Hyperbolic groups have solvable word problem

Fix the finite alphabet $S^{\pm 1}$ and finite relator list $R$ supplied by [L1]. Freely reduce the input word. At each stage enumerate its finitely many subwords and the finitely many cyclic conjugates of relators in $R^{\pm 1}$; if a subword is longer than half of one of those relators, replace it by the inverse of the complementary piece and freely reduce again. Choose the first match in a fixed finite ordering. Each replacement preserves the group element and strictly decreases length, so this effective procedure terminates.

If the algorithm stops at the empty word, then $w=1$ in $G$. Conversely, if $w=1$ in $G$ and the current freely reduced word is nonempty, [L1] supplies another enumerated shortening move, so the procedure cannot stop there. Thus it decides whether $w$ represents the identity.