The proof uses induction on the number $n$ of elements of $\mathbf H$. Suppose we are given the result for the case $n = 1$, that is, when $\mathbf H$ is a singleton. Suppose also that we are given the result for all sets $\mathbf H'$ with $n$ elements. Now, given a set $\mathbf H' = \left\{{\mathbf A_1, \ldots, \mathbf A_{n+1}}\right\}$ with $n+1$ elements. Let $T$ be a finite propositional tableau. By induction hypothesis, there is a finished finite propositional tableau $T'$ containing $T$ as a subgraph, and with root $\mathbf H \cup \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right)$. Now apply the case $n = 1$ to this resulting propositional tableau $T'$ and the set $\left\{{\mathbf A_{n+1}}\right\}$. This yields a finished finite propositional tableau $T''$ which: $(1):\quad$ has root $\mathbf H \cup \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right\} \cup \left\{{\mathbf A_{n+1}}\right\} = \mathbf H \cup \mathbf H'$; $(2):\quad$ contains $T'$ as a subgraph. But then $T''$ also contains $T$ as a subgraph, proving the result for $\mathbf H'$. It thus only remains to take care of the base cases $n = 0$ and $n = 1$. First, the case $n = 0$. Let $T$ be a finite propositional tableau. To find the finite propositional tableau $T'$ with the desired properties, we use some of the tableau construction rules, starting with $T$. Let $t$ be any leaf node of $T$, and let $\Gamma_t$ be the branch from Leaf of Rooted Tree is on One Branch. Let $n \left({\Gamma_t}\right)$ be the number of non-basic WFFs that were not used to add any of the nodes of $\Gamma_t$ to $T$. It is seen that for any application of the tableau construction rules on $t$: :If $s$ is added by the rule, then $n \left({\Gamma_s}\right) \le n \left({\Gamma_t}\right)$. Moreover, it is seen that any rule reduces the total count $m \left({\Gamma_t}\right)$ of logical connectives occurring in these non-basic, unused WFFs along $\Gamma_t$. In conclusion: :If $s$ is added by a rule, then $m \left({\Gamma_s}\right) < m \left({\Gamma_t}\right)$ By the Method of Infinite Descent applied to $m \left({\Gamma_t}\right)$, only finitely many rules can be applied, starting from $t$. Since $T$ has only finitely many leaves and corresponding branches, only finitely many rules can be applied to $T$ in total. Let $T'$ be the finite propositional tableau resulting from applying all these possible rules. By construction of $T'$, it follows that every branch of $\Gamma$ is either contradictory or finished. That is, $T'$ is finished. Finally, the last case, $n = 1$. Let $\mathbf A$ be a WFF of propositional logic. Let $T$ be a finite propositional tableau. First, using the case $n = 0$, extend $T$ to a finished finite propositional tableau $T'$. Again using the case $n = 0$, let $T_{\mathbf A}$ be a finished finite propositional tableau with root $\left\{{\mathbf A}\right\}$. Now add $\mathbf A$ to the root of $T'$. Then at every leaf $t$ of $T'$, $\mathbf A$ is the only WFF that is not used yet. As far as the rules for propositional tableaus are concerned, there is no difference between: :$t$ as a leaf of $T'$, and  :the tableau consisting only of a root and with hypothesis set $\mathbf A$. Therefore, the rules allow to "paste", as it were, the finished tableau $T_{\mathbf A}$ under every leaf $t$ of $T'$. Denote the resulting tableau with $T'_{\mathbf A}$. Then for any branch $\Gamma$ of $T'_{\mathbf A}$ and every non-basic WFF $\mathbf B$ along it: :$\mathbf B$ is on $T'$, or: :$\mathbf B$ is on a copy of $T_{\mathbf A}$. In either case, the finished nature of these tableaus implies that: :$\mathbf B$ is used at some node of $\Gamma$ Hence $\Gamma$ is contradictory or finished. In conclusion, $T'_{\mathbf A}$ is finished, and contains $T$ as a subgraph. The result follows from the Principle of Mathematical Induction. {{qed}}