Obfuscation of Abstract Data-Types - Rowan

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CHAPTER 5. SETS AND THE SPLITTING 101 For this proof tree, the cost and height are as follows: C (insert) = 5 + C (Prop 9) = 9 and H (insert) = 3 + H (Prop 9) = 6 5.4.2 Second Definition We prove Equation (5.6) for insert1 by induction on xs. Base Case Suppose that xs = [ ]. Then insert1 a (insert1 a [ ]) = {definition of insert1} insert1 a [a] = {definition of insert1} [a] = {definition of insert1} insert1 a [ ] Step Case Suppose that xs = y : ys and for the induction hypothesis, we assume that ys satisfies Equation (5.6). We have three subcases. Subcase 1 Suppose that a > y then insert1 a (insert1 a (y : ys)) = {definition of insert1} insert1 a (y : (insert1 a ys)) = {definition of insert1} y : (insert1 a (insert1 a ys)) = {induction hypothesis} y : (insert1 a ys) = {definition of insert1} insert1 a (y : ys) Subcase 2 Suppose that a = y then insert1 a (insert1 a (y : ys)) = {definition of insert1} insert1 a (y : ys) Subcase 3 Suppose that a < y then insert1 a (insert1 a (y : ys)) = {definition of insert1} insert1 a (a : y : ys) = {definition of insert1}
CHAPTER 5. SETS AND THE SPLITTING 102 a : y : ys = {definition of insert1} insert1 a (y : ys) The tree for this proof is: insert1 Base Step Case 1 Step Case 2 Step Case 3 where Base insert1 insert1 and insert1 Step Case 1 insert1 insert1 Step Case 2 insert1 Step Case 3 insert1 insert1 IH insert1 So, insert1 C (insert1) = 11 and H (insert1) = 5 5.4.3 Third Definition We prove Equation (5.6) for insert asp — so we let xs = 〈l,r〉 asp and (ly,lz) = span (< a) l (ry,rz) = span (< a) r We have four cases. Case 1 Suppose that member lz a. By (B.3 ⇒) (proved in Appendix B.1) |ly| = |ry| and so, insert asp a (insert asp a 〈l,r〉 asp ) = {definition of insert asp } insert asp a (〈ly,ry〉 asp + asp 〈lz,rz〉 asp ) = {definition of + asp with |ly| = |ry|} insert asp a (〈ly + lz,ry + rz〉 asp ) = {property of span (Property 9)} insert asp a 〈l,r〉 asp
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Obfuscation of Abstract Data-Types
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Contents I The Current View of Obfu
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5.4.3 Third Definition . . . . . .
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List of Figures 1 An example of obf
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8 public class c { d t1 = new d();
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10 • Obfuscations can be proved c
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Chapter 1 Obfuscation Scully: “No
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CHAPTER 1. OBFUSCATION 14 These thr
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CHAPTER 1. OBFUSCATION 16 by changi
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CHAPTER 1. OBFUSCATION 18 We could
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CHAPTER 1. OBFUSCATION 20 B 1 [0] =
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Chapter 2 Obfuscations for Intermed
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CHAPTER 2. OBFUSCATIONS FOR INTERME
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Chapter 3 Techniques for Obfuscatio
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CHAPTER 3. TECHNIQUES FOR OBFUSCATI
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CHAPTER 3. TECHNIQUES FOR OBFUSCATI
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Page 143 and 144: BIBLIOGRAPHY 142 [43] Simon Peyton
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APPENDIX A. LIST ASSERTION 152 1 +
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APPENDIX A. LIST ASSERTION 154 0 +
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APPENDIX A. LIST ASSERTION 156 The
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APPENDIX A. LIST ASSERTION 158 The
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APPENDIX B. SET OPERATIONS 160 Now
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APPENDIX B. SET OPERATIONS 162 Proo
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APPENDIX B. SET OPERATIONS 164 B.2
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Appendix C Matrices We prove the as
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APPENDIX C. MATRICES 168 Base Case
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APPENDIX D. PROVING A TREE ASSERTIO
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APPENDIX E. OBFUSCATION EXAMPLE 186
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Obfuscation of Abstract Data-Types - Rowan

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