Consider the predicates Martian (x): x is a Martian isGreen(x): x is green Use equivalence laws of first-order logic to identify the expression that is logically equivalent to ∀x (Martian(x) ˄ isGreen(x)) Group of answer choices ∃x (Martian(x)) ˅ ∃x (isGreen(x)) ∃x (¬Martian(x)) ˄ ∃x (¬isGreen(x)) ∀x (Martian(x)) ˄ ∀x (isGreen(x)) ∀x (¬Martian(x)) ˄ ∀x (¬isGreen(x))

a. 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…

Construct 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     /: .  ~Z

Logic, set and relations. Discrete math using Latex