Confluence to show equivalent terms have one common reduct
$S \leftrightarrow^* T$ does not mean that $S \rightarrow^* T$ and $T \rightarrow^* S$! It means that there is a chain of reductions $S = S_0 \rightleftharpoons_1 S_1 \rightleftharpoons_2 S_2 \rightleftharpoons_3 \cdots \rightleftharpoons_n S_n = T$ where each of the $\rightleftharpoons_i$ might be either $\rightarrow$ or $\leftarrow$. The directions can alternate any number of times.

The confluence property does imply generically that if $S \leftrightarrow^* T$ then there exists $W$ such that $S \rightarrow^* W \leftarrow^* T$, but it takes a bit more work to show it. You can combine reductions in $S \leftrightarrow^* T$ to group consecutive reductions in the same direction together: $S = T_0 \leftarrow^* U_1 \rightarrow^* T_1 \leftarrow^* U_2 \rightarrow^* T_2 \leftarrow^* \cdots \leftarrow^* U_n \rightarrow^* T_n = T$. By confluence on $T_{n-1} \leftarrow^* U_n \rightarrow^* T_n$, there exists $V_n$ such that $T_{n-1} \rightarrow^* V_n \leftarrow^* T_n$. So we have $T_{n-2} \leftarrow^* U_{n-1} \rightarrow^* T_{n-1} \rightarrow^* V_n \leftarrow^* T_n$.

$$
\begin{matrix}
    &amp;            &amp; U_1 &amp;            &amp;     &amp;        &amp;         &amp;            &amp; U_{n-1} &amp;            &amp;     &amp;            &amp; U_n &amp;            &amp;     \\
    &amp; _*\swarrow &amp;     &amp; \searrow^* &amp;     &amp; \cdots &amp;         &amp; _*\swarrow &amp;     &amp; \searrow^* &amp;     &amp; _*\swarrow &amp;     &amp; \searrow^* &amp;     \\
T_0 &amp;            &amp;     &amp;            &amp; T_1 &amp;        &amp; T_{n-2} &amp;            &amp;     &amp;            &amp; T_{n-1} &amp;            &amp;     &amp;            &amp; T_n \\
    &amp;            &amp;     &amp;            &amp;     &amp;        &amp;         &amp;            &amp;     &amp; 
           &amp;     &amp; \color{green}{\searrow^*} &amp;     &amp; \color{green}{_*\swarrow} &amp;     \\
    &amp;            &amp;     &amp;            &amp;     &amp;        &amp;         &amp;            &amp;         &amp;            &amp;     &amp;            &amp; \color{green}{V_n} &amp;            &amp;     \\
\end{matrix}
$$

Now apply the confluence property again to $T_{n-2} \leftarrow^* U_{n-1} \rightarrow^* V_n$, getting $V_{n-1}$ such that $T_{n-2} \rightarrow^* V_{n-1} \leftarrow^* V_n \leftarrow^* T_n$. Repeat until you get $W = V_1$ such that $T_0 \rightarrow^* V_1 \leftarrow^* T_n$. Formally, this is a proof by induction on the number of alternations of directions in the chain $S \leftrightarrow^* T$. Confluence lets you remove one alternation at a time. In more colorful language, confluence lets you “pop a crease”; each time you pop a crease, you join two depressions together, and if you keep doing that, you eventually get a single depression.