Reasoning

Proof of P<==>Q

(i) Show P=>Q (i) Show Q=>P Therefore P<==>Q. or Prove the statement using only <==>(equivalent statements).

Theorem 2.1.2

If A and B are sets with no elements then A=B.

Thereom 2.2.2

(A^c)^c=A A∪A^c=U A∩A^c={} A-B=A∩B^c (A∪B)^c=(A^c)∩(B^c) (A∩B)^c=(A^c)∪(B^c) A∩B={} iff A⊆B^c

Common Incorrect Deductions

(Ex) P (x)⇒(∀x) P(x) (∀x)[ P (x)∨ Q (x)]⇒[(∀x) P (x)∨(∀x) Q (x)] [(∀x) P (x)⇒(∀x) Q (x)]⇒(∀x)[ P (x)⇒ Q (x)] (∀y)(Ex) P (x, y)⇒(Ex)(∀y) P (x,y)

Theorem 2.1.1

(a) For every set A, {}⊆A (b) For every set A, A⊆A (c) For all sets A, B, and C, if A⊆B and B⊆C then A⊆C

Tautology

A tautology is a propositional form that is true for every assignment of truth values to its components.

Conjunction and Disjunction

Given propositions P and Q, the conjunction of P and Q, denoted is the proposition "P and Q." is true exactly when both P and Q are true. The disjunction of P and Q, denoted is the proposition"P or Q." is true exactly when at least one of P or Q is true *The English words but, while, and although are usually translated symbolically with the conjunction connective, because they have the same meaning as and.

Proposition

Is a sentence that has exactly one truth value

Valid Manipulations of quantifiers

1.(∀x)(∀y) P (x, y) ⇐⇒(∀y)(∀x) P (x, y). 2. (Ex)(Ey) P (x, y) ⇐⇒(Ey)(Ex) P (x, y). 3. [(∀x) P (x)∨(∀x) Q (x)]⇒(∀x)[ P (x)∨ Q (x)]. 4. (∀x)[ P (x)⇒ Q (x)]⇒[(∀x) P (x)⇒(∀x) Q (x)]. 5. (∀x)[ P (x)∧ Q (x)] ⇐⇒[(∀x) P (x)∧(∀x) Q (x)]. 6. (Ex)(∀y) P (x, y)⇒(∀y)(Ex) P (x, y)

Contradicition

A contradiction is a propositional form that is false for every assignment of truth values to its components.

Open Sentence/Predicate

A sentence that contains variables and becomes a proposition only when its variables are assigned specific values. P(x_1,x_2 . . ., x_n)

Set Axioms

Axiom of Pairing: collection of two sets is a set Axiom of Powers: the collection of all subsets of a set is a set.

Set Operation Definition

A−B={x: xϵA and xϵ'B}. A∩B={x: xϵA and xϵB}. A∪B={x: xϵA or xϵB}

Direct proof of A⊆B

A⊆B ⇐⇒(∀x)(xϵA⇒xϵB) Let x be any object. Suppose x ϵ A ... Thus x ϵ B. Therefore A⊆B.

Proof Methods for (∃x)P(x)

Constructive: just show one example Another strategy to prove is to show that there must be some object for which is true, without ever actually producing a particular object. Contradiction: Suppose ~(∃x)P(x). then (∀x)~P(x). [proceed with proof by contradiction]. sometimes it will be necessary to provide some variable object that holds the property.

Theroem 2.2.1

For all sets A, B and C: A⊆A∪B A∩B⊆A A∩{}={} A∪{}=A A∩A=A A∪A=A A-{}=A {}-A={} A⊆B iff A∪B=B A⊆B iff A∩B=A If A⊆B then A∪C⊆B∪C If A⊆B then A∩C⊆B∩C Commutative Laws: A∪B=B∪A A∩B=B∩A Associative Laws: A∪(B∪C)=(A∪B)∪C A∩(B∩C)=(A∩B)∩C Distributive Laws: A∩(B∪C)=(A∩B)∪(A∩C) A∪(B∩C)=(A∪B)∩(A∪C)

Universal Quantifier

For an open sentence P(x) the sentence (∀x)P(x) is read "For all x, P(x)" and is true iff the truth set of P(x) is the entire universe. The symbol is called the universal quantifier.

Existential Quantifier

For an open sentence P(x) the sentence (∃x)P(x) is read "There exists x such that P(x) or "For some x, P(x). The sentence (∃x)P(x) is true iff the truth set of P(x) is nonempty. Th symbol ∃ is called the existential quantifier.

Theorem 2.1.3

For any sets A and B, if A⊆B and A!={} then B!={}.

Biconditional

For propositions P and Q, the biconditional sentence P ⇐⇒ Q is the proposition "P if and only if Q." is true exactly when P and Q have the same truth values. We also write P Q to abbreviate P if and only if Q.

Conditional

For propositions P and Q, the conditional sentence P ⇒ Q is the proposition "If P, then Q." Proposition P is called the antecedent and Q is the consequent. The conditional sentence is true if and only if P is false or Q is true.

Thereom 1.2.2

For propositions P, Q, and R, (a) P⇒Q is equivalent to ~P ∨ Q. (b) P⇐⇒Q is equivalent to (P ⇒ Q) ∧ (Q ⇒ P). (c)~(P⇒Q) is equivalent to P ∧ ∼Q. (d) ~(P∧Q) is equivalent to P ⇒ ∼Q and to Q ⇒ ∼P (e) P ⇒ (Q ⇒ R) is equivalent to (P ∧ Q) ⇒ R (f) P ⇒ (Q ∧ R) is equivalent to (P ⇒ Q) ∧ (P ⇒ R) (g) (P ∨ Q) ⇒ R is equivalent to (P ⇒ R) ∧ (Q ⇒ R) P ⇒ (Q ∨ R) is equivalent to (P∧~Q)⇒R

Theorem 1.1.1

For propositions P, Q, and R, the following are equivalent: Double Negation Law (a) P and ~(~P) Commutative Laws (b) P ∨ Q and Q ∨ P (c) P ∧ Q and Q ∧ P Associative Laws (d) P ∨ (Q ∨ R) and (P ∨ Q) ∨ R (e) P ∧ (Q ∧ R) and (P ∧ Q) ∧ R Distributive Laws (f) P ∧ (Q ∨ R) and (P ∧ Q) ∨ (P ∧ R) (g) P ∨ (Q ∧ R) and (P ∨ Q) ∧ (P ∨ R) DeMorgan's Laws (h) ∼(P ∧ Q) and ∼P ∨ ∼Q (i) ∼(P ∨ Q) and ∼P ∧ ∼Q

Four rules of writing proofs

In any proof you may: At any time state an assumption, an axiom, or a previously proved result. At any time state a sentence equivalent to any statement earlier in the proof. At any time state a sentence whose symbolic translation is a tautology. At any time after P=>Q and P appear in a proof, state that Q is true.(modus ponens)

Contrapositive

Let P and Q be propositions.The contrapositive of P⇒ Q is (∼Q) ⇒ (∼P). A sentence and it's contrapositive are equivalent.

Converse

Let P and Q be propositions.The converse of P⇒ Q is Q ⇒ P. A sentence and it's converse are not equivalent.

Definition of complement

Let U be the universe and The complement of A is the set A^c. Where A^c=U−A

Direct proof of (∀x)P(x)

Let x be an arbitrary object in the universe. (The universe should be named or its objects described.) [Proceed with normal Direct Proof] ... Hence P(x) is true. Since x is arbitrary, (∀x)P(x) is true.

Null Set

Let {}={x|x!=x}. Then {} is a set, called the empty set or null set.

Some ways to say P⇐⇒Q.

P if and only if Q. P if, but only if, Q. P is equivalent to Q. P is necessary and sufficient for Q.

Some ways to say P⇒Q.

P ⇒ Q If P, then Q. If a>5 then a>3 P implies Q. a>5 implies a>3 P is sufficient for Q. a>5 is sufficient for a>3 P only if Q. a>5 only if a>3 Q, if P. a>3 if a>5 Q whenever P. a>3 whenever a>5 Q is necessary for P. a>3 is necessary for a>5 Q, when P. a>3 when a>5

Proof by Contradiciton

P<==>(~P=>(~Q and Q)) The logic behind such a proof is that if a statement cannot be false, then it must be true. Form: Suppose ~P ... Therefore Q ... Therefore ~Q. Hence Q and ~Q is a contradiction. Thus P. Note: This method of proof can be applied to any proposition P(not just statements).

Direct Proof

P=>Q Form: Assume P...Therefore Q. Thus P=>Q Strategy: 1. Determine precisely the hypotheses (if any) and the antecedent and consequent. 2. Replace (if necessary) the antecedent with a more usable equivalent. 3. Replace (if necessary) the consequent by something equivalent and more readily shown. 4. Beginning with the assumption of the antecedent, develop a chain of statements that leads to the consequent. Each statement in the chain must be deducible from its predecessors or other known results. Hint: Useful to try and work backwards from the consequent at the same time you are working to it from the antecedent, moving to a common point. P=>(Q and/or R) requires P=>Q and/or P=>R for D.Proof (or the proof of an equivalent statement!) (P and Q)=>R assume P and Q and work to R for D.proof NOTE: DON'T FORGET ANY CASE!

Proof by Contraposition

P=>Q <==> ~Q=>~P Form: Assume ~Q...Therefore ~P. Thus ~Q=>~P . Therefore by by contraposition P=>Q. When to use: when the denials of P and Q are easier to understand or when the statement of either P or Q is itself a negation.

Proof by Contradiction of (∀x)P(x)

Suppose ~(∀x)P(x) Then (∃x)~P(x) Let t be an object such that ~P(t). Therefore (Q and ~Q) Thus (∃x)~P(x) is false, so (∀x)P(x) is true

Truth Set

The collection of objects that may be substituted to make an open sentence a true proposition. *Before a truth set can be determined, we must be given or must decide what objects are available for consideration; that is, we must have specified a universe of discourse.

Negation

The negation of a proposition P, denoted ∼P, is the proposition "not P." The proposition ∼P is true exactly when P is false.

Equivalent

Two propositional forms are equivalent if and only if they have the same truth tables.

Equivalancy of Predicates

With a universe specified, two open sentences P(x) and Q(x) are equivalent iff they have the same truth set.