It would be much better to look at historical cases, the miners and Thatcher, perhaps. — Banno

I agree here. Historical cases demonstrate how people have overcome various problems, or failed at overcoming various problems. What they lack in conceptual clarity they gain in fidelity to the vagaries of political action.

https://en.wikipedia.org/wiki/Ford_sewing_machinists_strike_of_1968 is a good example because it also has a movie you can watch which I found satisfying. I like that it depicts strikes as scary things, because they are scary. You don't know the outcome of your actions, and you have the real possibility of losing.

But progress has only come from normal people being willing to do scary things.

But keep in mind that the worker of the example has problems in his family: the wife is already unemployed, one kid is sick and the other goes to college. Maybe the rest of the workers are covered up thanks to the incomes of their respective families... — javi2541997

Families are rarely in a state of perpetual comfort. There's always something to take care of. This is the common mantra of the scab: "I understand what you're doing, but I have to take care of my family" -- which is fear. The scab believes the boss is going to win, so the scab chooses the boss's side. This is to the detriment of the strikers, whose families are similarly unstable, have needs, and so forth. It's not a neutral act of duty, it's a person actively sabotaging the efforts of strikers in the name of their family: Family over Union.

A common act of social grace is to say something along the lines of "The spouse wouldn't allow it" or "My kid is at home sick", and I understand them to be bowing out of whatever it is we're doing. But they're usually speaking for themself, and sometimes the kid is at school and the parent just needs some time to themself and the only excuse people accept is some duty or other.

But when you're talking union the "my family" excuse is out the door -- buddy, we all have families. That's what we're doing this for.

Workers do face difficult decisions in supporting a union drive, becoming active in the union, and in striking, especially when the employer is hostile. The risks are not a pretense. Strikes do not always succeed, and a failed strike can leave the union members broke and out in the cold. — BC

I agree.

That's why I thought that duty doesn't leave a choice.

You may have a family. But do you think that the other strikers don't? They're already risking exactly what the OP sets up. Some cave to fear, or selfishness, and think of their own family in the moment. It's particularly difficult because a person has these connections, and that's a struggle.

But philosophically speaking the duty is clear: the strikers are risking their families already. You should do the same. Don't blame your fellow workers who are just asking for fairness to be able to take care of their families, blame the boss for not doing what's right!

Which should point out the importance of looking at, even if I'm wrong, fear and anger.

With that being said: I really like the last chapter because it's close to what I was writing out, once upon a time, when attempting to invent a purely spacial logic (that didn't work out, and so I threw it away as nonsense). Also I enjoyed working through a calculus that isn't the same as all the others, though there are parts I'm still uncertain about. And I enjoyed GSB's fusion of poetry with math, which is something I have a soft spot for. (EDIT: Might be a bit of a surprise given my approach so far -- but I wanted to treat the book as a logic in addition to its poetic value, and thought maybe that was the part I could contribute to for understanding the text)

Good recommendation @unenlightened, and thanks for the prodding and motivation. I would not have finished the book without your help.

Empathy. You see what happened to someone, you understand that you could have been the one, and so you leave something nice because that's what you'd like to be done if it were you.

Typing out the ending, because it's probably the most interesting philosophical part:

An observer, since he distinguishes the space he occupies, is also a mark.

In the experiments above, imagine the circles to be forms and their circumferences to be the distinctions shaping the spaces of these forms.

In this conception a distinction drawn in any space is a mark distinguishing the space. Equally and conversely, any mark in a space draws a distinction.

We see now that the first distinction, the mark, and the observer are not only interchangeable, but, in the form, identical

While I've admitted ignorance to certain parts of GSB's demonstration, I'm not sure about the conclusion here. Not that it's wrong, only that I'm uncertain that it's earned.

Observers and such haven't really shown up until this point. He's asking us to interpret ourselves as an "m" outside of a circle where the circle is the forms around us. But this would be the simple subject, if it can be reduced to an "m"? Or no? It's not clear, because "observer" shows up at the very end.

All the same I think I like "a distinction drawn in any space is a mark distinguishing the space" -- to mark a space one must mark. Even "the unmarked state" has been used so far as equal to variables, and so works, in a sense, as a marked space would (in a different way from the way space pervades expressions)

But I'd say that GSB sees something I don't, at the end. And I suspect it's because he's an idealist. He can see that the first distinction makes the observer interchangeable with the mark because he believes that, at base, this is all mind-stuff and the forms we see are mentally constructed? Or something along those lines.

But all I see is a mark, and a man who wants to be that mark.

I said I'd wait but I'm not :P :D -- I finished the book today. But that chapter on time isn't clicking.

I'm tempted at this point to just grant chapter 7. I checked T10 through the proof by example of J2 and it works out, but since I didn't get to C9 the same as GSB, and it looks close enough for me, I'm fine with just allowing it and moving on. After working through those examples the text is reading a lot easier and I can kind of see how it's good enough for me. :D -- it was fun to work through the nitty gritty puzzles, but I'm starting to think "OK, I'm ready for a point now"

Seventh Cannon is what made me think the above -- it pretty much lays out what we've been doing in the demonstrations, but allows you to use the algebra now.

The bits on time: we get the conclusion I was thinking of, which is interesting to me!, that there are undecidable expressions (now that we have functions that go to infinity).

One thing I'm thinking is you could just posit another space-dimension to accommodate GSB's "cross in a plane", but I'm ok with saying this is space-time instead.

The oscillator function makes me want to take back what I said earlier about negation. It seems like we're close to negation with it because of the relationship between the two spaces inverts, but it's never named.

But I can say I've officially lost the plot at page 62's "Time in finite expression" -- I don't understand Figure 3, or most of the figures after that. The only one I get right now is page 65's rendition of E1.

But also I kind of continued past when I normally would just because I was hooked. I've officially finished the book now, but without as much math ;). Still chewing on the ending.

The boss's weapon is fear, the union's weapon is anger.

I prefer anger to fear. But many prefer fear. I don't think it comes down to duties as much as these two emotions -- which one a person lets rule them is how they'll decide to strike or scab. Or, if I were to put it in terms of duties, I'd say there isn't even a choice. But I doubt that's surprising ;).

And let's see if I can finish it out now...


(C9.12) from above shows how C2 can be used to obtain C9.3 more easily.

Let C2's a= unmarked state, and b= r for the right-hand expression in C9.12

$\left. {\overline {\, ab \,}}\! \right|b$ = $\left. {\overline {\, a \,}}\! \right|b$ (C2)

Plugging the right hand expression of C9.12 into the right-hand side of C2, while keeping $\left. {\overline {\, xy \,}}\! \right|$ as is...

$\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|r$-->$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right| \,}}\! \right|r \,}}\! \right|$ (C9.13)

We remove the double cross with C1 and put back into the original expression to get C9.3:

$\left. {\overline {\, ba\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|\left. {\overline {\, rxy \,}}\! \right|$ (C9.3)

Looking at the differences between C9.3 and C9.4 I thought that only the left-hand cross was involved in the transformation, but then saw that we're using C2 and the only way to get r-x-y-cross underneath that cross is by setting C2's b = r-x-y-cross, and a is everything else underneath the left-hand cross. It's a transformation from the right-hand-side of C2 to the left-hand-side.

$\left. {\overline {\, ba\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|\left. {\overline {\, rxy \,}}\! \right| \,}}\! \right|\left. {\overline {\, rxy \,}}\! \right|$ (C9.4)


This last step took some guesses for me. What worked was using C7. I could see that we really just want to get rid of some of the terms underneath the left-hand cross of C9.4. I pretty much just guess-and-checked my way to a solution.

C7, but inverted to show the direction of transformation:

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

Let C7's a =xy, b =xy, and c = r. Ignoring, from C9.4, and maintaining b and a from C9.4 (but erasing the cross that all of this is contained in to make it less bulky)

$\left. {\overline {\, ab\left. {\overline {\, xyr \,}}\! \right|\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|\,}}\! \right|$ --> $\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|xy \,}}\! \right|r \,}}\! \right|ab \,}}\! \right|$ (C9.5)

Which is close to where we want to get to, we just want to remove the redundant part at depth 3 and 4.

For that I went to C2:

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

Taking just the expression:

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

Let C2's a = unmarked state
Let C2's b = xy

It becomes

$\left. {\overline {\, * \,}}\! \right|xy$ (C9.6)

Then take this expression into C3:

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

Let C3's a = xy and C9.6 reduces to a single cross.

Plug C9.6 reduced back into C9.5:

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

Using C1 we reduce the double crosses over the unmarked state and plug this back into C9.4 and rearrange we obtain C9's conclusion:

$\left. {\overline {\, \left. {\overline {\, r \,}}\! \right|ab \,}}\! \right|\left. {\overline {\, rxy \,}}\! \right|$

And so we have C9:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, x \,}}\! \right|r \,}}\! \right|\left. {\overline {\, \left. {\overline {\, y \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ = $\left. {\overline {\, \left. {\overline {\, r \,}}\! \right|ab \,}}\! \right|\left. {\overline {\, rxy \,}}\! \right|$ (C9)

Which, yup, I can get the conclusion, but it certainly didn't involve using C6 -- and I tried to do it that way a few times but concluded it couldn't be used because there's just not that form present in the expression of interest. So a bit of a guess-and-check based on what we're trying to obtain got me through the demonstrations.

There's something odd about letting b = xy and a = xy, but I don't really see a reason within the logic presented so far to not allow two different variables to be assigned to similar names at different parts of an expression so I just went with it. But it does seem kind of funny. (Also noting how Theorem 12 states that I should use T10 in place of J2, and I didn't really use J2... I was just looking for a way to make the equation work. So that derivation may be wrong, or at least wasn't what GSB had in mind -- but it seems to follow still)

I have now figured out what he meant by "integration". I missed that C3 was named integration, and so this is him pulling out one of the sub-steps in the demonstration of C3 and calling it "integration".

That leaves the bit where he says "Thus if we consider the equivalence of steps..." which I'm still scratching me head on. I understand that the steps have an order to them, and so reduce depending upon the order of transformations. But his conclusion that "therefore the unmarked state is equivalent to a step" doesn't make sense to me. He demonstrated that taking steps is inconsistent and so he concludes that the unmarked state is a step rather than not putting steps into the calculus -- which is what I'd do since steps had never been introduced as a member of the calculus. We've been working through the marked state, the unmarked state, equivalence, and transformation through substitution. The steps are semi-arbitrary in that we can set any part of an expression equal some other conclusion as long as the forms match up. Why on earth would I take a step as a part of the calculus and set it equal to the unmarked state?

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ (C9.1)

Noticing that the form closely matches C8... (Note, I thought I had it in typing it out but then noticed I'm making a mistake, so the first part is a failed attempt at demonstrating this step, but after the break I figured out my mistake and demonstrate the step from C9.1 to C9.2)

Let C8's...

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

Because that's the only expression which appears under both crosses in C8's transformation to the right-hand-side.

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

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

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

Then plugging these values into the right hand side of C8 we get:

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

And then we remove the double brackets using C1 and re-arrange the expressions to obtain (though it looks more like 5 times rather than thrice?) (Actually I'm making a mistake here... I'm noticing that there's an added cross in the final conclusion that I'm not accounting for... hrm, hrm, hrm...)

$\left. {\overline {\, ba \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|\,}}\! \right|$$\left. {\overline {\, r \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ (C9.2) (but not obtained by the above procedure)

**************

Something I like to demonstrate in my postings here is that I'm constantly changing my mind, or noticing mistakes -- my thought is that the finished product never looks like how you get there, and I think that this forum is at its best when sharing our process of thinking in all of its messiness, in all of its faults. So I'm keeping the mistake up above, while working out the correct demonstration here (and simply adding to the post so as not to clog the front page too much) (Also why I'm fine with repeating myself, or going over old ground again)

And I figured out my mistake. The reason I had extra crosses, and needed to perform C1 more than thrice, is that the solution should be--

Let C8's...

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

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

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

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

Then plugging into the right hand side of C8 we obtain:

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

And then we can remove the double crosses, three times (C1 thrice), to obtain:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|ba \,}}\! \right|\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|r \,}}\! \right|$ (C9.12)

Which easily re-arranges to:

$\left. {\overline {\, ba\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|\left. {\overline {\, r\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right|\,}}\! \right|$ (C9.2)

I said that, but I took a hiatus with the USian labor day weekend, and this morning I got stuck on the first transformation of the demonstration of C9. This afternoon I think I worked out the first step, which I'm going to present here because it took me a second to see it.

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, x \,}}\! \right|r \,}}\! \right|\left. {\overline {\, \left. {\overline {\, y \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ = $\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|$

Just looking for patterns I noticed how the only transformation occurs on the final two depth=2 crosses, so I simplified to looking at those final two crosses alone:

$\left. {\overline {\, \left. {\overline {\, x \,}}\! \right|r \,}}\! \right|\left. {\overline {\, \left. {\overline {\, y \,}}\! \right|r \,}}\! \right|$

Which to fit the form of C1, as the text suggests, I set this whole expression = a, and from C1 I add two crosses onto the expression:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, \left. {\overline {\, x \,}}\! \right|r \,}}\! \right|\left. {\overline {\, \left. {\overline {\, y \,}}\! \right|r \,}}\! \right| \,}}\! \right| \,}}\! \right|$

Which gets us to something close to J2. From J2 let p = $\left. {\overline {\, x \,}}\! \right|$ and let q = $\left. {\overline {\, y \,}}\! \right|$, then the expression resembles the left hand side of J2. Converting to the right hand side of J2, but substituting back into the original expression we obtain:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, \left. {\overline {\, x \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, y \,}}\! \right| \,}}\! \right| \,}}\! \right|r \,}}\! \right|$

Which from C1, but this time going from the left hand side to the right hand side to remove two crosses above both X and Y, and plugging this expression back into its place from the original expression we get the first step:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, r \,}}\! \right| \,}}\! \right|\left. {\overline {\, \left. {\overline {\, xy \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ (C9.1)

EDIT:

About an hour later, some random thoughts I'm having while working through these:


One of the things I'm thinking of is how we're showing, from the rules set out so far, how one expression equals another expression. But even though we're using variables I remain uncertain that we have really marked out the domain of expressions such that these are proofs. And further it seems that we could show some other expression could equal one of the other conclusions, like C1. Or, at least, it's not clear to me that these hold as proofs in the same way that numerical expressions with variables have proofs in them, or that other logical systems have proofs in them, like De Morgan's Laws.

To make a list of what we are able to do so far: substitution, the marked state, the unmarked state, variables, equality, and step-wise transformation from J1/J2 (and all demonstrations from J1/J2). The demonstrations are claiming to be a calculus of the marked or unmarked state, but how to delimit that space such that these demonstrations are proofs, in that they hold for the whole space of all expressions? Are there no other expressions other than the marked/unmarked state, or is there a value in-between marked/unmarked? Or is the law of the excluded middle an assumption of the calculus such that we also can conclude that?

One of the differences I'm seeing between this logic and the other two I listed (algebra, Boolean logic) is the lack of negation. There is no negation in this system. There is marked/unmarked, but no negation of the marked/unmarked, which I wonder if that somehow ties into making a consistent system of symbol manipulation. It makes me think that we can think of the unmarked state as more in analogue with 1/1, like it can perpetuate anywhere within an expression in the same way that (1/1) can always be added to either side of an algebraic expression.

Still meandery. One thing these exercises is providing me is a vantage from which to see how logical systems work in the abstract, or at least a vantage to reach for that perspective. ALSO, back of the mind thought, if negation never shows up then perhaps Godel's Incompleteness Theorem will not apply here. (Back of the mind for so many reasons... but I've noticed that the system may not be powerful enough to represent arithmetic, and that's why I have the thought)

Also interesting to note how the proofs of J1/J2 work by showing all cases: under the assumption that p,q, or r is such and such we show the whole expression is equal in all possible cases. This is important, I think, because it may be the case that at some additional variable point we would be unable to check by the method of all cases (which reminds me of truth-tables' check for validity, actually), and so one wonders if a multitude of variables could lead to undecidable, or multiply decidable expressions such that they could lead to both the marked/unmarked state. I think this is the thing that would have to be secured to count these demonstrations as proofs -- we have the demonstration, but is it possible for the demonstration to turn out the opposite value? Like with C1, is it possible to come up with an expression that reduced a-cross-cross to the unmarked state without a from the transformation rules? It's a niggling thought at the back of my mind, and it would be hard to find such an expression, and I may just be completely off base. But hey, sharing the thoughts in the spirit of the thread.

Good stuff.

I especially like the connection between vulnerability and transparency: forthrightness can be a boast, but if you're really at your limit of certainty then it's a good idea to let go of the desire for certainty -- especialy the certainty that you'll win the argument ;).

Transparency, though, is a way to subject yourself to the criticism of philosophy.

Also true.

:)

I'm pleased to see you thinking through the problem. It's something I tend to think about, once and again.

The reason I say the guide is the senses has to do with my time as an activist. I am no longer an activist, so don't trust this as a forever-truth -- one of the things I learned as an activist is that it's too easy to think you have a forever-truth, at least for the philosophically inclined.

But then the world you live within has a way of interrupting thoughts and forever-truths and even certain truths.

****

A dark note -- "violates human rights" in the universal sense is something the United States does. Not in an "on occasion" basis, but of course this will involve a discussion on what human rights are, and when they are worth violating (like war).

But if "human rights" has any meaning outside of the nation, at least, then the U.S. is really good at killing people -- sometimes in the name of war, but sometimes in the name of selling weapons.

My response was a joke, of course, but a joke with a point -- the question needs refinement, else the little modus ponens I offered is an answer: touch is possible because I'm touching something, and in order for me to be able to touch something it must be possible for me to touch it. (Or, if we want to be sticklers about form, I should have started with the implication)

Note that the original question didn't mention anything of physics, but rather was speaking in terms of spaces. After @T Clark pointed out that touching is not when two objects occupy the same space, and @elucid saying it makes sense I read the problem as being resolved.

Still -- the argument I offered wasn't entirely vacuous. For one, you know I'm touching a keyboard because I'm typing out to you (or, at least, you could infer that it likely, given that I have no reason to lie about it even though it's possible I could be using a talk-to-text program), and so the argument points out a relationship between possibility and true statements -- to demonstrate that something is possible all you need is one true statement of the case.

For your argument you'd need to somehow introduce physics and electrons as worthwhile entities in the domain of discourse about touching.

Thanks for the tip -- that would have been a frustrating discovery to make on my own. I'll still check it with C6 first to see, but C9 looked like quite a doozy already.

Sometimes it feels like the demonstrations are purposefully harder than need be -- to get you in the habit of switching out variables for one another. After getting this far the substitution rules made more sense upon reading them -- they were formal statements of what we're doing to check Brown's work that were needed to give meaning to "equality".

And I checked out what comes after C9, and can say that I find it confusing. This is the first appearance of "integration" that I could find by checking the names of each transformation from before, and I don't understand what the part with the series of "is changed to" symbols are arranged means. "The unmarked state is changed to the unmarked state is changed to the unmarked state is equal to the unmarked state is changed to the unmarked state" is the literal translation of the first example, and I don't know what he's getting at with it.

I'll probably at least work out C9 by the time you get back, but probably wait from there. Have a good week!

What are your thoughts? — elucid

I'm touching a keyboard.

If I'm touching a keyboard then it is possible to touch something.

Therefore, it is possible to touch something.

Hell yeah. Finally managed to prove a truth.

C8

$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, br \,}}\! \right|\left. {\overline {\, cr \,}}\! \right| \,}}\! \right|$ (1)

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

Let C1's a = $\left. {\overline {\, br \,}}\! \right|\left. {\overline {\, cr \,}}\! \right|$

Reflect from the right hand side to the left hand side to place two crosses:

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

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

Let J2's:
p = b, q = c, and r = r and collect r from the left hand side of J2 to the right hand side of J2:

$\left. {\overline {\, \left. {\overline {\, a \,}}\! \right|\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, b \,}}\! \right|\left. {\overline {\, c \,}}\! \right| \,}}\! \right|r \,}}\! \right| \,}}\! \right|$ (3)

Call C7:

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

This one took me several guesses. What helped me was to see that the form of C7 has c on both sides of the two separate crosses on its right hand side, and so C7's c must equal (3)'s $\left. {\overline {\, a \,}}\! \right|$ since the conclusion has $\left. {\overline {\, a \,}}\! \right|$ collected into two separate crosses.

Once I saw that then I Let C7's a = $\left. {\overline {\, b \,}}\! \right|\left. {\overline {\, c \,}}\! \right|$, and I transposed (3)'s $\left. {\overline {\, a \,}}\! \right|$ to the right hand side so that it fit the form of C7 more apparently. Then plugging it in sure enough I got C8:

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

If you use the keyboard shortcut Ctrl+C after highlighting it copies to your clipboard and can be pasted here. (At least, that worked for me when I tested it)

I think I figured out C7 this morning. (and the others prior -- I figured out 5 when it clicked that the unmarked state was taking the place of "b" in using C4)

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

By C1 -- $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right| = a$, which we apply to the token $\left. {\overline {\, b \,}}\! \right|$ to obtain

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

Then by J2:

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

Let p = a, q = $\left. {\overline {\, b \,}}\! \right|$ and r = c then distribute from the Right-hand side to the left hand side.

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, ac \,}}\! \right|\left. {\overline {\, \left. {\overline {\, b \,}}\! \right|c \,}}\! \right| \,}}\! \right| \,}}\! \right|$

Then by C1, Let a = $\left. {\overline {\, ac \,}}\! \right|\left. {\overline {\, \left. {\overline {\, b \,}}\! \right|c \,}}\! \right|$ and reflect from the left-hand side to the right hand side to remove the top two crosses to obtain... well, exactly what I just wrote.

End of demonstration for C7.

A modal definition - it's a slab if it has slabbyness in every possible world? Or is it enough for it to have slabbiness in this possible world? Or it's a slab IFF it's width is greater than it's height...

Or it's a slab if the builder places it horizontally, a block if he places it vertically... — Banno

Heh. I thought my response clever but upon inspection, not so much. I think the act of pointing has a place in the definition of "slab!", for the initiate. Or the act of the other builder bringing a slab such that the initiate sees what a slab is without an essence.

How it happens, so I'd maintain @Leontiskos, is not known to the Aristotelian, though the Aristotelian could probably derive a complicated enough description to satisfy the definition of "slab" that fits for all cases thus far seen.

But then I might give you a slab of steak.

Well, that answers that question. Cool. Then I'm tracking!

And actually I had that thought, but then I thought -- well of course we can Let p = whatever we want. It's the form that matters. If I wanted to make sure I was tracking things correctly I could introduce another variable, like s, and let it equal $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ and the form would still work out.

Thanks for working that with me.

OK I've worked my way through to the last step with that help. I think you're right about the presentation being confusing. When I started thinking of the equality symbol as more like the --> symbol that helped, and then when I started looking at the initials like things I could plug into my starting point in a step-wise fashion then that also helped.

The last step I'm struggling with because it seems like I have to Let p = $\left. {\overline {\, \left. {\overline {\, a \,}}\! \right| \,}}\! \right|$ -- is it allowed to switch what p equals in the middle of a demonstration?

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

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

"It's the slabby one. The one with the essence of slab"

Fun idea.

What is…

Being — Mikie

Good question.

Awareness

I presume it's the same for most humans in this respect: the focus of our consciousness -- not in a collective sense, but rather I think most humans have an individual focus on their consciousness, whereas you can still feel a pain even if you're not focused on it.

Consciousness

The phenomenal "feeliness" of the world. The taste of pizza isn't just salty-spicy-sweet, but the particular combination of your bodily make-up and its bodily make-up in conjunction -- if you want a cognitive answer -- or what it tastes like, if you don't.

Thinking

I'm not sure.

Time

Not sure.

Sensation

I think this one can't have an answer. The other topics are more general than "sensation".

Perception

Discrimination.

Mind

Body

Good

All uncertain for me.

Happiness

Justice

Interlinked. Happiness is ataraxia, and ataraxia is only achievable by living in a just society.

Truth

...is embedded in language.

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?

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.

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.

Let's suppose some news article.

This was the article at the top of google news for me this morning: https://www.cnn.com/2023/08/29/weather/tropical-storm-idalia-florida-tuesday/index.html

Here we have some claims that are written. Some of them have already happened. Some of them are predicted to happen, like the peak storm surge forecast. Now if you follow my original suggestion you couldn't really verify any of what's written down unless you're in Florida. Or you could verify it if you believe that the published weather reports tend to report true things that have happened, but are a little less reliable when it comes to what is going to happen.

Notice how what we're reading is an important part of judging whether we should believe it or not. This is from CNN, it is a weather report, they have a history of having accurate weather reports mostly because they rely upon government agencies and trained individuals. Since it's the weather, rather than an election, there's less of a reason to lie or generate an alternative story to what the other news organizations are saying.

It's that latter bit -- when news is political, or propagandized -- which undermines trust. Or, in some cases, reinforces trust if they're selling the truth you want to hear. But that kind of truth you want to hear isn't usually related to the senses, is it?

And that's where I'd say we have a guide to choosing what to believe.

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.

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

...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.

In a society where govenments try to tell you what is true and raise you into believing what you believe, in a world that is ever more dividing, when we're looking at news or whatever is going on around us, how do we know what to believe in? — Hailey

I'd say start with believing your senses.

But this is a beginning, and a guess. The trouble you raise is we do not know what to believe in, but we do know that there's a fair bit of false beliefs which seem true. In fact I'd go further and say that we don't know that it's the governments, or any one culprit, which is the culprit in spreading false beliefs. And I'd go further to note that I couldn't answer the question for you -- how do you know I'm not from the government, spreading false beliefs about believing your senses first? The government could be an empiricist, in this silly universe I'm proposing, which wants its people to believe that knowledge comes from the senses.

But then remember the suggestion -- you don't have to believe me. You can believe your senses, and work from there, even if you're following the empiricist's shadow-government ;)

Fair point.

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

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.

Yup.

I know @SophistiCat added the SEP article, but it's worth noting the formalization of supervenience in this thread I think --

A weakly supervenes on B if and only if necessarily, if anything x has some property F in A, then there is at least one property G in B such that x has G, and everything that has G has F, i.e., iff

□∀x∀F∈A[Fx → ∃G∈B(Gx & ∀y(Gy → Fy))]
A strongly supervenes on B if and only if necessarily, if anything x has some property F in A, then there is at least one property G in B such x has G, and necessarily everything that has G has F, i.e., iff

□∀x∀F∈A[Fx → ∃G∈B(Gx & □∀y(Gy → Fy))]
(Kim 1984)

Which still is hard for me to read through.

Heh, yes. Undoubtedly.

The obscure and the strange is one of those things that just nabs my attention. Also I had some notions back when learning baby logic that this book seems to run parallel to. Notions which after writing them down I threw out because they seemed nonsensical, but hey -- there was something interesting about how the calculus managed to deal with the notion of the philosophy of philosophy as an unmarked state rather than a marked state.

Originally I wanted to actually put the fourth cannon example underneath a bracket of its own, but I found it difficult to stack multiple bracketed maths within a single bracketed math so there's a bit of a limit there. The only difference, though, would have been that there would have been another step of elimination where the deepest space's value for a was the unmarked state rather than the marked state.

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.

There's something similar to this and using nested sets as representatives of numbers, I think. But then the value isn't numerical, but is rather the marked or unmarked state at its simplest. The first theorem of Chapter 4 points out that these initials are a starting point for building more complicated arrangements and the simple arithmetic of the crosses is what's needed to make sense of the calculus of the crosses.



I'm going to try and work out the proof here by arbitrarily using this arrangement as "a" --

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, s(sub(d)) \,}}\! \right| \left. {\overline {\, * \,}}\! \right| \,}}\! \right| \,}}\! \right|$

s is contained in a cross.

All the crosses in which s(sub(d)) is within are empty other than the space in which s(sub(d)) is in. ("*" counting as the unmarked space)

The arrangement chosen uses both cases --

Case 1 -- there are two crosses that are empty underneath a cross next to one another such that s(sub(d)) could have been in either cross. They're equivalently deep.

Case 2 -- the crosses surrounding the two deepest crosses are alone within another cross

So using the steps of condensation and elimination:

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, s(sub(d)) \,}}\! \right| \left. {\overline {\, * \,}}\! \right| \,}}\! \right| \,}}\! \right|$ --> $\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, s(sub(d)) \,}}\! \right| \,}}\! \right| \,}}\! \right|$ Condensation

$\left. {\overline {\, \left. {\overline {\, \left. {\overline {\, s(sub(d)) \,}}\! \right| \,}}\! \right| \,}}\! \right|$ --> $\left. {\overline {\, s(sub(d)) \,}}\! \right|$ Elimination

And by the definition of Expression from chapter 1: "Call any arrangement intended as an indicator an expression" we can draw the conclusion that any arrangement of a finite number of crosses can be taken as the form of an expression. (since we're indicating the marked or the unmarked state)

The argument could also be read syllogistically, in which case 'anything' makes more sense:

All appearances are known mediately
No first-person actions are known mediately
Therefore, no first-person actions are appearances

Of course this is also valid. — Leontiskos

True! And that'd be more appropriate for the source material.