TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Fixed finite semigroup has undecidable terminal equality

Statement

There is one fixed finite positive semigroup presentation, with one state letter on each side of every relation, for which deciding $C (w) = q$ on encoded initial words is impossible. In particular both unrestricted equality to $q$ and unrestricted word equality in this same semigroup are undecidable.

Facts & Assumptions

Given: The fixed machine and semigroup already constructed; all inputs $w$ are binary finite words.

[F1]

The fixed $T$ stops on $v (w)$ exactly for $w \in H A L T_{T M}$ , an undecidable language. (Normalized fixed halting machine for boone simulation)

[F2]

For this fixed $T$ , the construction gives one finite positive semigroup presentation $Γ$ , its fixed terminal letter $q$ , and the computable initial words $C (w) = h q_{s t a r t} v (w) h$ . Every defining relation has exactly one state letter on each side. (Boone machine semigroup and augmented configurations)

[F3]

$C (w) = q$ in $Γ$ exactly when $T$ stops on $v (w)$ . (Boone semigroup histories detect halting)

Proof

1.1

Fix the table of $T$ and hence the finite relation list of $Γ$ once. On input $w$ , test emptiness, write the encoding of $s_{0}$ in the empty case and otherwise copy $w$ , and prefix $h q_{s t a r t}$ and suffix $h$ . This computes $C (w)$ by finite-symbol scans; no part of the presentation depends on $w$ .

F1F2construct

2.1

A decider answering whether this $C (w)$ equals $q$ would answer yes exactly when $T$ stops on $v (w)$ , hence exactly when $w \in H A L T_{T M}$ . Running it after the terminating encoding in step 1.1 would be a total decider of that language, contradicting [F1].

F1F3step 1.1

3.1

An unrestricted terminal-equality decider could be run on these initial words; an unrestricted two-word equality decider could be run on $(C (w), q)$ . Each would give the impossible decider in step 2.1. The reductions include $C (ε) = h q_{s t a r t} s_{0} h$ .

step 1.1step 2.1∎

Source locator

Rotman, printed pp.428–430, Theorem 12.5 and Corollary 12.6. The direct composition above fixes the presentation before varying the input.

Depends on

Used by

Novikov boone undecidability of the word problem Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Joseph J. Rotman, An Introduction to the Theory of Groups, Chapter 12, pp.428–430, Theorem 12.5 and Corollary 12.6 (standard reference, not scraped)