Inquiry Into Inquiry

Re: Logical Graphs • Formal Development
Re: Laws of Form • Lyle Anderson

LA:

What does it mean to assign a label or name to a node of the Logical Graph?

In LoF, the variables of the algebra represent unknown expressions of the arithmetic.  There are two tokens in the expressions for Logical Graphs, the node and the edge.  You assign different symbols to the naked node of the outside and the node representing the inside, since the edge between them represents the boundary of a distinction.  When you put a letter “a” next to the naked node, what does that mean?  If “a” represents another Logical Graph of uncertain arrangement, then how is it attached to the naked node?

Dear Lyle,

By now we’ve seen quite a few ways to represent Peirce’s logical graphs and Spencer Brown’s formal arrangements in various styles of formal languages and concrete media.  A fairly detailed discussion of how to translate among the more common representations we’ve been using, along with those I found useful in computing logical graphs, was given in the post linked below and serialized in the fourteen posts which follow it.

• Logical Graphs • First Impressions

As a general consideration, we need to remember one of the first lessons we learned in geometry, and never confuse the drawing, the representation, with the mathematical object it represents.  Despite their name, “graphs” in the sense of mathematical graph theory are mathematical objects, not to be found on the page, screen, or in the state of any concrete system, whether cognitive or computational.  The same goes for Spencer Brown’s formal arrangements.  Among other things, that is one of the reasons Peirce’s pragmatic semiotics is so critical to understanding logical graphs, laws of form, and logic in general.

Regards.

Jon

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Re: Logical Graphs • Formal Development
Re: Laws of Form • Armahedi Mahzar

AM:

GSB took J1 : (a(a)) =   as the first algebraic primitive and the second one is transposition so he only need only 2 primitives for the primary algebra.

Reflexion ((a)) = a is proven without using Cancellation (( )) =   .

In fact, he can prove cancellation from C1 reflexion.

Condensation ( )( ) = ( ) can be derived from C4 iteration.

So, his algebra is simpler from your Cactus Calculus.

Dear Arma,

I had a feeling we’ve discussed this before, and probably in a lot more detail than I have time for at the moment, so I hunted up the previous discussion — turns out it was on the old Yahoo Group — there’s a copy of that below for whatever memory‑jogging it may be worth.

To my way of thinking, what you say about reducing the primary arithmetic to the primary algebra shows a lack of appreciation for the fundamental nature of that distinction.  Indeed, the recognition and clarification of that distinction is one of the most important upgrades Spencer Brown added to Peirce’s initial systems of logical graphs.

As far as the other score goes, the advantages of handling label changes and structure changes separately in one’s syntactic operations is just one of those things I learned in the hard knocks way of programming theorem provers for logical graphs, and I all I can do is keep recommending it on that account.

Regards,

Jon

From: Jon Awbrey
To:   Yahoo Laws Of Form
Date: 8/15/2017, 2:10 PM
Re:   Peirce's Law

Arma, All ...

Re: https://intersci.ss.uci.edu/wiki/index.php/Logical_graph#Axioms

Let me try a few ascii graphics to see how this site treats them.

It's clear that the two systems are equivalent, since we have:

 a   a
 o---o
 |
 @
 =======J1′ [delete]
 o---o
 |
 @
 =======I2  [cancel]
 @
 =======QED J1

For my part, I am less concerned with small differences in the lengths
of proofs than I am with other factors.  It's hard for me to remember
all the reasons for decisions I made 40 years ago — as a general rule,
Peirce's way of looking at the relation between mathematics and logic
was and still is very influential and the other main impact came from
the nuts @ and bolts | requirements of computational representation.

But looking at the choice with present eyes, I think I would continue
to prefer the I1, I2, J1′, J2 system over the alternative simply for
the fact that it treats two different types of operation separately,
namely, changes in formal or graphical structure versus changes in
the occurrence or placement of variables.

Regards,

Jon

On 8/14/2017 11:01 PM, armahedi@yahoo.com [lawsofform] wrote:
> Hi Jon
>
> With your answer
>     'Recall that I am using “p(p)=( )” for my J1.
>      I can go back to calling it J1′ if need be.',
> I realize that your axiom system is different from 
> Brownian Primary Algebra.
>
> Here is your axioms
> I1   ()()=()
> I2   (())=
> J1'  (a)a=( )
> J2   ((ab)(ac))=a((b)c)
> and these are Spencer-Brown axioms
> J1   ((a)a)=
> J2   ((ab)(ac))=a((b)(c))
>
> From these axioms you derived three theorems which are 
> identical to the first Spencer-Brown Primary Algebra 
> consequences: Reflection, Generation and Integration.
>
> Here are your proofs in parentheses notation with 
> my critical observations.
>
> You prove Reflection ((a))=a in 8 steps
> ((a))
> =((a))((                 ))           I2
> =((a))((     (a))(     a)))        J1'
> =     ((((a))(a))(((a))a)))      J2
> =     ((        )(((a))a)))         J1'
> =     ((( a ) a )(((a))a)))       J1'
> =    a((( a )   )(((a)) )))        J2
> =    a((                 ))            J1'
> =    a                                  I2
> while in Brownian Primary Algebra the proof is only 6 steps
> because I2+J1'=J1'+I2=J1
>
> You proved Generation (a)b=(ab)b in 5 steps
> (a)b
> =(((a)b))          C1
> =(((a)b))((   ))   I2
> =(((a)b))((b)b)) J1'
> =((a))((b)))b      J2
> =(ab)b               C1
> while in Brownian Primary Algebra the proof is only 4 steps
> because I2+J1'=J1.
>
> You proved Integration a( )=( ) in 2 steps
> a( )
> =a(a)            C2
> =( )               J1'
> While in Brownian Primary Algebra the proof is 3 steps
> because it needs another step C1
> a( )
> =a(a)            C2
> =((a(a)))        C1
> =( )                J1
>
> Comparing the proofs it seems to me that Brownian proof is better 
> because it does not need the arithmetical primitives I1 and I2.
>
> The beauty is that the arithmetical primitives become derived 
> theorems in Brownian Primary Algebra. I2 is J1 with x=   dan 
> I1 is the 4th consequence xx=x with x=( ). So Brownian Primary 
> algebra is more economic than your algebra.
>
> However, it is just my subjective aesthetical preference, your 
> logical superstructure (cactus calculus, differential logic and 
> dynamical logic) is still intact fortunately.
>
> Please correct me if I am wrong.
> Thanks
> Arma

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