What makes you think that a guy (i.e. Lacan) who didn't know the difference between an irrational number and an imanginary number would know anything about symbolic logic?
Jim Farmelant
On Wed, 03 Nov 1999 14:53:46 -0800 (PST) Chuck Grimes <cgrimes at tsoft.com>
writes:
>
>Ken,
>
>This will be fun. You might recognize some of this from a long ago
>post I did to Bad on the link between structuralism and group
>theory--a link created by Levi-Strauss and Jean Piaget--that was
>originally suggested by Ernst Cassirer.
>
>You wrote out the following truth table for a model of scientific
>truth:
>
>
> A B A and B
> ---------------------------
> T T T
> T F F
> F T F
> F F F
>
>
>This is then jutxaposed to Lacan's system:
>
>Lacan adopts the term "right polariation" for clockwise directions and
>"left polarization" for counterclockwise directions, terms used to
>describe the "orientation" of knots like his Borromean knot.
>
> Real (R)
>
> Imaginary (I) Symbolic (S)
>
>Lacan never provides a detailed account of all the discourses covered
>by this particular combinatory. He mentions only two: religious
>discourses: RSI and psychoanalytic discourse: IRS.
>
>----------
>
>Okay, both tables are isomorphic to finite abstract groups.
>
>The first is build from the cyclic product group on two elements: 2 x
>2 x 2 x 2 = 16. One representative element is (1010) and another is
>(0101). The total number of permutations are 16 and the elements of
>the group are:
>
>0. (0000) 1. (0001) 2. (0010) 3. (0011) 4. (0100) 5. (0101) 6. (0110)
>7. (0111) 8. (1000) 9. (1001) 10. (1010) 11. (1011) 12. (1100)
>13. (1101) 14. (1110) 15. (1111)
>
>These are isomorphic to the binary operators on two elements:
>
>F0 = 0, null
>F1 = x . y and, x and y
>F2 = xy' = x/y inhibition, x but not y
>F3 = x the function is equal to x
>F4 = x'y = y/x inhibition, y but not x
>F5 = y the function is equal to y
>F6 = x'y + xy' = x o y exclusive-or, x or y but not both
>F7 = x + y or, x or y
>F8 = (x + y)' = x v y nor, not-or
>F9 = x'y' + xy = (x = y) equivalence, x equals y
>F10 = y' complement, not y
>F11 = x + y' = x u y implication, if y then x
>F12 = x' complement, not x
>F13 = x' + y = x n y implication, if x then y
>F14 = (xy)' = x ^ y nand, not-and
>F15 = 1 identity
>
>
>This amounts to the entire logic system available to the
>computer. Take the elements like (0000)-(1111) and write them out as
>the top row and first column of a multiplication table. Use the rules
>F0-F15 as the rule of combination and generate sixteen tables of 256
>elements each. (I know, fat chance, chuck).
>
>What emerges from that exercise is a demostration that the only two
>symmetric operators are F6 and F9, and these generate the only tables
>with non-repeating rows and columns. This means that the only group
>theoretic operators are F6 and F9. The diagonal elements in a 16 x 16
>array using F6 are the null element (0000), and therefore the operator
>is equivalent to addition and group is Abelian. Using F9, the
>diagonal
>elements are (1111) or the identity element and the operator is
>equivalent to multiplication. The other operators are set-theoretic
>and generate asymetric and/or repeating tables of elements.
>
>Experiment with truth tables using the operators (F0-F15) so
>that T and F are mapped to x and y and see the results.
>
>Now, this is the part that would really piss-off Lacan. His system of
>three elements (r,i,s) is isomorphic to a dihedral group of order 3,
>the equilateral triangle. You gave (r,s,i) and (i,r,s).
>
>Take an equilateral triange and label the vertices anything, say
>(a,b,c). There will six different permutations, or 1 x 2 x 3 = 6 or 3!
>The permutations that you gave are two of three possible rotations
>that form a subgroup, or the elements cycled in order. The remaining
>permutations are called the reflections and these will appear as
>interchanges between two of the three elements. The reflections are
>found by flipping the triangle to exchange the position of two
>vertices
>
>The collection of six permutations is called S3, the symmetric group
>on three elements. It is isomorphic to an automorphic subgroup of the
>cyclic two group laid out above.
>
>To show the isomorphism, draw the same triangle and connect the
>vertices to a center point. Label the vertices (01), (11), (10) with
>the center as (00). Now write out the path from the center to one
>vertex say, (00 10). Write the same path going the other way
>(10 00). Each of the other possible paths going in both directions
>will
>generate all the other sixteen elements, just as they appear
>above. This happens because you are taking a combination of two
>elements, two at a time and combining them with a binary
>operator. That is you are doing the same thing as before.
>
>In other words under some linear transformation, Lacan's group is
>nothing more than a subgroup of the above, just as the first truth
>table is a subgroup! If you arrange the labeling in a certain way, you
>will find that all three are merely different representations of the
>same group! That is the first truth table, an automorphism of
>the symmetric operators on two elements, and the last group S3.
>
>And it only gets worse. This little three group is the logical model
>for an incident geometry and forms the elementary linear space group
>(three non-colinear points = a plane).
>
>I don't know what matheme Lacan thought he was communicating, but this
>is the one he was using.
>
>Chuck Grimes
>
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