SolutionsModalLogic - CSE 541 - Logic in Computer Science...

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Unformatted text preview: CSE 541 - Logic in Computer Science Sample Solutions for Selected Exercises on Modal Logic Exercise 5.2.1 a Answer key . (iv) false; (v) true; (vi) true; (vii) false; (viii) true; (ix) true; (x) true; (xi) false; (xii) true. Exercises 5.2.1 b,c i. Only worlds c , d and e satisfy the given formula. ii. Only world b satisfies the given formula. iii. Only worlds a , b and d satisfy the given formula. iv. Only worlds a , b and d satisfy the given formula. v. Only worlds c , d and e satisfy the given formula. vi. All worlds satisfy the given formula. Exercise 5.2.2 Let M = ( W, R, L ) be a model with W = { s } , R = ∅ , and L ( s ) = ∅ . The formula scheme φ → ψ is not valid in M (e.g., ¬ p → p is false), but has true instances, such as p → p . Exercise 5.2.3 b i. World c satisfies the given formula. ii. Worlds a and b satisfy the given formula. iii. Worlds a , b and e satisfy the given formula. iv. Worlds b , c , d and e satisfy the given formula. v. All worlds satisfy the given formula. Exercise 5.2.5 a. Let M = ( W, R, L ) be a model with W = { s , s 1 , s 2 } , R = { ( s , s 1 ) , ( s 1 , s 2 ) } , and L ( s ) = L ( s 2 ) = ∅ and L ( s 1 ) = { p } . Then M , s | = 2 p but M , s 6| = 22 p . c. The formulas 2 ( p ∧ q ) and 2 p ∧ 2 q are equivalent; see the remarks on page 269. d. Let M = ( W, R, L ) be a model with W = { s , s 1 , s 2 } , R = { ( s , s 1 ) , ( s , s 2 ) } , and L ( s ) = ∅ , L ( s 1 ) = { p } , and L ( s 2 ) = { q } . Then M , s | = 3 p ∧ 3 q but M , s 6| = 3 ( p ∧ q ). e. Taking the same model M , we also find that M , s | = 2 ( p ∨ q ) whereas M , s 6| = 2 p ∨ 2 q . f. The formulas 3 ( p ∨ q ) and 3 p ∨ 3 q are equivalent; see the next exercise. g. Take again the same model M as before and note that M , s | = ( 2 p → 2 q ) but M , s 6| = 2 ( p → q ). h. Let M = ( W, R, L ) be a model with W = { s } , R is the empty relation, and L ( s ) = ∅ . Then M , s | = > but M , s 6| = 3 > . Exercise 5.2.6 a. We prove that 2 ( φ ∧ ψ ) and 2 φ ∧ 2 ψ are true in the same worlds. Consider first a model M = ( W, R, L ) and a world x ∈ W , such that x | = 2 ( φ ∧ ψ ). By the definition of the 2 operator, we have y | = ( φ ∧ ψ ), for all y ∈ W with xRy . By the semantics of conjunction, we obtain (i) y | = φ , for all y ∈ W with xRy , and (ii) y | = ψ , for all y ∈ W with xRy . This implies both x | = 2 φ and x | = 2 ψ , and hence x | = ( 2 φ ∧ 2 ψ ), which completes the first part....
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SolutionsModalLogic - CSE 541 - Logic in Computer Science...

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