Ribbon Proofs - A Proof System for the Logic of Bunched Implications

... O’Hearn, Reynolds, Yang and others ([24, 29, 40]) in which the resources concerned are computational, such as memory cells; this provides a powerful practical motivation for studying the logic. However the two inference systems in [32] do not directly reflect these semantic intuitions about BI, and ...

... O’Hearn, Reynolds, Yang and others ([24, 29, 40]) in which the resources concerned are computational, such as memory cells; this provides a powerful practical motivation for studying the logic. However the two inference systems in [32] do not directly reflect these semantic intuitions about BI, and ...

Image-Space Decomposition Algorithms for Sort

... are likely to pass through the same cells involving similar computations. In addition, since decomposition is done irrespective of input data, it is still possible that some regions of the screen are heavily loaded and some processors may perform substantially more work than the others. In the dynam ...

... are likely to pass through the same cells involving similar computations. In addition, since decomposition is done irrespective of input data, it is still possible that some regions of the screen are heavily loaded and some processors may perform substantially more work than the others. In the dynam ...

Handling Exceptions in nonmonotonic reasoning

... Example 3.2. Let τ = (∅, {g1 = P –( Q, g2 = R–( ¬Q, g3 = Q–( , g4 = ¬Q–( }). There are two complete candidates, namely: γ1 = {P –( Q, ¬Q–( }; γ2 = {R–( ¬Q, Q–( }. Note that γ1 is complete because g2 is rejected and g3 is excluded by γ1 . γ2 is complete because g1 is rejected and g4 is excluded by γ2 ...

... Example 3.2. Let τ = (∅, {g1 = P –( Q, g2 = R–( ¬Q, g3 = Q–( , g4 = ¬Q–( }). There are two complete candidates, namely: γ1 = {P –( Q, ¬Q–( }; γ2 = {R–( ¬Q, Q–( }. Note that γ1 is complete because g2 is rejected and g3 is excluded by γ1 . γ2 is complete because g1 is rejected and g4 is excluded by γ2 ...

Predicate logic

... Definition: integer a is odd iff a = 2m + 1 for some integer m Let a, b ∈ Z s.t. a and b are odd. Then by definition of odd a = 2m + 1.m ∈ Z and b = 2n + 1.n ∈ Z So ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 and since m, n ∈ Z it holds that (2mn + m + n) ∈ Z, so ab = 2k + 1 for s ...

... Definition: integer a is odd iff a = 2m + 1 for some integer m Let a, b ∈ Z s.t. a and b are odd. Then by definition of odd a = 2m + 1.m ∈ Z and b = 2n + 1.n ∈ Z So ab = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 and since m, n ∈ Z it holds that (2mn + m + n) ∈ Z, so ab = 2k + 1 for s ...

abdullah_thesis_slides.pdf

... Given d,t ∈ N, we can define the concept of type signatures of radius d with threshold t such that the values (#Type1 ,...,#Typen ) are counted only upto a threshold t and anything ≥ t is considered ∞. Two structures A and B, are said to be d-equivalent with threshold t if their type signatures with ...

... Given d,t ∈ N, we can define the concept of type signatures of radius d with threshold t such that the values (#Type1 ,...,#Typen ) are counted only upto a threshold t and anything ≥ t is considered ∞. Two structures A and B, are said to be d-equivalent with threshold t if their type signatures with ...

Computer Science Foundation Exam

... (a) (5 pts) Consider a relation defined by people who know each other. For example, (Arup Guha, Ben Douglass) would be in the relation because the first person listed in the ordered pair knows the second person listed. We assume that a person knows their self and that if person A knows person B, the ...

... (a) (5 pts) Consider a relation defined by people who know each other. For example, (Arup Guha, Ben Douglass) would be in the relation because the first person listed in the ordered pair knows the second person listed. We assume that a person knows their self and that if person A knows person B, the ...

Propositional Logic

... In a split, both of the new sequents must be proved. Axiom: A sequent in which any proposition symbol occurs at top level on both the left and right sides. e.g., P, P v Q P CSE 415 -- (c) S. Tanimoto, 2008 Propositional Logic ...

... In a split, both of the new sequents must be proved. Axiom: A sequent in which any proposition symbol occurs at top level on both the left and right sides. e.g., P, P v Q P CSE 415 -- (c) S. Tanimoto, 2008 Propositional Logic ...

Piecewise Tri-linear Contouring for Multi

... The key idea behind these approaches are that the signed grid can be stored as a 3D texture and that a single texture fetch can be used to evaluate f (x, y, z) via tri-linear interpolation at an arbitrary point. In practice, this tri-linear boundary surface provides better normals (for shading) and ...

... The key idea behind these approaches are that the signed grid can be stored as a 3D texture and that a single texture fetch can be used to evaluate f (x, y, z) via tri-linear interpolation at an arbitrary point. In practice, this tri-linear boundary surface provides better normals (for shading) and ...

Modal Logics of Submaximal and Nodec Spaces 1 Introduction

... out in [3] that for a space X the following three conditions are equivalent: (i) X is an I-space; (ii) X is nodec and (weakly) scattered; (iii) X is submaximal and (weakly) scattered. Examples of I-spaces that are not door are the ordinals α ∈ [ω2 + 1, ω 2 ]. For examples of door spaces that are not ...

... out in [3] that for a space X the following three conditions are equivalent: (i) X is an I-space; (ii) X is nodec and (weakly) scattered; (iii) X is submaximal and (weakly) scattered. Examples of I-spaces that are not door are the ordinals α ∈ [ω2 + 1, ω 2 ]. For examples of door spaces that are not ...

Regular Languages and Finite Automata

... Sections 4 and 6.5. But the present theory would work equally well with more than 2 states. Nothing would be gained, however, as p cells each admitting 2 states could be used to replace one cell admitting any number q (2 ≤ q ≤ 2p ) of states 0, 1, . . . , q − 1, where if q < 2p we could either consi ...

... Sections 4 and 6.5. But the present theory would work equally well with more than 2 states. Nothing would be gained, however, as p cells each admitting 2 states could be used to replace one cell admitting any number q (2 ≤ q ≤ 2p ) of states 0, 1, . . . , q − 1, where if q < 2p we could either consi ...

Satisfiability is Decidable for a Fragment of AMSO Logic on Infinite

... difficulty is to prove completeness, saying that using some other fancy pictures one cannot obtain more base languages than we obtain using pictures generated from our schemas. Let us now define the two kinds of schemas generating new base languages: product schemas and diagonal schemas. Let ρ1 , ρ2 ...

... difficulty is to prove completeness, saying that using some other fancy pictures one cannot obtain more base languages than we obtain using pictures generated from our schemas. Let us now define the two kinds of schemas generating new base languages: product schemas and diagonal schemas. Let ρ1 , ρ2 ...

PDF

... retinal cells was formed superficial to telencephalic and diencephalic compartments, but it was split into left and ...

... retinal cells was formed superficial to telencephalic and diencephalic compartments, but it was split into left and ...

1 Introduction - Institute for Logic, Language and Computation

... 2.3.1 From pre to peri (puncturing neighborhoods). The following construction produces all peritopological spaces. Start with any pretopological space (E, (V(x))x∈E ). Choose a subset G ⊂ E and let U(x) = V(x) for x ∈ G, U(x) = V∗ (x) = {V \{x} : V ∈ V(x)} ∪ V(x) for x 6∈ G. Then, (E, (U(x))x∈E ) is ...

... 2.3.1 From pre to peri (puncturing neighborhoods). The following construction produces all peritopological spaces. Start with any pretopological space (E, (V(x))x∈E ). Choose a subset G ⊂ E and let U(x) = V(x) for x ∈ G, U(x) = V∗ (x) = {V \{x} : V ∈ V(x)} ∪ V(x) for x 6∈ G. Then, (E, (U(x))x∈E ) is ...

Cell Simile Poster

... Task: Your task is to design a simile between a plant or animal cell, and some other system you are familiar with. A simile is when you use something that is compared to something else usually using “like” or “as.” An example would be: A cell is like a school with all of the organelles in the cell w ...

... Task: Your task is to design a simile between a plant or animal cell, and some other system you are familiar with. A simile is when you use something that is compared to something else usually using “like” or “as.” An example would be: A cell is like a school with all of the organelles in the cell w ...

JavaAPI

... How to use the Java class libraries Brief documentation of how to do this all with Java. ...

... How to use the Java class libraries Brief documentation of how to do this all with Java. ...

Flowchart Thinking

... So far, you have been writing explanations as paragraph proofs. Example A in your book is an example. Read this example and make sure you understand the proof. When a logical argument is complex or includes many steps, a paragraph proof may not be the clearest way to present the steps. In such cases ...

... So far, you have been writing explanations as paragraph proofs. Example A in your book is an example. Read this example and make sure you understand the proof. When a logical argument is complex or includes many steps, a paragraph proof may not be the clearest way to present the steps. In such cases ...

Chapter 1 Section 2

... B says “The two of us are of opposite types.” Example: What are the types of A and B? Solution: Let p and q be the statements that A is a knight and B is a knight, respectively. So, then p represents the proposition that A is a knave and q that B is a knave. If A is a knight, then p is true. S ...

... B says “The two of us are of opposite types.” Example: What are the types of A and B? Solution: Let p and q be the statements that A is a knight and B is a knight, respectively. So, then p represents the proposition that A is a knave and q that B is a knave. If A is a knight, then p is true. S ...

Chapter1_Parts2

... If the system is working correctly (all assumables are true), the observations and the knowledge base are consistent (i.e., satisfiable).! The augmented knowledge base is clearly not consistent if the assumables are all true. The switches are both up, but the lights are not lit. Some of the assumabl ...

... If the system is working correctly (all assumables are true), the observations and the knowledge base are consistent (i.e., satisfiable).! The augmented knowledge base is clearly not consistent if the assumables are all true. The switches are both up, but the lights are not lit. Some of the assumabl ...

full text pdf

... intelligent? Let us return briefly to Cope’s definition: “any activity or thing doing that activity can be creative if it ‘associates two ideas heretofore not considered related but now revealed as logically connected.’” Is it creative? No: it is an information retrieval system which judges possible ...

... intelligent? Let us return briefly to Cope’s definition: “any activity or thing doing that activity can be creative if it ‘associates two ideas heretofore not considered related but now revealed as logically connected.’” Is it creative? No: it is an information retrieval system which judges possible ...

A Play on Words: Using Cognitive Computing as a

... intelligent? Let us return briefly to Cope’s definition: “any activity or thing doing that activity can be creative if it ‘associates two ideas heretofore not considered related but now revealed as logically connected.’” Is it creative? No: it is an information retrieval system which judges possible ...

... intelligent? Let us return briefly to Cope’s definition: “any activity or thing doing that activity can be creative if it ‘associates two ideas heretofore not considered related but now revealed as logically connected.’” Is it creative? No: it is an information retrieval system which judges possible ...

User Guide - bioimagerie

... III - Adjust the plugin parameters: 1. Select the input directory, which is the directory of the file containing your images. 2. Select the output directory, which is the directory of the file that will contain all the results of the analysis. 3. Select the “image type”, which is the filename exten ...

... III - Adjust the plugin parameters: 1. Select the input directory, which is the directory of the file containing your images. 2. Select the output directory, which is the directory of the file that will contain all the results of the analysis. 3. Select the “image type”, which is the filename exten ...

Chapter 1, 2 Pictions

... Chapter 1&2 Pictions (picture definitions): Skim through these chapters. For each of the following terms below; provide a definition in your own words, give an example from American history/government and create a picture to represent the important aspects of the term Federal Form of Gov’t (Chpt 1) ...

... Chapter 1&2 Pictions (picture definitions): Skim through these chapters. For each of the following terms below; provide a definition in your own words, give an example from American history/government and create a picture to represent the important aspects of the term Federal Form of Gov’t (Chpt 1) ...

Generating Sokoban Puzzle Game Levels with Monte

... One of the challenges for generating Sokoban puzzles is ensuring solvability of the generated levels. Since solving Sokoban has been shown to be PSPACE-complete, directly checking whether a solution exists for a candidate puzzle becomes intractable with increasing puzzle size. To overcome this chall ...

... One of the challenges for generating Sokoban puzzles is ensuring solvability of the generated levels. Since solving Sokoban has been shown to be PSPACE-complete, directly checking whether a solution exists for a candidate puzzle becomes intractable with increasing puzzle size. To overcome this chall ...

Cell Project

... traveling from organelle to organelle. Incorporate the function of the organelles into your game. ...

... traveling from organelle to organelle. Incorporate the function of the organelles into your game. ...

Nonograms, also known as Hanjie, Picross or Griddlers, are picture logic puzzles in which cells in a grid must be colored or left blank according to numbers at the side of the grid to reveal a hidden picture. In this puzzle type, the numbers are a form of discrete tomography that measures how many unbroken lines of filled-in squares there are in any given row or column. For example, a clue of ""4 8 3"" would mean there are sets of four, eight, and three filled squares, in that order, with at least one blank square between successive groups.These puzzles are often black and white, describing a binary image, but they can also be colored. If colored, the number clues are also colored to indicate the color of the squares. Two differently colored numbers may have a space in between them. For example, a black four followed by a red two could mean four black boxes, some empty spaces, and two red boxes, or it could simply mean four black boxes followed immediately by two red ones.Nonograms have no theoretical limits on size, and are not restricted to square layouts.