Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Set up truth tables for the statements
P ⇒ (P ⇒ Q), (P ⇒ Q) ⇒ Q, P ∧ (∼ Q), (∼ P) ∨ (∼ Q)
Are any of these statements equivalent?
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- Use laws of logic to prove that (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ p ∧ ¬r.arrow_forward1. Prove or refute: (∀x, y, z, w)(B(x, y, z) ∧B(w, y, z) →B(x, w, y)).2. Prove or refute: (∀x, y, z, w)(B(x, y, z) ∧B(x, w, y) →B(w, y, z)).arrow_forwardSet up an explicit one-to-one matching between the rows of a truth tablefor a statement form in n variables, say, p1, p2, . . . ,pn, and the set of all subsetsof the set {PI, P2, . . ,Pn}.arrow_forward
- Prove ⊢ (¬A → A) → A in Hilbert deductive system. Note: In addition to the axioms and rule of inference of H, you may use any of the derived rules and/or theorems 3.20-3.30 (as numbered in the textbook). You may not use theorem 3.31, as this is precisely that theorem.book Mordechai Ben-Ari Mathematical Logic for Computer Science Third Edition Prove {¬A} ⊢ (¬B → A) → B in Hilbert deductive system. Note: In addition to the axioms and rule of inference of H, you may use any of the derived rules and/or theorems 3.20-3.30 (as numbered in the textbook). book Mordechai Ben-Ari Mathematical Logic for Computer Science Third Edition PLEASE solve these with the help of 3 axioms and 1 rule of inference with the derived proof 3.20-3.30.arrow_forwarda. Correctness of dynamic programming algorithm: Usually, a dynamic programming algorithm can be seen as a recursion and proof by induction is one of the easiest way to show its correctness. The structure of a proof by strong induction for one variable, say n, contains three parts. First, we define the Proposition P(n) that we want to prove for the variable n. Next, we show that the proposition holds for Base case(s), such as n = 0, 1, . . . etc. Finally, in the Inductive step, we assume that P(n) holds for any value of n strictly smaller than n' , then we prove that P(n') also holds. Use the proof by strong induction properly to show that the algorithm of the Knapsack problem above is correct. b. Bounded Knapsack Problem: Let us consider a similar problem, in which each item i has ci > 0 copies (ci is an integer). Thus, xi is no longer a binary value, but a non-negative integer at most equal to ci , 0 ≤ xi ≤ ci . Modify the dynamic programming algorithm seen at class for this…arrow_forwardConstruct proof for the following argument within the system of sentential logic: 1. (A & B) ⊃ (C V D) Premise2. ~(C V (B ⊃ X)) Premise3. ~[D ≡ ~(X & Y)] Premise4. ~A ⊃ ~Z Premise /: . ~Zarrow_forward
- Discrete Math Construct a truth table for: ∼ (p → q) ∧ p Is the statement ∼ (p ∨ q) ∧ p a tautology, contradiction, or neither?arrow_forwardWhich connective in the following formula has the narrowest scope? Which has the widest? What does the negation have scope over? A ⟶ (B ∧ ¬(C ⟷ (D ∨ E)))arrow_forwardWrite the compound proposition (¬p) → (q ∨ (r ∧ ¬s)) in a. Postfix notationb. Prefix notationc. Infix notationarrow_forward
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