1 Modal logic. 2 Tableaux in modal logic
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1 1 Modal logic Exercise 1.1: Let us have the set of worlds W = {w 0, w 1, w 2 }, an accessibility relation S = {(w 0, w 1 ), (w 0, w 2 )} and let w 1 p 2. Which of the following statements hold? a) w 0 p 2 b) w 0 p 2 c) w 1 p 1 d) w 1 p 1 e) w 0 p 1 f) w 0 p 1 Solution 1.1: The only statement which does not hold is b). 2 Tableaux in modal logic Contradictory tableaux in modal logic are constructed in a similar way as in predicate logic. To prove that a formula ϕ is a tautology (i.e. it holds for all worlds of all Kripke frames over the used modal-logic language), it is necessary to construct a contradictory tableau with the root Fw ϕ. In addition to predicate logic it is necessary to consider the world in which the formula should be true or false (it is captured in w ). A path in a tableau is contradictory when it contains both Tv ϕ and for the same world v and a formula ϕ. Our modal logic language is supposed not to contain equivalence connectives and function symbols. When nodes of the form and Fv xϕ(x) are expanded, only constants are used (not ground terms). Only constants that belong to the particular world or to its predecessors can be used. When nodes Tv xϕ(x) or Fv xϕ(x) are expanded, a new constant (which is not present in any node of the tableau yet) should be used. Nodes with toplevel operators and are reduced in the following way: when reducing Tv ϕ or Fv ϕ we first adjoin the node vsw to the end of the path (w is a new world that has not been used in the tableau yet). Then the node Tw ϕ or Fw ϕ is adjoined. Nodes of the form Tv ϕ or Fv ϕ are expanded into Tw ϕ or Fw ϕ where w is an arbitrary world for which there is a node vsw on the expanded path. If it is not possible to get such a world on the path, we consider the nodes to be reduced. 1
2 Nodes of the form, Fv xϕ(x), Tv ϕ and Fv ϕ should be always copied when reduced! Exercise 2.1: Using tableaux prove that the following formulas are tautologies. a) Φ 1 ( xϕ(x)) ( x ϕ(x)) b) Φ 2 ( (ϕ ψ)) ( ϕ ψ) c) Φ 3 ( (ϕ xψ(x)) x(ϕ ψ(x))), x is not free in the formula ϕ d) Φ 4 x(ϕ(x) ψ) ( xϕ(x) ψ), x is not free in the formula ψ Solution 2.1: See Figures 1 and 2. Fw ( xϕ(x)) ( x ϕ(x)) Fw ( (ϕ ψ)) ( ϕ ψ) Tw xϕ(x) Tw (ϕ ψ) Fw x ϕ(x) Fw ( ϕ ψ) Fw ϕ(c) new c Tw ϕ Fw ψ Tw xϕ(x) Fv ψ Tw ϕ Tv ϕ Tw (ϕ ψ) Tv ϕ ψ Tv ψ Figure 1: Finished contradictory tableaux for Φ 1 (left) and Φ 2 (right). 2
3 Fw ( (ϕ xψ(x)) x(ϕ ψ(x))) Fw x(ϕ(x) ψ) ( xϕ(x) ψ) Tw ( (ϕ xψ(x)) x(ϕ ψ(x))) Tw x(ϕ(x) ψ) Fw ( xϕ(x) ψ) Tv (ϕ xψ(x)) x(ϕ ψ(x)) Tv (ϕ xψ(x)) Tv x(ϕ(x) ψ) Tv x(ϕ ψ(x)) ψ new c Tv ϕ ψ(c) new c Fw ( xϕ(x) ψ) Tv ϕ Fv ( xϕ(x) ψ) Tv ψ(c) xψ(x) Fv ψ Fv xψ(x) vsu Fv xψ(x) Fu ψ Fv ψ(c) Tv ψ Tv ψ Tu ψ Figure 2: Finished contradictory tableaux for Φ 3 (left) and Φ 4 (right). 3
4 Fw x ϕ(x) xϕ(x) Tw x ϕ(x) Fw xϕ(x) Fv xϕ(x) new c Tw x ϕ(x) Tw ϕ(c) Tw ϕ(c) Figure 3: Exercise 2.2: Consider the tableau with the root Fw x ϕ(x) xϕ(x) given in Figure 3. Decide whether the tableau is correct or not. Explain your decision. Solution 2.2: The tableau is not correct (in our modal logic) and it does not prove that the formula is a tautology. When we expand the node ( ) it is not possible to replace the variable x with the constant c. The constant c appeared in the world v, however, we are in the world w in the node ( ). The expansion would be correct if the world w was a successor of world v, not its predecessor. Exercise 2.3: Prove the following logical consequences: a) {ϕ} = ϕ b) { xϕ(x)} = xϕ(x) 4
5 c) { xϕ(x)} = x ϕ(x) d) {ϕ ϕ} = ϕ ϕ Solution 2.3: The tableau for the logical consequence S = ϕ starts with and at any time we can add to the end of any path P a node of the form Tw α, where w is any world that appears somewhere on the path P and α S. Finished contradictory tableaux for the consequences (a),(b) and (c) are in the picture 4. The tableau for the consequence (d) is left as an exercise. Fw ϕ Fw xϕ(x) Fw x ϕ(x) Fw ϕ(c) new c Fv xϕ(x) Tv ϕ new c Figure 4: Finished contradictory tableaux for the logical consequences (a),(b) and (c). Nodes that correspond to the addition of premises are marked with an asterisk. 5
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