MATH.6:5 - Archetypal Grounding

MATH.6:5.1 - A relation that cannot serve as the proposed equivalence

The claim is that every reflexive, symmetric relation is transitive. On X={a,b,c}, assign the following relation; 1 means the pair is in the relation.

The diagonal establishes reflexivity. Matching entries across the diagonal establish symmetry. But a R b and b R c hold while a R c fails. This is a countermodel with the required three witnesses.

With at most two elements, every reflexive symmetric relation is transitive: in a two-link sequence x R y R z, either x=z, or one adjacent pair is an equal pair and the other link already supplies x R z. The third element makes the failure possible.

If the work needs the smallest equivalence relation containing this R, its transitive closure adds the missing pairs and gives the single class {a,b,c}. It answers whether elements are connected through R-links. A question about the original direct relation still needs the original table. For classes on which further operations must be defined, continue with MATH.2’s operation-preservation condition.

MATH.6:5.2 - A left inverse and the elements it does not recover

Consider the claim: if f:A→B has g:B→A with g(f(a))=a for every a, then f(g(b))=b for every b.

Choose A={u}, B={0,1}, f(u)=0, and g(0)=g(1)=u. Both maps are total. The premise holds at the only element of A, but f(g(1))=0. The claim fails because the left-inverse condition constrains recovery of source elements while B also contains an element outside the image of f.

Now require A and B to be the same finite set. The premise makes f injective: equality f(a)=f(a') gives a=a' after applying g. An injective self-map of a finite set is surjective. Write any b as f(a); then f(g(b))=f(g(f(a)))=f(a)=b. This supplies a proof for the changed domain.

For the same infinite set N={0,1,2,...}, take f(n)=n+1, g(0)=0, and g(n+1)=n. Then g(f(n))=n for every n, but f(g(0))=1. Every finite self-map search can miss this failure because the finite claim is true. The symbolic construction locates the lost premise: finiteness supplied the step from injection to surjection.

A receiving construction can retain the original left inverse for source recovery, require surjectivity for recovery of every target element, or restrict the target to the image. Which result is useful depends on the proposed representation.

MATH.6:5.3 - A separate answer for each input and one answer for all inputs

Let X=Y={0,1} and let R(x,y) mean x≠y. For every x there is a y satisfying R: choose y=1-x. The proposed stronger conclusion is that one y works for every x.

To defeat that conclusion, take any proposed y and choose x=y. Then R fails. This gives the required argument for both possible choices of y. It does not replace the premise’s input-dependent choice with a uniform one.

The useful result can instead be a function h(x)=1-x, satisfying R(x,h(x)) for every x. MATH.4 supplies inductive witness construction when a comparable task has finite inductively formed inputs and suitable base and constructor clauses. The countermodel identifies which input dependence the requested result must retain.