Substructures

Hello, Bonjour, Salut

The dytran-threetran technology
We want to show that if a binary relation is trasitive (ab&bc-->ac), it is also "3-transitive", meaning ab&bc&cd-->ad.

1. define the dytran

 * Let M be a set and D a part of M × M having the property :
 * D1: for all x,y in M, xy & yz → xz (2-transitivity)

2. define the threetran and the counterthreetran

 * Let T be the dytran D above satisfying an extra property :
 * D1: for all x,y in M, xy & yz → xz (2-transitivity, as in the dytran )
 * T1: for all x, y, z, t in M, xy & yz & zt → xt (the extra 3-transitivity)

A dytran with the negation of T1, say CT1, is called counter-threetran.

3. remember the target : all dytrans are threetrans

 * By construction, all threetrans and all counter-threetrans are dytrans. A dytran is either a trheetran, or a counter-threetran. We want to show that all threetrans are dytrans or, equivalenttlly, that there are no counter -threetrans.

4. define sub-counter-threetrans
Given a subset N of M, CT restricted to N is said to be a sub-counter-threetran of CT.

5. remark that all sub-counter-threetrans are counter-threetrans

 * Remark : all sub-counter-threetrans are counter-threetrans ; their sub-dytrans are dytrans.

6. describe the counter-threetran

 * Suppose by r.a.a. one has listed a counter-threetran. In this hyphotetic example, D1 and CT1 hold. This counter-threetran contains a, b, c, d such that ab, bc, cd but not ad.

7. build the smaller sub-counter-threetran

 * CT' = CT / {a, b, c, d}, defined by the constants of CT1, is a sub-counter-threetran. By the remark above, it is also a counter-threetran, i.e. a dytran that is not a threetran.

8. the verification of a small number of sub-dytrans

 * It suffice to verify by listing that all small dytrans not richer than { 1, 2, 3, 4 } are threetrans.

9. conclude : all dytrans are threetrans

 * If the hypothetic example were true, one should find a small counter-example, built from the original. The explicit verification of all cases showed that this is not the case. In conclusion, all dytrans are threetrans.

10. explain to the reader what has been proved

 * We have proved that a transitive relation is also 3-transitive : ab, bc, cd → ad. Normally one shoud aplly twice the D1 axiom to deduce T1. On the other hand, teacher Boole showed that thinking has a mechanical component in its core. Why think when is so mechanical ?

The mathematic of cycles
I have an example of actually using a cyc.

McKay's proof of Cauchy's theorem (1959, AMM) uses a p-tuples of elements of G

( x1, x2,....,xp ) and x1.x2......xp = 1

Well, this is a cyc ! (cyclic order does not matter) as usual, when species are present it turns spectaculary.

Huntington's cyclic order (1914)

 * Take as dytran a subset of K×K×K together with the axioms.
 * Take as threetran a dytran together with a theorem, e.g.
 * T1) AXB & BYC & CZA → XYZ.
 * the counter threetran would present {a, b, c, x, y, z} such that axb, byc, cza but not xyz (CT1).
 * Take the sub-counter threetran CT /{a, b, c, x, y, z} with the induced structure, that is also a counter-threetran.
 * Check all counter-threetrans with no more than 6 elements.
 * there are no small counter-threetrans, hence there are no counter-threetrans.
 * The theorem must be true if it is true on cyclic orders with no more than 6 elements.

limits
• The method is limited to the axioms like:
 * For all x in N, there is x' in (the same N) that dosomething.
 * The existential function must be mapped on different sets.


 * example Putnam 1965 A-4 ; applying both rule, it occurs :
 * for each boy there is a "succesor", b≠g-b'
 * for each girls there is a "succesor" g-b≠g'