Tseitin formula on 2-connected graph

Question

How can we prove that for $\\\\$ every $\\\\$ 2-connected graph G with an odd number of vertices, the unsatisfiable Tseitin formula for it is minimally unsatisfiable, that is, if we remove even a single clause, it becomes satisfiable?

Answer 1

Clearly, whether a formula is minimally unsatisfiable depends on the exact details of how it is formulated, so let me give a specific definition of Tseitin formulas. Given a graph $G=(V,E)$ and a vertex labelling $l\colon V\to\{0,1\}$, the CNF $T_{G,l}$ uses variables $\{p_e:e\in E\}$, and it includes clauses expressing for each vertex $v$ that the parity of its incident edges $\bigoplus_{e\ni v}p_e$ is $l(v)$: $$\tag{$C_{v,\vec c}$}\let\LOR\bigvee\LOR_{e\ni v}p_e^{1-c_e}$$ for each $v\in V$ and each $\{0,1\}$-vector $\langle c_e:e\ni v\rangle$ such that $\bigoplus_{e\ni v}c_e=\overline{l(v)}$, where $p_e^1=p_e$ and $p_e^0=\overline{p_e}$. The formula $T_{G,l}$ is unsatisfiable if $\bigoplus_{v\in V}l(v)=1$, because any satisfying assignment would have to satisfy $$1=\bigoplus_{v\in V}l(v)=\bigoplus_{v\in V}\bigoplus_{e\ni v}p_e=\bigoplus_{e\in E}\bigoplus_{v\in e}p_e=\bigoplus_{e\in E}0=0.$$

Now, let us try to show that for any given $v$ and $\langle c_e:e\ni v\rangle$, $T_{G,l}\let\bez\smallsetminus\bez\{C_{v,\vec c}\}$ is satisfiable. To guide our search, notice that since $T_{G,l}$ is unsatisfiable, any satisfying assignment of $T_{G,l}\let\bez\smallsetminus\bez\{C_{v,\vec c}\}$ must refute $C_{v,\vec c}$, i.e., it must extend the partial assignment $a$ such that $a(p_e)=c_e$ for all edges $e$ incident to $v$. But it is easy to see that if we restrict $T_{G,l}\bez\{C_{v,\vec c}\}$ with $a$, we end up with another Tseitin formula $T_{G',l'}$, where $G'$ is $G$ with $v$ and all its incident edges deleted, and $$l'(u)=\begin{cases}l(u)\oplus c_{\{u,v\}}&\text{if }\{u,v\}\in E,\\ l(u)&\text{otherwise.}\end{cases}$$ Thus, it suffices to show that $T_{G',l'}$ is satisfiable. Crucially, observe that $$\bigoplus_{u\ne v}l'(u)=\bigoplus_{u\ne v}l(u)\oplus\bigoplus_{e\ni v}c_e=\bigoplus_{u\ne v}l(u)\oplus(l(v)\oplus1)=\bigoplus_{u\in V}l(u)\oplus1=0.$$ Moreover, if we assume $G$ is $2$-connected, then $G'$ is connected. Thus, it remains to prove the following:

Lemma. If $G'=(V',E')$ is connected and $l'\colon V'\to\{0,1\}$ satisfies $\bigoplus_{v\in V'}l'(v)=0$, then $T_{G',l'}$ is satisfiable.

To prove this, we may assume without loss of generality that $G'$ is a tree (we may take an arbitrary spanning tree of $G'$, which is still connected, and set the remaining edges to $0$). Then the $\mathbb F_2$-linear system $\{\bigoplus_{e\ni v}p_e=l'(v):v\in V'\}$ has a unique solution that can be determined by working inwards from the leaves; the condition $\bigoplus_vl'(v)=0$ ensures that it works out in the root.

In fact, the unique satisfying assignment $a$ has a simple explicit description: any given edge $e$ splits the tree in two trees with vertex sets $V_0$ and $V_1$ (that is, $V_0$ and $V_1$ are the two components of the graph $(V',E'\bez\{e\})$), and we put $$a(p_e):=\bigoplus_{v\in V_0}l'(v)=\bigoplus_{v\in V_1}l'(v).$$ To see that $a(T_{G',l'})=1$, consider any vertex $v\in V'$, let $\{e_i:i<d\}$ enumerate the edges incident to $v$, and for each $i<d$, let $V_i$ be the component of $(V',E'\bez\{e_i\})$ that does not contain $v$. Then by definition, $a(p_{e_i})=\bigoplus_{u\in V_i}l'(u)$, and $V'\bez\{v\}$ is a disjoint union of the sets $V_i$ for $i<d$, thus $$0=\bigoplus_{u\in V'}l'(u)=l'(v)\oplus\bigoplus_{i<d}\bigoplus_{u\in V_i}l'(u)=l'(v)\oplus\bigoplus_{i<d}a(p_{e_i}),$$ that is, $\bigoplus_{e\ni v}a(p_e)=l'(v)$.