Solve the exercise on p. 24, items 1, 4, 5, and 8.
We prove that the tableaux method terminates.
We generate an analytic tableaux for the unsigned formula in the following manner. At each stage, we have an ordered dyadic tree , In the first stage, we construct the tree , a one-point tree whose origin is . In the stage, we select from the left-most branch that is unclosed and incomplete. From this branch, we consider all on such that either or does not occur on and all on such that neither nor occur on . We select the or node that dominates all others. If we select an node such that does not occur on , then extend by to form . If we select an node such that does not occur on , then extend by to form . If we select node, then extend by , to form .
We show that the above method either closes or completes any branch
by induction on the ``size'' of the branch. Define
as follows:
Suppose . Then is closed or complete.
Suppose . Suppose an node is selected. In either one or two steps, we must close or complete . Note because . Hence, by the induction hypothesis, the method closes or completes the branch. Suppose a node is selected. Then we must close or complete and . Note because ; likewise, because . Hence, by the induction hypothesis, the method closes or completes both branches.
Hence, the method closes or completes any branch. In particular, the method closes or completes the one-point tree whose origin is . Thus, the tableaux method terminates.
Let be a downward closed set satisfying for all signed formulas , iff . We show that satisfies the laws of a truth set:
Let be a signed formula. If , then , by the definition of . If , then , by the definition of . Hence, exactly one of , belongs to .
Consider . If , then and , by the definition of downward closed. Suppose and . By way of contradiction, assume . By the definition of , . By , is some . Hence or , by the definition of downward closed. Furthermore, by , is and is . Thus, or . If , then , by definition of , but contrary to the assumption. If , then , by the definition of , but contrary to the assumption. Hence .
Consider . If , then or , by the definition of downward closed. Suppose or . By way of contradiction, assume . By the definition of , . By , is some . Hence and , by the defintion of downward closed. Furthermore, by , is and is . Thus, and . Hence and , by the defintion of , but contrary to the assumption. Hence, .
Form = var: Var + neg: Form + and: Form * Form + or: Form * Form + imp: Form * Form Sign = t: Unit + f: Unit Tableaux = index: Nat * sign: Sign * form: Form * next: (closed: Int + complete: Unit + alpha: Int * Tableaux + beta: Int * Tableaux * Tableaux)