Homework 3 Solutions

Homework 3 Solutions - 1 i S t = cfw s 1 s 3 a 1s0,s2,s4 T...

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1) i. Sat( ϕ _1) = {s_0, s_1, s_2, s_3, s_4} TS ⊨ϕ ii. ϕ _2) = {s_4} ¬TS ⊨ϕ iii. ϕ ⊨ϕ 2) ϕ _1 = ∃◇∀ □c Subformulas: □c, c Sat(c) = {s_2, s_3, s_4} □c) = {s_2, s_3, s_4} ∃◇∀ □c) = {s_0, s_1, s_2, s_3, s_4} Therefore TS ⊨ϕ _1 ϕ _2 = (aU ∀◇ c) ∀◇ c, c ∀◇ c) = {s_0, s_1, s_2, s_3, s_4} ∀◇ c)) = {s_0, s_1, s_2, s_3, s_4} Or alternatively normalizing the formula for the algorithm: (aU¬ □¬c) Subformulas: c, ¬c, □¬c, ¬ □¬c Sat(¬c) = {s_0, s_1} □¬c) = {} Sat(¬ □¬c) = {s_0, s_1, s_2, s_3, s_4} And we get the same result. ⊨ϕ
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3) (TS ⊨ ∃ ( ϕ Uψ)) = (TS' ⊨ ∃◇ ψ). Theorem Since TS has more or equal transitions than TS' and everything else the same, for any formula ϕ we have that (TS' ⊨ ϕ ) (TS ⊨ ϕ ) ­­­­­­­­­­­­­­­­ assump_0 I formalize the removal of outgoing transitions π:TS'∙ i∙ j∙j<i (π_j ¬(¬ ϕ∨ ψ) π_i=π_j) ­­­­ assump_1 Proving (TS ⊨ ∃ ( ϕ Uψ)) ⊨ ∃◇ ψ): TS ⊨ ∃◇ ψ def of = TS ⊨ ∃ (TUψ) base law
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Homework 3 Solutions - 1 i S t = cfw s 1 s 3 a 1s0,s2,s4 T...

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