Step 1. Denote the whole expression by $E$ , and let $n=0$ and the set of translations $\{T_{i}\}$ be initially empty.

Step 2. If $E$ is a sentence symbol or a symbol $T_{n}$ then this is the root of the tree, stop(+).

Step 3. Search for the first right parenthesis, if there is no, stop(-). Search for the rightmost left parenthesis that is to the left of the found right parenthesis. If there is no, stop(-). Therefore, $E=...(\alpha)...$ where $\alpha$ has no parentheses in it. Increase $n$ by 1, set $T_{n}=(\alpha)$ , form a vertex $T_{n}$ , and substitute $T_{n}$ for $(\alpha)$ in $E$ . If $\alpha=\neg A_{i}$ then form a leaf with $A_{i}$ and set its parent to $T_{n}$ , if $\alpha=T_{k}$ for some $k<n$ , then set the vertex $T_{k}$ created earlier as the only child of $T_{n}$ , if $\alpha=\beta\square\gamma$ , where each of $\beta$ and $\gamma$ is a sentence symbol or $T_{k}$ for some $k<n$ , then form new vertices corresponding to sentence symbols if necessary and set vertices corresponding to $\beta$ and $\gamma$ as children of $T_{n}$ , otherwise stop(-). Go to Step 2.

stop(+): the tree is formed, recursively substitute translations $T_{i}$ ’s into their parents

stop(-): the tree cannot be formed

Example: $((A\wedge(\neg B))\rightarrow(C\vee D))$ .

$E$	$\alpha$	$T_{n}$	children of $T_{n}$
$((A\wedge(\neg B))\rightarrow(C\vee D))$	$\neg B$	$T_{1}=(\neg B)$	$B$
$((A\wedge T_{1})\rightarrow(C\vee D))$	$A\wedge T_{1}$	$T_{2}=(A\wedge T_{1})$	$A$ , $T_{1}$
$(T_{2}\rightarrow(C\vee D))$	$C\vee D$	$T_{3}=(C\vee D)$	$C$ , $D$
$(T_{2}\rightarrow T_{3})$	$T_{2}\rightarrow T_{3}$	$T_{4}=(T_{2}\rightarrow T_{3})$	$T_{2}$ , $T_{3}$
$T_{4}$

After substituting all $T_{i}$ ’s, the tree formed is as follows: $T_{4}=((A\wedge(\neg B))\rightarrow(C\vee D))\rightrightarrows$ [$T_{2}=(A\wedge(\neg B))\rightrightarrows$ [$A$ ][$T_{1}=(\neg B)\rightrightarrows$ [$B$ ]]][$T_{3}=(C\vee D)\rightrightarrows$ [$C$ ][$D$ ]].