Denote with $\bar{\mathbf A}$ and $\bar U$ the logical complement of a WFF $\mathbf A$ and set $U$, respectively. === Necessary Condition === We aim to prove this result using the Second Principle of Mathematical Induction, applied to the minimal length of a formal proof of $U$. Suppose first that $U$ is an axiom of $\mathscr G$. Then $U$ contains a complementary pair of literals. Hence, so does $\bar U$, by definition of logical complement. Therefore, the semantic tableau comprising only a root node labeled $\bar U$ is closed. Next, suppose that the last step in proving $U$ was an instance of the $\alpha$-rule. That is, for some $\alpha$-formula $\mathbf A$ and corresponding $\mathbf A_1, \mathbf A_2$: :$U = U_1 \cup U_2 \cup \left\{{\mathbf A}\right\}$ where $U_1 \cup \left\{{\mathbf A_1}\right\}$ and $U_2 \cup \left\{{\mathbf A_2}\right\}$ are $\mathscr G$-theorems. By induction hypothesis, the logical complements: :$\bar U_1 \cup \left\{{\bar{\mathbf A}_1}\right\}$ :$\bar U_2 \cup \left\{{\bar{\mathbf A}_@}\right\}$ of these theorems have closed tableaus. From the Soundness Theorem for Semantic Tableaus, it follows that these sets are unsatisfiable. Now by Superset of Unsatisfiable Set is Unsatisfiable, so are: :$\bar U' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$  :$\bar U'' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$ By the Completeness Theorem for Semantic Tableaus, these have closed tableaus. Since $\mathbf A$ is an $\alpha$-formula, it is equivalent to $\mathbf A_1 \land \mathbf A_2$. By De Morgan's Laws, it follows that: :$\bar{\mathbf A}$ is equivalent to $\bar{\mathbf A}_1 \lor \bar{\mathbf A}_2$. Therefore, $\bar{\mathbf A}$ is a $\beta$-formula, and $\bar{\mathbf A}_1, \bar{\mathbf A}_2$ correspond to it as in the table of $\beta$-formulas. This allows the Semantic Tableau Algorithm to expand a leaf labeled $\bar U$ into two leaves labeled $\bar U'$ and $\bar U''$. Because $\bar U'$ and $\bar U''$ have closed tableaus, so does $\bar U$. Finally, suppose that the last step in proving $U$ was an instance of the $\beta$-rule. That is, for some $\beta$-formula $\mathbf B$ and corresponding $\mathbf B_1, \mathbf B_2$: :$U = U_1 \cup \left\{{\mathbf B}\right\}$ where $U' := U_1 \cup \left\{{\mathbf B_1, \mathbf B_2}\right\}$ is a $\mathscr G$-theorem. By induction hypothesis, the logical complement: :$\bar U' = \bar U_1 \cup \left\{{\bar{\mathbf B}_1, \bar{\mathbf B}_2}\right\}$ of this $\mathscr G$-theorem has a closed tableau. Since $\mathbf B$ is a $\beta$-formula, it is equivalent to $\mathbf B_1 \lor \mathbf B_2$. By De Morgan's Laws, it follows that: :$\bar{\mathbf B}$ is equivalent to $\bar{\mathbf B}_1 \land \bar{\mathbf B}_2$. Therefore, $\bar{\mathbf B}$ is an $\alpha$-formula, and $\bar{\mathbf B}_1, \bar{\mathbf B}_2$ correspond to it as in the table of $\alpha$-formulas. This allows the Semantic Tableau Algorithm to expand a leaf labeled $\bar U$ into a leaf labeled $\bar U'$. Because $\bar U'$ has a closed tableau, so does $\bar U$. The result follows by the Second Principle of Mathematical Induction. {{qed|lemma}} === Sufficient Condition === Let $T$ be a closed tableau for $\bar U$. We prove that $U$ is a $\mathscr G$-theorem by the Second Principle of Mathematical Induction, applied to the number of nodes of $T$. Suppose $T$ has only one node. Then $\bar U$ has to contain a complementary pair of literals for $T$ to be closed. But then $U$ also contains a complementary pair. Hence, it is an axiom of $\mathscr G$. If $T$ has more than one node, the first iteration of the Semantic Tableau Algorithm must have selected either an $\alpha$-formula $\mathbf A$ or a $\beta$-formula $\mathbf B$ from $\bar U$. First the case that an $\alpha$-formula $\mathbf A$ was selected. Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf A}\right\}$. Then the rest of $T$ is a semantic tableau for $\bar U' \cup \left\{{\mathbf A_1, \mathbf A_2}\right\}$. This tableau has fewer nodes than $T$. For $T$ to be closed, this subtree must also be a closed tableau. By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf A}_1, \bar{\mathbf A}_2}\right\}$ is a $\mathscr G$-theorem. Now, since $\mathbf A$ is an $\alpha$-formula, it is equivalent to $\mathbf A_1 \land \mathbf A_2$. Hence by De Morgan's Laws, $\bar{\mathbf A}$ is equivalent to $\bar{\mathbf A}_1 \lor \bar{\mathbf A}_2$. That is, $\bar{\mathbf A}$ is a $\beta$-formula, so that the $\beta$-rule may be applied to it. Hence, $U = U' \cup \left\{{\bar{\mathbf A}}\right\}$ is a $\mathscr G$-theorem, as desired. Now the case that a $\beta$-formula $\mathbf B$ was selected. Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf B}\right\}$. Then the rest of our closed tableau $T$ comprises two semantic tableaus for $\bar U' \cup \left\{{\mathbf B_1}\right\}$ and $\bar U' \cup \left\{{\mathbf B_2}\right\}$, respectively. Each of these has fewer nodes than $T$. For $T$ to be closed, these subtrees must also be closed tableaus. By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf B}_1}\right\}$ and $U' \cup \left\{{\bar{\mathbf B}_2}\right\}$ are $\mathscr G$-theorems. Now, since $\mathbf B$ is a $\beta$-formula, it is equivalent to $\mathbf B_1 \lor \mathbf B_2$. Hence by De Morgan's Laws, $\bar{\mathbf B}$ is equivalent to $\bar{\mathbf B}_1 \land \bar{\mathbf B}_2$. That is, $\bar{\mathbf B}$ is an $\alpha$-formula, so that the $\alpha$-rule may be applied to it. Hence, $U = U' \cup \left\{{\bar{\mathbf B}}\right\}$ is a $\mathscr G$-theorem, as desired. The result now follows from the Second Principle of Mathematical Induction. {{qed}}