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ISBN-10: 9048133181

ISBN-13: 9789048133185

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E. there should be no tuples H, ν, ι such that H, ν, ι |= ϕ and H, ν, ι |= ψ. Our main result, leading immediately to decidability of entailments, is that, if ϕ is of the form ∃x1 . . ∃xn Q1 l1 . . Qm lm . θ(x, l), with Qi ∈ {∃N , ∀N } and θ is a boolean combination of predicates with ¬, ∧ and ∗, all models H, ν, ι of ϕ can be represented using a finite number of (finite) structures called symbolic graph representations (SGR). The decision procedure for the validity of a QSL entailment ϕ ⇒ ψ is based on the following idea.

E. the sum over the componentwise products. If the two vectors involved in an operation are of different length, the shorter one is padded with 0s (as in Obua’s treatment of matrices [18]). We can prove all the algebraic properties we need, like xs + ys,zs = xs,zs + ys,zs . 1. 2 I R (r = cs) xs = (r = cs,xs ) Ferrante and Rackoff Ferrante and Rackoff [6], inspired by Cooper [5], avoided DNF conversions by the test point method explained in §4. 2. If you replace y↓z in (1) by (y + z)/2 you almost obtain their algorithm.

Except that −∞, ∞ and y↓z are not proper values. ∀y ≤ x. ∀y ≥ x. φ(∞) = φ(y) y < z ∧ φ(y↓z) =⇒ ∀x ∈ (y, z). φ(x) For the only-if-direction assume φ(x) and not φ(−∞) ∨ φ(∞) ∨ y∈E φ(y). We have to show that φ(y↓z) for some y ∈ L and z ∈ U . From the assumptions it follows by induction on φ that there must be y0 ∈ L and z0 ∈ U such that x ∈ (y0 , z0 ). Now we show (by induction on φ) the lemma that innermost intervals (y, z) completely satisfy φ: Lemma 3. If x ∈ (y, z), x ∈ / E, (y, x) ∩ L = ∅ and (x, z) ∩ U = ∅, then φ(x) implies ∀u ∈ (y, z).

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Automated Reasoning International Joint Conference IJCAR2008


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