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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29
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The word problem for a finitely generated free group is solvable by free reduction

Statement

Let F(X) be a free group on a finite set X. The word problem in F(X) is solvable: a word on XX1 represents the identity if and only if its free reduction is the empty word.

Facts & Assumptions

Given: A finite basis X and a word w on XX1.

[L1]

A word is reduced exactly when no adjacent inverse pair remains, and free reduction is obtained by repeatedly deleting such pairs. (Words in an alphabet with formal inverses, elementary cancellation, and reduced words)

[L2]

Every class in the reduced-word model of the free group contains exactly one reduced word. (Every class in W(X)/ contains exactly one reduced word)

Proof

technique · direct
1.1

Repeatedly apply the elementary cancellations of [L1] until no adjacent inverse pair remains. Because each cancellation shortens the word by two letters, the process halts after finitely many steps with a reduced word r.

L1given
2.1

The free group element represented by w is the same as that represented by r, because step 1.1 used only the free-equivalence moves of [L1]. By [L2], the identity class has exactly one reduced representative, namely the empty word. Therefore w represents the identity if and only if r is empty.

L1L2step 1.1
3.1

The halting free-reduction procedure of step 1.1 therefore decides the word problem in F(X).

step 2.1algebra

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