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P. Let, X. , X. ?. , and X. ??-?-l, Assume that X ? X ? . According to the compactness axiom, it holds that CN(X ) = Y ? f X CN(Y )

C. Assume and . Cn, According to the expansion axiom Moreover, X ? X ? X ?? thus CN(X ) ? CN(X ? X ?? ) (a') Similarly From (a), it follows that, )). Finally, by the idempotence axiom, the inclusion CN(X ? ? X ?? ) ? CN(X ? X ?? ) holds. To show that CN(X ? X ?? ) ? CN(X ? ? X ?? ), the same reasoning is applied by starting with X instead of X ?

P. Let and X. , Assume that X is consistent, then CN(X ) = L (1)

?. ?. Assume, X. Such, and . Inconsistent, This means that CN(X ? ) = L. However, since X ? ? X then CN(X ? ) ? CN(X ) (according to the monotonicity axiom, Thus, L ? CN(X ). Since CN(X ) ? L, then CN(X ) = L. Thus, X is inconsistent. Contradiction

T. Since and . Concs, Thus, there exists x ? ? C such that x ?, GE ConcsGE

?. Then and . Ext, Base(E) is consistent. Consequently, T satisfies consistency. From Amgoud (2012), T is also closed under sub-arguments