By Jack Minker (auth.), Esra Erdem, Joohyung Lee, Yuliya Lierler, David Pearce (eds.)
This Festschrift released in honor of Vladimir Lifschitz at the social gathering of his sixty fifth birthday provides 39 articles via colleagues from worldwide with whom Vladimir Lifschitz had cooperation in a variety of respects. The 39 contributions replicate the breadth and the intensity of the paintings of Vladimir Lifschitz in common sense programming, circumscription, default good judgment, motion concept, causal reasoning and resolution set programming.
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Extra info for Correct Reasoning: Essays on Logic-Based AI in Honour of Vladimir Lifschitz
A Decidable Subclass of Finitary Programs. Journal of Theory and Practice of Logic Programming (TPLP) 10(4-6), 481–496 (2010) 5. : Adding Weak Constraints to Disjunctive Datalog. In: Proceedings of the 1997 Joint Conference on Declarative Programming APPIA-GULPPRODE (1997) 6. : Functional Answer Set Programming. Journal of Theory and Practice of Logic Programming (TPLP) 11, 203–234 (2011) 7. : Enhancing ASP by Functions: Decidable Classes and Implementation Techniques. In: Proceedings of the Twenty-Fourth Conference on Artificial Intelligence, pp.
Intuitively, a non-transitive verb has a category “S\N P ”, meaning that if a noun phrase is added on the left, we get a sentence. Similarly, the category “N P/N P ” means that a phrase of that category concatenated on the right by a phrase of category “N P ” results in a phrase of the category “N P ”. The phrase “Texas” has a category “N P ” and it is being applied from the right to “borders a state”. [state(x)∧ borders(x, y)]. 42 C. A. Gonzalez, and A. Gottesman Now we can take H as the meaning of “borders a state” and G as the meaning of “borders”, which has category (S\N P )/N P , and we can can ﬁnd the meaning of F , for “a state”.
G = 3. : f = 2. and let us compute its reduct. For r1 , first we have to check if S3 |= body(r1 ) \ pos(r1 ), that is if S3 |= not g = 1, not h = 0. Extended literal not g = 1 is satisfied by S3 only if S3 |= g = 1. Because g = 1 is a seed literal, it is satisfied by S3 if g = 1 ∈ S3 . Since g = 1 ∈ S3 , we conclude that S3 |= g = 1 and thus not g = 1 is satisfied by S3 . In a similar way, we conclude that S3 |= not h = 0. Hence, S3 |= body(r1 ) \ pos(r1 ). Therefore, the reduct of r1 is p ← f = 2.