WEBVTT

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We now construct a different finite object.

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Its vertices are indices of selected tests, not the valuations themselves.

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It records which groups of tests can all be true at one

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valuation.

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Label the carrier type before drawing a triangle.

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Select three indexed tests.

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Test one is X.

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Test two is Y.

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Test three is the negation of the whole conjunction X and Y.

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It is not a new variable Z.

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Their regions are A and B; B and C; and A, C

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and D.

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Each test is satisfiable, so all three indices become vertices.

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The first two tests can be jointly true at B.

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Draw the edge between their test indices and retain B as its

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witness.

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This edge concerns two available formulas, not a physical connection between state

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cards.

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X is compatible with the third test at A, and Y with

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it at C.

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Each pair therefore has an edge.

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The three edges form the boundary of a triangle, with a different

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valuation witnessing each pair.

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Requiring all three tests together is contradictory.

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No valuation witnesses their triple intersection.

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We therefore do not fill the triangle.

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Three pairwise witnesses do not supply a single witness satisfying all three

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formulas at once.

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A simplicial complex contains every subset of any included simplex.

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Joint satisfiability has this property: any subset of a jointly satisfiable test

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group is also satisfiable.

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This construction is the observation-side relation nerve of the satisfaction table.

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Following the three edges gives a loop, with no triangular face filling

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it.

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In this elementary example the first homology detects one independent one-dimensional

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hole.

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The companion note gives a small algebraic check; no general homology calculation

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is required here.

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The nerve is connected, but the valuation observation space for the same

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menu has components B and A, C and D.

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These objects even disagree about connectedness.

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Do not infer the topology or homology of one from the other'

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s drawing.

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The observation topology is generated from selected true-row regions by unions

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and finite intersections.

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Symmetric distance adds weights at differing slots.

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Directed loss adds weights only at true-to-false changes from the

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source.

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We can check each construction separately when a new test is added.

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A tautology is true at every valuation.

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Its region was already the whole open carrier, so adding it changes

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no generated topology.

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Its constant coordinate adds no disagreement or directed loss, so both signature

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distances also remain unchanged.

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The tautology can accompany every previously satisfiable test group.

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In the nerve it adds a vertex and a cone over the

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old complex, filling the old loop through new faces.

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The resulting cone has no reduced homology, unlike the triangle boundary.

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Does preserving the observation topology and both distance functions force the indexed

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test nerve to preserve its hole?

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State what the tautology changes and what it leaves fixed.

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The topology and distances remain the same, while the nerve becomes a

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cone and loses its reduced homology.

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Therefore the nerve's homology is not determined by the observation topology.

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Any invariant claim needs additional restrictions and a stated presentation policy.

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The nerve usefully records higher-order compatibility of indexed tests.

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Its value does not require pretending it is a topological invariant of

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the valuation space.

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State its vertices, simplices, presentation and intended comparison before interpreting a hole.