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Summary Hunter college Discrete structures study guide

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  • Summary Hunter College Discrete Structures Study G

Summary Hunter college Discrete structures study guide

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  • September 25, 2024
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  • Summary Hunter college Discrete structures study g
  • Summary Hunter college Discrete structures study g
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Summary Hunter college Discrete structures study guide

power set - ANSWER The set of all subsets of a set A, denoted P(A). If A = {a, b},
then P(A) = {{a}, {b}, {a, b}, ∅}. The cardinality of P(A) is 2 raised to the cardinality of
A.

cardinality - ANSWER Denoted | A |, it is the number of distinct elements in A.

disjoint sets - ANSWER A ∩ B = ∅.

set difference - ANSWER The set of elements {x∈A | x∉B}.

implication - ANSWER Equivalent to ∼p ∨q and is false only when p is true and q is
false.

tautology - ANSWER A proposition that is always true, like p ∨ ~p.

contradiction - ANSWER A proposition that is always false, like p ∧ ~p.

disjunctive normal form (DNF) - ANSWER A form of writing propositions with a
disjunction of conjunctions where (1) every variable or its negation appears once in
each conjunction, called a minterm, and (2) each minterm appears only once.

dual - ANSWER A related to a proposition S, denoted S*, where
(1) ∧ becomes ∨
(2) ∨ becomes ∧
(3) T becomes F
(4) F becomes T.
Note that (S*)* = S. If S = T, then S* = T*.

functional complete set - ANSWER Every proposition is logically equivalent to
another using only the operators in the set. Examples of these sets include {~, ∧},
{NAND}, {NOR}, etc.

universal quantifier - ANSWER Denoted by ∀. The statement ∀x P(x) reads "P(x) for
all x in the universe of discourse."

existential quantifier - ANSWER Denoted by ∃. The statement ∃x P(x) reads "There
exists an x in the domain of discourse such that P(x)."

unique existential - ANSWER Denoted ∃!. The statement ∃! x P(x) reads "There is a
unique x such that P(x)."
Equivalently, ∃x∀y (P(x) ∧ P(y) → x = y).

De Morgan's Law for Quantifiers - ANSWER Negating a predicate forces ∀ to
become ∃ and vice versa.

, (1) ~∀x P(x) ≡ ∃x ~P(x)
(2) ~∃x P(x) ≡ ∀x ~P(x)

modus ponens - ANSWER 1. p→q
2. p
∴q

modus tollens - ANSWER 1. p → q
2. ~q
∴ ~p

direct proof - ANSWER For p → q, assume that p is true, then prove that q must be
true.

indirect proof - ANSWER For p → q, apply direct proof on the contrapositive ~q →
~p.

proof by contradiction - ANSWER To prove a statement p, assume ~p and show
that it leads to a contradiction.

constructive proof of existence - ANSWER 1. P(a)
∴ ∃x P(x).
Finding this P(a) can be an example or a general solution.

non-constructive proof of existence - ANSWER Proof using theorems to guarantee
existence or use of indirect proof like proof by contradiction. However, no actual
example is ever provided.

proof of unique existence - ANSWER To prove ∃! x P(x), prove existence first, then
assume a proposition that contradicts uniqueness and disprove it. It is identical to
proving these two parts:
∃x (P(x) ∧ ∀y (P(y) → y = x))

mathematical induction - ANSWER A method of proof in two parts:
1. basis step, P(a)
2. inductive step, P(k) → P(k+1)
∴ ∀n P(n)

strong induction - ANSWER Another form of induction:
1. basis step, P(a)
2. inductive step, P(1) ∧ ... ∧ P(k) → P(k+1)
∴∀n P(n)

structural induction - ANSWER A method of proof on recursively defined structures:
1. basis step, proof that P(x) for all x in the basis step of recursion
2. inductive step, proof that all future iterations preserve the property from before
∴ ∀n P(n)

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