Exam 1 Flashcards by Alexey Filatov | Brainscape

Q

!(p ^ q)

A

!p v !q (De Morgan)

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Q

conjecture of p and q

A

p ^ q

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Q

p ^ T == p

A

Identity Laws

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Q

p v F == P

A

Identity Laws

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Q

p v T == T

A

Domination Laws

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Q

p ^ F ==F

A

Domination Laws

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Q

p v !p == T

A

Negation Laws

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Q

p v p == p

A

Idempotent Laws

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Q

p ^ p == p

A

Idempotent Laws

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Q

!(!p) == p

A

Double Negation Laws

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Q

p v q

A

!p -> q

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Q

p v q == q v p

A

Commutative Laws

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Q

p ^ q == q ^ p

A

Commutative Laws

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Q

p ^ !p == F

A

Negation Laws

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Q

(p v q) v r == p v (q v r)

A

Associative Laws

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Q

p v (q ^ r) == (p v q) ^ (p v r)

A

Distributive Laws

Q

p v (p ^ q) == p

A

Absorption Laws

Q

p -> q

A

!p v q

Q

!(p v q)

A

!p ^ !q (De Morgan)

Q

p ^ q

A

!(p -> !q)

Q

!q ^ (p ->q) -> !p

A

Modus Tollens

Q

p -> q

A

!q -> !p

Q

!ExP(x)

A

Ax!P(x)

Q

!AxP(x)

A

Ex!P(x)

Q

p ^ (p -> q) -> q

A

Modus Ponenss

Q

((p->q) ^ (q->r ))-> (p->r)

A

Hypothetical syllogism

Q

p->(p v q)

A

Addition

Q

(p ^ q) -> P

A

Simplification

Q

((p) ^ (q)) -> (p ^ q)

A

Conjunction

Q

((p v q) ^ (!p v r)) -> (q v r)

A

Resolution

Q

even integer

A

if there exists an integer k such tat n=2k

Q

odd integer

A

if there exists an integer such that n = 2k + 1

Q

perfect square

A

if there exists an integer p such that a=b^2

Q

Proof by contradiction

A

p->q == !q -> !p show that contrapositive is true

Q

rational number

A

if there exists an integers p and q with q != 0 such that r = p/q

Q

irrational number

A

if a number can not be made via p/q