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DISCRETE MATH STUDY EXAM TEST BANK QUESTIONS WITH CERTIFIED ANSWERS

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DISCRETE MATH STUDY EXAM TEST BANK QUESTIONS WITH CERTIFIED ANSWERS Logical Equivalences (Conditionals) - Answer-p → q ≡ ¬p ∨ q p → q ≡ ¬q →¬p p ∨ q ≡ ¬p → q p ∧ q ≡ ¬(p →¬q) ¬(p → q) ≡ p ∧¬q (p → q) ∧ (p → r) ≡ p → (q ∧ r) (p → r) ∧ (q �...

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  • October 22, 2024
  • 5
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

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  • discrete math
  • discrete math study exam test bank questions

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DISCRETE MATH STUDY EXAM TEST
BANK QUESTIONS WITH CERTIFIED
ANSWERS

Logical Equivalences (Conditionals) - Answer-p → q ≡ ¬p ∨ q p
→ q ≡ ¬q →¬p
p ∨ q ≡ ¬p → q
p ∧ q ≡ ¬(p →¬q)
¬(p → q) ≡ p ∧¬q
(p → q) ∧ (p → r) ≡ p → (q ∧ r)
(p → r) ∧ (q → r) ≡ (p ∨ q) → r
(p → q) ∨ (p → r) ≡ p → (q ∨ r)
(p → r) ∨ (q → r) ≡ (p ∧ q) → r

Logical Equivalances (Biconditionals) - Answer-p ↔ q ≡ (p → q) ∧ (q → p) p
↔ q ≡ ¬p ↔¬q
p ↔ q ≡ (p ∧ q) ∨ (¬p ∧¬q)
¬(p ↔ q) ≡ p ↔¬q

Demorgan Law extensions - Answer-¬(p1 ∨ p2 ∨ · · · ∨ pn) ≡ (¬p1 ∧¬p2 ∧ ··· ∧ ¬pn)
¬(p1 ∧ p2 ∧ · · · ∧ pn) ≡ (¬p1 ∨¬p2 ∨ ··· ∨ ¬pn)

Propositional Satisfiability - Answer-A compound proposition is satisfiable if there is an
assignment of truth values to its variables that
makes it true. When no such assignments exists, that is, when the compound
proposition is false
for all assignments of truth values to its variables, the compound proposition is
unsatisfiable.

Predicate, Quantifier - Answer-x = y + 3, "computer x is under attack by an intruder"
x and y are variables. The part other than variables is predicate.

PRECONDITIONS AND POSTCONDITIONS - Answer-The statements that describe
valid input are known
as preconditions and the conditions that the output should satisfy when the program has
run
are known as postconditions.

, Universal quantifier - Answer-For all values of a variable in a particular domain, called
the domain of discourse (or
the universe of discourse)
"for all", "for every," "all of," "for each," "given any," "for arbitrary," "for each," and "for
any."
Ex: ∀xN(x)
Existential quantifier - Answer-We form a proposition that is true if and only if P(x) is true
for at least one value of x in the domain.
"There exists a such that ".
∃xP(x)

∃, ∀ - Answer-"∀" applies the variable over every values of the domain (conjunction)
"∃" applies the variable over at least one of the values of the domain (disjunction)

Quantifiers with Restricted Domains - Answer-∀x < 0 (x2 > 0) {∀x(x < 0 → x2 > 0)}
∃z > 0 (z2 = 2) {∃z(z > 0 ∧ z2 = 2)}

Precedence of Quantifiers - Answer-∀ and ∃ have higher precedence than all logical
operators from propositional calculus.
∀xP(x) ∨ Q(x) ≡ (∀xP(x)) ∨ Q(x)

Bound, Free Variables - Answer-∃x(x + y = 1)
x is bound because a quantifier is used over it. But y is free.

Logical Equivalences Involving Quantifiers - Answer-Statements involving predicates
and quantifiers are logically equivalent if and only if they
have the same truth value no matter which predicates are substituted into these
statements
and which domain of discourse is used for the variables in these propositional functions.
Notation used: S ≡ T

Negating Quantified Expressions - Answer-¬∀xP(x) ≡ ∃x ¬P(x)
¬∃xQ(x) ≡ ∀x ¬Q(x)

Equivalence of compound functions - Answer-∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x)
∃x(P(x) ∨ Q(x)) ≡ ∀xP(x) ∨ ∀xQ(x)
¬∀x(P(x) → Q(x)) ≡ ∃x(P(x)∧¬Q(x))

Nested Quantifiers - Answer-∀x∀yP(x, y) ≡ ∀y∀xP(x, y)
∃x∃yP(x, y) ≡ ∃y∃xP(x, y)
∀x∃yP(x, y) =! ∃y∀xP(x, y) {remember x+y=0 example the RHS makes no sense}
∃y∀xP(x, y) → ∀x∃yP(x, y)

Proposition - Answer-A proposition is a declarative sentence (that is, a sentence that
declares a fact) that is either true
or false, but not both. It is a basic building block of logic.

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