& ~r0, p0, q0, ~q0} -------------------------- (id) ---------------------- (id) closed closed The left hand tableau closes after only one rule application while the right hand one misses the mark and takes a lot longer to close. Clearly, we would prefer to always find the shortest closed tableaux but it can be shown that one single algorithm that finds the shortest closed tableaux for all input sets of formulae cannot exist. The three rules , and given above are then enough to decide if a given set of formulae in negated normal form are jointly satisfiable: Just apply all possible rules in all possible orders until we find a closed tableau for or until we exhaust all possibilities and conclude that every tableau for is open. In the first case, is jointly unsatisfiable and in the second the case the leaf node of the open branch gives an assignment to the atomic formulae and negated atomic formulae which makes jointly satisfiable. Classical logic actually has the rather nice property that we need to investigate only (any) one tableau completely: if it closes then is unsatisfiable and if it is open then is satisfiable. But this property is not generally enjoyed by other logics. These rules suffice for all of classical logic by taking an initial set of formulae X and replacing each member C by its logically equivalent negated normal form C' giving a set of formulae X' . We know that X is satisfiable if and only if X' is satisfiable, so it suffices to search for a closed tableau for X' using the procedure outlined above. By setting we can test whether the formula A is a tautology of classical logic: If the tableau for closes then is unsatisfiable and so A is a tautology since no assignment of truth values will ever make A false. Otherwise any open leaf of any open branch of any open tableau for gives an assignment that falsifies A. Conditional Classical propositional logic usually has a connective to denote material implication. If we write this connective as ⇒, then the formula A ⇒ B stands for "if A then B". It is possible to give a tableau rule for breaking down A ⇒ B into its constituent formulae. Similarly, we can give one rule each for breaking down each of ¬(A ∧ B), ¬(A ∨ B), ¬(¬A), and ¬(A ⇒ B). Together these