And fuck knows what is happening in chapter eleven, where moving out of a plane is equated with bending time... or something. — Banno

I haven't read past the introduction, but perhaps this video conveys something of relevance?

Very much a guess.

Heh. You gotta read along with us!

I'm guessing I'll be skeptical when I get to those passages, but no matter the text it's a good idea to read it with multiple people.

I'm not seeing it.

Yeah, it's like there might be something interesting but just below the surface. Trouble is, so often such intuitions end in disappointment.

I'm not seeing it. — Banno

Ah, ok. Like I said, it was very much a guess.

I thought there might be some relevant analogies.

Cool.

Heh. You gotta read along with us! — Moliere

It's been feeling too much like coming home from work to go back to work. I talk to electrical engineers all day long. :joke:

Fair point.

Though that's interesting that the book is close enough to work to actually feel like work.

Wouldn't it be better to spend your time learning a more widely used version of predicate calculus? — Banno

Of course it would, if application is what you are looking for. My sister used to work for the electricity board on their very early computer that ran their pay-roll and bill producing accounts system, as a programmer in machine language. Not much call for that these days. But it's still how the machines operate. This book ends at the point where it links up with all the familiar systems of boolean algebra and predicate logic and set theory. If your philosophy is "shut up and calculate", a perfectly reasonable position, this book and this thread are not for you.

Don't waste your time telling us we're wasting our time with it.

Chapter 3 feels like a set up for chapter 4, which is what I said about 1 and 2 so I may just be in that habit. But I felt like it was all a set up for the final paragraph to make sense -- we have the initials of number and order for the calculus of indications, and Chapter 4 begins to actually write out some proofs from what has been written thus far. — Moliere

Yes. somewhere in the introduction/preface he says that this all developed backwards to the way it was written as a way of trying to understand why what they were already doing in practice worked. It's quite usual in philosophy: you build your castle in the air, and then go back afterwards to grub around for some foundations for it.

So we should kind of do the same; pass lightly for now over chapter 3, and I am going to pass lightly over the first 4 theorems too, as GSB satisfying himself that the rules and notation does what he wants and doesn't do what he doesn't want.

We have now shown that the two values which the forms of the calculus are intended to indicate are not confused by any
step allowed in the calculus and that, therefore, the calculus does in fact carry out its intention.

Again, I find it helpful to think of the left margin as a power source, and the right side as a light that will be either on or off. Thus an empty cross is a switch that is on, and...
$\left. {\overline {\, a \,}}\! \right|$ is a switch that is on if 'a' is not on, and off if 'a 'is on.

T5, T6 and T7 are more housekeeping; necessary but boring.

For T8, I am going to start with...
$\left. {\overline {\, a \,}}\! \right| a$ which we can think of as two circuits in parallel on one circuit 'a' operates a switch, and on the other it is the circuit. So if 'a' is on, it turns the switch off and connects via the direct route, and if 'a' is off it connects via the switch.

In T8, this identical arrangement which is always on, turns the switch it controls off. Light goes off!

T9 is also important. T8 and T9 together form the basis of everything that follows, so I'm going to give T9 a post of its own post, later.

Sure. Whatever gets you through the night.

T9.

$\left. {\overline {\, \left. {\overline {\, pr \,}}\! \right|\left. {\overline {\, qr \,}}\! \right| \,}}\! \right|$ = $\left. {\overline {\, \left. {\overline {\, p \,}}\! \right|\left. {\overline {\, q \,}}\! \right| \,}}\! \right| r$

Using my circuit analogy, on the left, p & r are parallel paths, and so are q & r. So if r = $\left. {\overline {\, * \,}}\! \right|$ then p & q are redundant, and 'light is on'. On the other hand if r is empty, it can disappear, leaving the expression on the right. So we have the parallel circuits on the right, of the p&q expression and a solitary r to cover both possibilities.

Not as complicated as it looks.

Chapter 6. The Primary Algebra.

J1. $\left. {\overline {\, \left. {\overline {\, p \,}}\! \right|p \,}}\! \right|$= .

J2. $\left. {\overline {\, \left. {\overline {\, pr \,}}\! \right|\left. {\overline {\, qr \,}}\! \right| \,}}\! \right|$ = $\left. {\overline {\, \left. {\overline {\, p \,}}\! \right|\left. {\overline {\, q \,}}\! \right| \,}}\! \right| r$

T8 & T9 now become J1 &J2 the foundations for some new developments, after a bit more housekeeping.

C1. $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right| = a$

I had to struggle to follow with this one. I found the condensed version clearer, and by going through the steps and noting down the substitutions for each line, I just about got there. Except I don't understand why not use the two substitutions for a, as was done for earlier proofs? Anyone?

I haven't read all the discussion but like the topic. I've written a bit about Spencer Brown and once talked to him on the phone.for a hour. A terrifying experience.

I can't help much with the details of the calculus he presents in LoF but have deeply explored its metaphysical implications. His calculus describes the logic of 'non-dualism', hence his quoted references to Lao Tzu. This is a fundamental description of reality and, equivalently, a fundamental theory of sets, and as such it solves Russell' and Cantor's problem of self-reference.

Basically, it states that there is no such thing as the 'set of all sets'. Rather, sets would reduce to the blank sheet of paper on which the Venn diagram is drawn. For an information theory this would be the
information space, whether in psychology or cosmology. ,.

In his way the 'Perennial' philosophy solves all metaphysical problems, including the reduction of the many to the one. You could say Brown's book explains thee reason why problems of self-reference do not arise for the philosophy of the Upanishads, Buddhism, Taoism and so forth, allowing it to be fundamental without giving rise to paradoxes.

Later in life Brown became a close friend of Wei Wu Wei, the renowned nonduality teacher, and perhaps this indicates that he knew his stuff. In his phone call he stated he was a buddha and I had no reason to doubt him other than the fact he mentioned it.

Now this is sounding like an esoteric cult.

Now this is sounding like an esoteric cult. — Banno

Make up your mind whether you think it is too boring or too interesting.

If you can't keep quiet, get involved! Uninformed and self-contradictory criticism sounds like mere prejudicial insult.

Thanks for that. You have summarised many of the points of interest, that hopefully we will eventually get to. That it seems to make a connection between East and West, and science and non duality is what interests me too, but I want to get there armed with as clear an understanding as possible of the systematic backing for those things you indicate about set theory etc.

I know what you mean but I feel it's misleading to speak of East vs West. Brown is a Westerner and his mate Wei Wu Wei was born Terence Grey, an Irishman with a love of fine wines and racehorses. I'd;even want to argue about the distinction between non-duality and science but that's a trickier topic. Minor quibbles. Brown is underappreciated imho so it's great that you've raised his profile a little.

Good luck with LoF. When I was getting started on it I found this essay useful (by the president of the Jungian Society in the USA). Robin Robertson, SOME-THING FROM NO-THING: G. SPENCER-BROWN’S LAWS OF FORM http://www.angelfire.com/super/magicrobin/lof.htm

This extract makes the connection between Brown's approach and philosophy. (I suspect that by 'consciousness' here Robertson means intentional consciousness, since for Brown consciousness is not emergent but is the birthplace of form.). . .

“Anyone who thinks deeply about anything eventually comes to wonder about nothingness, and how something (literally some-thing) ever emerges from nothing (no-thing). A mathematician, G. Spencer-Brown (the G is for George) made a remarkable attempt to deal with this question with the publication of Laws of Form in 1969. He showed how the mere act of making a distinction creates space, then developed two “laws” that emerge ineluctably from the creation of space. Further, by following the implications of his system to their logical conclusion Spencer-Brown demonstrated how not only space, but time also emerges out of the undifferentiated world that precedes distinctions. I propose that Spencer-Brown’s distinctions create the most elementary forms from which anything arises out of the void, most specifically how consciousness emerges.”



.

:ok:

when we're done with this book, we can maybe look at
http://homepages.math.uic.edu/~kauffman/VarelaCSR.pdf
And perhaps it might start to convince @Banno that we are not a cult.

$\left. {\overline {\, a \,}}\! \right|a$

...which we can think of as two circuits in parallel on one circuit 'a' operates a switch, and on the other it is the circuit. So if 'a' is on, it turns the switch off and connects via the direct route, and if 'a' is off it connects via the switch. — unenlightened

Hrm I'm not following the analogy here for T8 very well. How would the analogy work for the worked example of T8:

$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|a \,}}\! \right|$

?

Two circuits in parallel on a single circuit I follow. So "a" is an arrangment of wires between a battery with a switch on the circuit such that the lights which are wired in parallel both turn off in the worked example of T8, as you say.

So just visualizing a simple circuit diagram, 'a' is on when it turns to switch off -- does that mean the switch is not connected to the parallel wiring? Where is the switch in the diagram, in parallel with the lightbulbs or on the outer circuit?

Or am I just breaking the analogy in trying to concretize your rendition here?

EDIT: Mostly thinking through the analogy here. No need to reply. The below post serves better as a question since it has a diagram.

Using my circuit analogy, on the left, p & r are parallel paths, and so are q & r. So if r = — unenlightened

$\left. {\overline {\, * \,}}\! \right|$

then p & q are redundant, and 'light is on'. On the other hand if r is empty, it can disappear, leaving the expression on the right. So we have the parallel circuits on the right, of the p&q expression and a solitary r to cover both possibilities. — unenlightened

OK so "r" is the switch on the outer ring -- and if it is marked, or reduces to the marked state in the arithmetic, then the light is on because the switch is closed. And if it is not marked, then the light is off because the switch is open, but the marking of p and q is still there to be the wires or something like that.


I think I'm getting lost on the map between the arithmetic and the circuit diagram. I can stick with the arithmetic so far, though -- in the abstract.

EDIT: Outer/inner ring diagram, with ASCII -- for fun and profit:


___+/-___
r00000000|
!00000000|
------p------
!00000000|
!00000000|
------q------

?

(you'll have to read "0" as empty space, and "r" is that first little squiggly on the upper left hand side -- it's supposed to be a switch in my hypothetical)

Also -- I can just move on with the text itself. I realize this is an analogy.

when we're done with this book, we can maybe look at
http://homepages.math.uic.edu/~kauffman/VarelaCSR.pdf
And perhaps it might start to convince @Banno that we are not a cult

Regrettably, this is the kind of article that goes over my head. I have to leave the technical details to mathematicians and stick with basic principles. .

By the way, can you tell me why I often don't get a 'quote' option and have to copy/paste replies? Occasionally I do but usually not and it seems odd.

I'm not following the analogy here for T8 very well. How would the analogy work for the worked example of T8: — Moliere

"a" is a circuit, that operates the cross (switch) it is under. If "a" is live, it switches the circuit it is under off. This is how a cross under a cross cancels out - the inner switch switches the outer switch off and there is no circuit. That is the situation if "a" = unmarked - we ignore it and are left with a switch that has turned off a switch. So no circuit. But if "a" connects, it switches off the switch it is under in both cases, so both switches are turned off. either way the whole is off.

This is more than just an analogy, it is the application which he was working on when he developed the system. I think it's worth trying to get hold of, particularly when it comes to the really difficult section that introduces time. If you are at all familiar with such things, it is quite commonplace for an electrical switch to be electrically operated, for example by means of an electromagnet physically pulling a lever.

In the formalism, a cross is a switch that might or might not be switched off by a circuit 'inside it, and might or might not switch off a switch it is 'inside', if it is on. All crosses are on unless something (an inner cross) turns them off.

Regrettably, this is the kind of article that goes over my head. — FrancisRay

I know how you feel. It looks pretty daunting. But I'm hoping to at least get a feel for how the formal system can apply to living systems. Maybe...

You should get a quote button whenever you select some text in a post. I don't know why you wouldn't, unless you are on a phone and the button is coming up off-screen somewhere. You can do it the hard way: [ quote=aDude] some text [ /quote] without the space after the open brackets, but it's not as good because it doesn't have the post number that can take the reader to the original post.

This is more than just an analogy, it is the application which he was working on when he developed the system. I think it's worth trying to get hold of, particularly when it comes to the really difficult section that introduces time. If you are at all familiar with such things, it is quite commonplace for an electrical switch to be electrically operated, for example by means of an electromagnet physically pulling a lever. — unenlightened

Cool. I'm more familiar with the Physics 2 stuff than the practical stuff, and it's been more than a minute since I've studied that. I think I'm tracking better now with your explanation, and I had a gander at this website to get a grasp on the concrete side a bit better.

Aha. Thanks. I get it now.

Finally caught up to here. I'm struggling to follow the demonstration as well, so I'm going to type it out and see where it takes me.

It's the use of R1 that's confusing me. I understand that having derived an expression which is equivalent to the unmarked state we can substitute the unmarked state for said expression, but when I do so it seems like there should still be an "a" left over.

Or re-reading the use of R2 I'm not following again. It seems we have to

Let p = $\left. {\overline {\, a \,}}\! \right|$

And by R2 that means the initial J1 becomes

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|\left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ = . (2)

Then we start with the conclusion in the next step?

So we start with C1:

$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ = $a$

And substitute the unmarked state from (2) into C1 --

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|\left. {\overline {\, a \,}}\! \right| \,}}\! \right|$$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ = $a$$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|\left. {\overline {\, a \,}}\! \right| \,}}\! \right|$

And then subsitute $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ for $a$ in the next step? (that seems obviously fatal, but I'm not sure how else to do it)

EDIT: I really feel like that can't be it. I mean I get that we're making a logic, but a logic that assumes its own conclusions to demonstrate relationships is usually only done in a reductio or something like that. (though we haven't gotten to negation or truth yet, so...) It just seems kinda squirrely.

Actually hold.... C1 can be derived from axiom 2 as well. So I'm even more lost. :rofl: What is going on with C1? (EDIT: Maybe it's a demonstration of substitution rules?)

EDIT2:

(that seems obviously fatal, but I'm not sure how else to do it) — Moliere

Actually.... then they'd be exactly equal in form too. There is something very confusing about substituting for the unmarked state*. I did it on both sides of the equation, like you'd do for a variable in algebra, but I think maybe Brown did it only on one side of the equation. This relates to another confusion I had put aside, but the notion of the unmarked cross maybe relates?


*Like, if we can do that can we constantly substitute any amount of crosses which equate to the unmarked state into any unmarked part of an expression?

I think it's the layout that's confusing you, along with the strange fact that we already did this by substitution of a, which was easy. This time we only use J1 or J2.

Go to the bottom of page 31 Where it says,

We repeat this demonstration, and give subsequent demonstrations, with only the key indices to the procedure.

We are going to change the left side, $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ at the top into the right hand side at the bottom via the steps shown, using J1 and J2 and nothing else.
———————————————————————————————
the first step is to put p= $\left. {\overline {\, a \,}}\! \right|$ into the J1 formula. and stick it in front of $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ which we are allowed to do because it sums to the unmarked state, and so changes nothing.
—————————————————————————————————
We now have something that looks like the right hand side of J2 if we set r =$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ You should be able to see what the p and q substitutions are, and the result is what is written. (This is the most difficult line to follow)
———————————————————————————————————
Step 3 uses J1 again to remove the left side of the expression, leaving just the right hand half, which is:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|a \,}}\! \right| \,}}\! \right|$


That's half way through the proof. With me so far?

That's half way through the proof. With me so far? — unenlightened

That helps. Thanks. I'm with you up to this point now.