Two finite presentations define isomorphic groups if and only if a finite sequence of Tietze transformations and inverses connects them

If a finite sequence of Tietze transformations connects $\mathcal P$ to $\mathcal Q$, composing the isomorphisms supplied by [L1] along that sequence gives an isomorphism between the groups they present; the zero-move case is the identity isomorphism.

Conversely, fix an isomorphism $\phi:G_{\mathcal P}\to G_{\mathcal Q}$. If $X\cap Y\neq\varnothing$, first apply one renaming transformation to $\mathcal Q$, replacing $Y$ by a finite set disjoint from $X$, and compose $\phi$ with the induced isomorphism. Write $\mathcal Q=\langle Y\mid S\rangle$ for this renamed presentation; after connecting $\mathcal P$ to it, the inverse renaming returns to the original $\mathcal Q$. By surjectivity in [L3], for each $x\in X$ choose a word $v_x(Y)$ representing $\phi([x])$, and for each $y\in Y$ choose a word $w_y(X)$ representing $\phi^{-1}([y])$; only the finitely many choices indexed by $X\cup Y$ are made, successively by [L4].

Starting from $\mathcal P$, add every $y\in Y$ by the dictionary relation $d_y:=y^{-1}w_y(X)$. In the resulting presentation, $v_x(Y)$ and $x$ represent the same element because eliminating the new letters sends $v_x(Y)$ to the representative of $\phi^{-1}(\phi([x]))=[x]$; hence [L2] makes $d_x:=x^{-1}v_x(Y)$ a redundant relator. Add every $d_x$, and then add every $s\in S$, which is redundant because eliminating $Y$ evaluates it as $\phi^{-1}([s])=1$. This is a finite legal sequence from $\mathcal P$ to $\mathcal C:=\langle X\cup Y\mid R\cup S\cup\{d_x:x\in X\}\cup\{d_y:y\in Y\}\rangle$.

Starting from $\mathcal Q$, add every $x\in X$ by the dictionary relation $d_x=x^{-1}v_x(Y)$. In that presentation, $w_y(X)$ and $y$ represent the same element because eliminating $X$ evaluates $w_y(X)$ as $\phi(\phi^{-1}([y]))=[y]$, so [L2] licenses adding every $d_y$; each $r\in R$ is then redundant because eliminating $X$ evaluates it as $\phi([r])=1$. Thus another finite legal sequence runs from $\mathcal Q$ to the same presentation $\mathcal C$.

Reverse the sequence of step 2.2. Each relator is deleted in reverse order while the earlier relators that originally forced it remain, so the redundant-relator inverse condition is satisfied. Each dictionary generator is deleted only after every later-added relator containing it has been removed, leaving that generator in its dictionary relation alone, so the dictionary inverse condition is satisfied. Hence there is a finite legal sequence from $\mathcal C$ to the renamed $\mathcal Q$. Concatenate it with step 2.1 and, when step 1.2 used a renaming, append that renaming's legal inverse. The resulting finite sequence connects the original $\mathcal P$ to the original $\mathcal Q$.

Step 1.1 proves the forward implication and steps 1.2 through 3.1 construct the reverse implication, so the two conditions are equivalent.