction. That is, if we attempt to include J in this standard way, then the whole system will collapse!

**George. ** You are right, and I realize that one way to handle this
situation would be to give up some of the algebraic rules that hold for finite
expressions. However, there is another way due to Jim Flagg. I am sure you know
it. In the Flagg Resolution we say that while , there is only one J and so if you change J to J with a box around it
somewhere, then you must make this change everywhere! With this principle in
mind, you cannot make the step

.

The best you can do is

and this will hardly lead to any contradictions!

The Flagg resolution lets us keep the basic algebra and still include temporal expressions like J. Saying that there is only one J and that substitutions must be performed everywhere is way to include the temporality in the algebra.

**Lou. ** I like the Flagg Resolution. We should explore the world of
infinite expressions using it.

**George. ** Thanks. Lets do that.