# Comparing the Reasoning Capabilities of Equilibrium Theories and Answer Set Programs
## Appendix A Proofs of Results
Notice that \([\tilde{Z}]\in\langle\!\langle\,G\wedge\varphi\,\rangle\!\rangle^{Z}\), so  
\[[\tilde{Z}]=\mathcal{H}_{1}\cap\llbracket\,\varphi\,\rrbracket^{\tilde{Z}}\]  
for some \(\tilde{Z}\)-support \(\mathcal{H}_{1}\in\langle\!\langle\,G\,\rangle\!\rangle^{Z}\). This implies that  
\[\tilde{Z}\cap A\in(\mathcal{H}_{1})_{A}\cap\llbracket\,\varphi\,\rrbracket^{ \tilde{Z}}_{A}\subseteq\llbracket\,\varphi\,\rrbracket^{\tilde{Z}\cap A}.\]  
Then, \(\tilde{Z}\cap A=Y\cap A\) and so \(\tilde{Z}\in D(Y)\). Now we will show that \(\mathcal{H}_{1}\) is \(A\)-feasible. Suppose that \(H\in\mathcal{H}_{1}\) satisfies \(H\cap A=\tilde{Z}\cap A\). Then, \(H\in\llbracket\,\varphi\,\rrbracket^{2}\) because \(\varphi\in\mathcal{L}_{A}\) and \(\tilde{Z}\cap A\in\llbracket\,\varphi\,\rrbracket^{Z}_{A}=\llbracket\,\varphi \,\rrbracket^{\tilde{Z}\cap A}\). Then, \(H\in\mathcal{H}_{1}\cap\llbracket\,\varphi\,\rrbracket^{Z}=[\tilde{Z}]\). Finally, suppose that \((\mathcal{H}_{1})_{A}\
ot\subseteq\mathcal{H}_{A}\), so there exists \(H\in\mathcal{H}_{1}\) such that \(H\cap A\
eq Z\cap A\) and \(H\cap A\in(\mathcal{H}_{1})_{A}\cap\overline{\mathcal{H}_{A}}\subseteq \mathcal{H}_{0}\). This would imply that \(H\in\mathcal{H}_{1}\cap\llbracket\,\varphi\,\rrbracket^{Z}\), so \(H=\tilde{Z}\) which is a contradiction. (Right to left)  
Suppose that \(T\in E(F\wedge L)|B\) for some fork \(L\in\mathcal{L}_{A}\). Then, \(T=Z\cap B\) with \(Z\in E(F\wedge L)\). Take \(Y:=Z\cap(A\cup B)\). Because:  
\[[Z]=\mathcal{H}\cap\mathcal{H}^{\prime}\]  
for some \(\mathcal{H}\) in \(\langle\!\langle\,F\,\rangle\!\rangle^{Z}\) and \(\mathcal{H}^{\prime}\in\langle\!\langle\,L\,\rangle\!\rangle^{Z}\), which we can suppose maximal in \(\langle\!\langle\,L\,\rangle\!\rangle^{Z}\) and then \(A\)-respectful by Lemma 6, we can say that \(\mathcal{H}\) is \(A\)-feasible because if \(H\in\mathcal{H}\) satisfies \(H\cap A=Z\cap A\); then, \(H\in\mathcal{H}^{\prime}\cap\mathcal{H}\) and \(H=Z\). This implies that \(\mathcal{H}_{A}\in\langle\!\langle\,F\,\rangle\!\rangle^{Y}_{A,B}\subseteq \langle\!\langle\,G\,\rangle\!\rangle^{Y}_{A,B}\), so there exists \(Z^{\prime}\in D(Y)\) and \(\mathcal{H}_{1}\in\langle\!\langle\,G\,\rangle\!\rangle^{Z^{\prime}}\) being \(A\)-feasible such that \(\mathcal{H}_{A}\preceq(\mathcal{H}_{1})_{A}\) or \((\mathcal{H}_{1})_{A}\subseteq\mathcal{H}_{A}\). We can apply Lemma 7 to deduce that there exists \(\mathcal{H}^{\prime\prime}\in\langle\!\langle\,L\,\rangle\!\rangle^{Z^{\prime}}\) such that \(\mathcal{H}^{\prime}_{A}\preceq\mathcal{H}^{\prime\prime}_{A}\). In order to finish the proof, we only have to show that \(Z^{\prime}\in E(G\wedge L)\) or, equivalently that:  
\[[Z^{\prime}]=\mathcal{H}_{1}\cap\mathcal{H}^{\prime\prime}\in\langle\!\langle \,G\wedge L\,\rangle\!\rangle^{Z^{\prime}}\]  
Take \(H\in\mathcal{H}_{1}\cap\mathcal{H}^{\prime\prime}\). Then:\[ \begin{array}{l}H\cap A\in(\mathcal{H}_{1}\cap\mathcal{H}^{\prime\prime})_{A}\quad \quad\text{implies}\\ H\cap A\in(\mathcal{H}_{1})_{A}\cap(\mathcal{H}^{\prime\prime})_{A}\quad \quad\text{implies}\\ H\cap A\in\mathcal{H}_{A}\cap(\mathcal{H}^{\prime})_{A}\quad\quad\text{implies}\\ H\cap A=Z\cap A=Z^{\prime}\cap A\quad\quad\text{implies}\\ H=Z^{\prime}\end{array} \]  
Notice that we used Lemma 6.