Unformatted text preview: nequality holds for n = 1. (2 pts)
Inductive hypothesis: Assume for an arbitrary positive integer n = k that
(2 pts)
Inductive step: Prove for n = k + 1 that
(2 pts)
(2 pts)
, using I. H. (3 pts)
(0 pts) , since using the inequality given it follows that
(4 pts)
This completes proving the inductive step. It follows that the original assertion holds for
all positive integers n. Page 2 of 4 Spring 2012 Discrete Structures Exam, Part A 2) (15 pts) PRF (Logic)
Prove that the following logical expression is a tautology using the laws of logic equivalence
and the definition of the conditional statement only. Show each step and state which rule is
being used. Note: You may combine both associative and commutative laws in a single step, s o
long as you do this properly. Solution )
(
T
T defn of implication
Double Negation
DeMorgan’s Law
Commutative Law
Inverse Law (using
Domination Law (3 pts)
(2 pts)
(3 pts)
(2 pts)
) (3 pts)
(2 pts) Grading: Give positive points for steps in the correct direction, roughly the steps are
outlined above. Students may use mo...
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 Spring '09
 Computer Science, Addition, pts, Natural number, Peano axioms

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