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**Sample text**

3. M j= F i M f:F g is inconsistent. Proof. 4 (Ex falso quodlibet). 5 (Compactness theorem, 2nd version). If M j= F, then there is already a nite subset M0 M such that M0 j= F: Proof. If M j= F , then M f:F g is inconsistent. By the compactness theorem we nd already a nite subset M0 M f:F g which is inconsistent. 6 (Deduction theorem). M G j= F i M j= G ! F: Proof. Inconsistency of M fG :F g is equivalent to the inconsistency of M fG^:F g which because of :(G ! F ) S G ^:F is the same as the inconsistency of M f:(G !

9xGi for i = 1 : : : k Hj = 9xGj ! Gjx(c9xGj ) for j = k + 1 : : : m and Hk+1 : : : Hm are ordered in such a way that we have degH (Gj ) degH (Gj +1): Therefore we can be sure that c9xGj does not occur in Hj +1 : : : Hm : Now we replace any Henkin constant successively by a new variable. The result F1 ! : : : ! Fn ! H10 ! : : : ! Hm0 ! F is still boolean valid. Because of Fi 2 M for i = 1 : : : n we have M ` Fi and get M ` H10 ! : : : ! Hm0 ! F: Since we have for i = 1 : : : k M ` Hi0 by an 9-axiom we have using a boolean inference M ` Hk0 +1 !

Fx(t) (8-axiom) and the dual rule M ` G ! F and x 2= FV(M fGg) ) M ` G ! 8xF (8-rule) though, due to the restriction of the logical symbols to : _ 9 they are not really basic axioms or rules. 2 (Soundness theorem). M ` F entails M j= F: Proof. We prove the theorem by induction on the de nition of M ` F: Cases 1. and 2. 1 and 3. 9 and 4. and 5. 11, respectively. 2, which is the real aim of this section, still needs some preparation. 50 I. 3. Let (F1 ! : : : ! (Fn ! G) : : :) be a boolean valid formula.