A recursive Dehn function yields a solution to the word problem

Let $n=\mathrm{\mid }w\mathrm{\mid }$ and let $M:=\max (\{0\}\cup \{\mathrm{\mid }r\mathrm{\mid }:r\in R\})$, so $M=0$ when $R=\mathrm{\varnothing }$. Because ${\delta }_{\mathcal{P}}$ is recursive, one can compute the bound $B={\delta }_{\mathcal{P}}(n)$. If $w$ is null and $B>0$, [L2] gives a relator expression of area at most $B$; choose one of minimal area and, among those, with minimal total conjugator length. Then each conjugator may be taken of length at most $n+BM$: otherwise an initial segment that never survives the free reduction to $w$ could be shortened, contradicting the chosen minimality.

Step 1.1 reduces the search for a certificate of triviality to finitely many possibilities: at most $B$ relator factors, each chosen from the finite set $R^{\pm 1}$, and, when $B>0$, each conjugator drawn from the finite set of words of length at most $n+BM$. When $B=0$, the only candidate certificate is the empty product. Enumerate these possibilities and use [L3] to test in the free group whether any of them equals $w$.

If the search in step 2.1 succeeds, then [L1] says $w$ is trivial. If it fails, then no relator expression of area at most $B$ exists, so by the definition of the Dehn function $w$ cannot be null. Thus step 2.1 decides whether $w{=}_{\mathcal{P}}1$.

Therefore a recursive Dehn function gives a solution to the word problem.