Discrete Mathematics Using a Computer - 06 - Propositional Logic Notes

8/7/2019 Discrete Mathematics Using a Computer - 06 - Propositional Logic Notes

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DiscreteMathComputer 06 - Propositional Logic

Kanak Kshetri

1 Chapter Outline

• Difficulties with informal logic

• Propositional logic (one type of formal logic)

• Three mathematical systems for reasoning about propositions:

– Truth Tables: easy to understand but unwieldy for large expressions; “semantic

technique”

– Natural Deduction: “syntactic technique”, uses inference rules

– Boolean Algebra: axiomatic approach

2 Need for Formalism

• Natural language has ambiguities and subtleties

– e.g. “and” can sometimes suggest an implication as in “The sun is shining and Iam happy”

• Proposition: variable whose value must be either True or False

• Logical Operators: and, or, not, implies . Operate on propositions

• Statements that can’t be represented by propositional logic (due to lack of context):

– e.g. “That cloud looks like a motor bike”

– context required: which cloud? who thinks that it looks like a motor bike?

3 Basic Logical Operators

3.1 Logical And (∧)• aka Conjunction

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3.1.1 Truth Table

A B A ∧ B

False False FalseFalse True TrueTrue False False

True True True

3.2 Inclusive Or (∨)

• aka Disjunction

3.2.1 Truth Table

A B A ∨ B

False False FalseFalse True TrueTrue False True

True True True

3.3 Exclusive Or (⊕)

• Either one or the other, but not both

3.3.1 Truth Table

A B A ⊕ B

False False FalseFalse True TrueTrue False True

True True False

3.4 Logical Not (¬)

3.4.1 Truth Table

A ¬A

True FalseFalse True

3.5 Logical Implication (⇒)

• Use to represent conditional phrases

• e.g. “If it is sunny today, then there will be a picnic.”

• Implication does not say anything about cause-and-effect

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3.5.1 Truth Table

A B A ⇒ B

False False TrueFalse True TrueTrue False False

True True True

3.5.2 Understanding the truth table

• Why is an implication true even if A is False?

• One way to think about is that an implication is a statement that if something happensthen something else must also happen

– So, if the first event happens, but the second doesn’t, the implication must havebeen faulty

– But the first event doesn’t happen at all, then we’re indifferent about the implica-

tion

• Rephrasing the above point, “If A then B” is like saying that “A” is sufficient for B tooccur

– So, if we see an A but not a B, clearly A isn’t sufficient for B

– But if we see a B on its own, maybe something other than A caused it.

– e.g. Matchsticks are sufficient for fire. But just because you use a lighter to makea fire doesn’t mean matchsticks can’t cause fire. However, if you have a matchstickand still can’t produce fire, it clearly isn’t sufficient.

• Book is a little handwavy and just says “this is the way it is”

3.6 Logical Equivalence (⇔)

• Use to make a claim that two propositions always have the same value, or are equivalent

• aka “If and only if”, “necessary and sufficient”

3.6.1 Truth Table

A B A ⇔ B

False False TrueFalse True False

True False FalseTrue True True

3.6.2 "If and only If"

• A ⇔ B can be written as (A ⇒ B) ∧ (B ⇒ A)

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4 The Language of Propositional Logic

4.1 Well-Formedness

• Have a grammar for propositional logic

•

If a formula doesn’t follow this grammar, it isn’t well formed• e.g. Well-Formed Formula: A ⇒ (B ∧ (¬A))

• e.g. not Well-Formed Formula: ∧AB¬C

• Well-formedness is only about the syntax and says nothing about the semantics

of the formula

– e.g. “if it compiles”, then it is well-formed. Just because something compiles doesn’tmean it’ll pass all the tests

4.2 Syntax of Well-Formed Formulas

• Constants False and True are WFF.

• Any propositional variable is WFF.

• Negation: If a is WFF, then ¬a is WFF

• Conjunction: If a and b are WFF, then a ∧ b is WFF

• Disjunction: If a and b are WFF, then a ∨ b is WFF

• Implication: If a and b are WFF, then a ⇒ b is WFF

• Equivalence: If a and b are WFF, then a ⇔ b is WFF

4.2.1 Example: P ⇒ (Q ∧ R) is a WFF

• P,Q,R are WFF because they’re propositional variables

• So, Q ∧ R is a WFF (conjunction rule)

• So, P ⇒ (Q ∧ R) is a WFF (implication rule)

4.3 Order of Precedence

• Negation (¬) is the most tightly binding operator

• Conjunction (∧)

• Disjunction (∨)

• Implication (⇒)

• Equivalence (⇔) is the least tightly binding operator

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4.4 Object Language and Meta Language

• WFFs of propositional language will be called object language

– They refer to sentences in propositional logic

• Language of equations, substitutions, and justifications will be called the meta lan-

guage

– The surround the object language

– They’re used to talk about propositions

4.5 Computing with Boolean Expressions

• ¬x is writtenn o t x

• a ∧ b is writtena & & b

• a ∨ b is writtena | | b

• a ⇒ b is writtena = = > b

• a ⇔ b is written a < = > b

4.6 (Exercise 1) : Lying Bus Driver

4.6.1 Scenario

• Want to go to the airport, are at the bus stop

• Either this bus or the next is going to the airport, but not both

• Can only ask one “yes-or-no” question

• Some drivers will answer “yes” when the correct answer is “no”

– You don’t know if this bus driver is one of those people

• Ask one “yes or no” question that can help you decide whether to take this bus or thenext

4.6.2 Solution 1

• I first gave names to the situations:

– A: this bus goes to the airport– So, ¬A means the next one goes to the airport

– L: this driver is a liar

– So, ¬L means this driver answers truthfully

• I want to formulate my question so that if the answer is “True”, then A.

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• I then made a table of the possible situations:

L A Desired Answer

True True TrueTrue False FalseFalse True TrueFalse False False

• So, clearly just asking A gives me the desired answer regardless of what L is

4.7 (Exercise 2) : Evaluating propositional formulas

• False ∧ True

– False

• True ∨ (¬True)

– True

• ¬(False ∨ True)

– ¬True

– False

• (¬(False ∧ True)) ∨ False

– (¬False) ∨ False

– True ∨ False

– True

• (¬True) ⇒ True

– False ⇒ True

– True

• True ∨ False ⇒ True

– True ⇒ True

– True

• True ⇒ (True ∧ False)

– True ⇒ False

– False

• False ⇒ False

– True

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• ¬False ⇔ True

– True ⇔ True

– True

• True ⇔ (False ⇒ False)

– True ⇔ True

– True

• False ⇔ (True ∧ (False ⇒ True))

– False ⇔ (True ∧ True)

– False ⇔ True

– False

• ¬(True ∨ False) ⇔ False ∧ True

– ¬True ⇔ False

– False ⇔ False

– True

4.7.1 Verification using the Book’s Stdm Library

∗Stdm> False /\ True

False

∗Stdm> True \ / ( not True)True

∗

Stdm> not ( False \ / True )False

∗Stdm> ( not ( False / \ True ) ) \ / False

True

∗Stdm> ( not True) ==> True

True

∗Stdm> True \ / False ==> True

True

∗Stdm> True ==> ( True / \ False )False

∗Stdm> False ==> False

True

∗Stdm> ( not False ) <=> True

True

∗Stdm> True <=> ( False ==> False )True

∗Stdm> False <=> ( True / \ ( False ==> True ) )False

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∗Stdm> ( not ( True \ / False ) ) <=> False / \ True

True

∗Stdm>

5 Truth Tables: Semantic Reasoning

• Table

• Column headers are all the variables (and expressions)

• Add a row for each possible setting of values for variables

– 3 variables means 23 = 8 rows

– because 2 possible settings for each of the 3 variables

5.1 Example: Calculating using a Truth Table

• Suppose we want to know when the proposition ((A ⇒ B) ∧ ¬B) ⇒ ¬A is true

• Make a Truth TableA B A ⇒ B ¬B (A ⇒ B) ∧ ¬B ¬A ((A ⇒ B) ∧ ¬B) ⇒ ¬A

F F T T T T TF T T F F T TT F F T F F TT T T F F F T

5.2 Definition: Tautology

• Proposition that is always True, regardless of its variables

5.3 Definition: Contradiction

• Proposition that is always False, regardless of its variables

5.4 Definition: Entailment ()

• P 1, P 2, · · · , P n Q means that if all the propositions P 1, P 2, · · · , P n are True, then Q isalso True

5.4.1 Discussion

• makes a statement about actual meanings of propositions

• Model of a logic system: Set of truth values along with the method for calculatingmeaning of any well-formed formula

• Different values of n

– n = 1 gives P Q, which has the same semantics as ⇒

– n = 0 gives Q, which means that Q is always True, or is a tautology

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5.5 Limitations of Truth Tables

• Size is exponential in number of variables

• Other techniques (natural deduction, Boolean algebra) give much better insight intowhy a proposition is true

5.6 Exercise 3: Determine which formulas are tautologies using truth ta-bles

5.6.1 (True ∧ P ) ∨ Q

P Q True ∧ P (True ∧ P ) ∨ Q

T T T TT F T TF T F TF F F F

• not a tautology

• without truth tables: (True ∧ P ) ∨ Q becomes P ∨ Q, which is False when both P and Q

are False

5.6.2 (P ∨ Q) ⇒ (P ∧ Q)

P Q P ∨ Q P ∧ Q (P ∨ Q) ⇒ (P ∧ Q)

T T T T TT F T F FF T T F FF F F F T

• not a tautology

5.6.3 (P ∧ Q) ⇒ (P ∨ Q)

P Q P ∧ Q P ∨ Q (P ∨ Q) ⇒ (P ∧ Q)

T T T T TT F F T TF T F T TF F F F T

• Tautology

5.6.4 (P ∨ Q) ⇒ (Q ∨ P )

P Q P ∨ Q Q ∨ P (P ∨ Q) ⇒ (P ∧ Q)

T T T T TT F T T TF T T T TF F F F T

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• Tautology

• Without Truth Table: by commutativity of ∨

5.6.5 ((P ∨ Q) ∧ (P ∨ R)) ⇔ (P ∧ (Q ∨ R))

P Q R P ∨ Q P ∨ R LHS Q ∨ R RHS LHS ⇔ RHSF F F F F F F F TF F T F T F T F TF T F T F F T F TF T T T T T T F F

T F F T T T F F FT F T T T T T T TT T F T T T T T TT T T T T T T T T

• Not a tautology

5.7 Exercise 4: Determine Well-formedness, and build a truth table

5.7.1 (True ∧ P ) ∨ Q

• Truth table from Ex 3.1

• See that it is satisfiable, but not a tautology

5.7.2 ((P ∨ Q) ∧ (P ∨ R)) ⇔ (P ∧ (Q ∨ R))

• Truth table from Ex 3.5

•

Satisfiable, but not a tautology

5.7.3 ((P ∧ ¬Q) ∨ (Q ∧ ¬P )) ⇒ ¬(P ⇔ Q)

P Q A=P ∧ ¬Q B=Q ∧ ¬P A ∨ B C=(P ⇔ Q) ¬C (A ∨ B) ⇒ C

T T F F F T F TT F T F T F T TF T F T T F T TF F F F F T F T

• Tautology

5.7.4 (P ⇒ Q) ∧ (P ⇒ ¬Q)

P Q LHS=P ⇒ Q RHS=P ⇒ ¬Q LHS ∧ RHS

T T T F FT F F T FF T T T TF F T T T

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• Satisfiable, but not a tautology

5.7.5 (P ⇒ Q) ∧ (¬P ⇒ Q)

P Q LHS=P ⇒ Q RHS=¬P ⇒ Q LHS ∧ RHS

T T T T T

T F F T FF T T T TF F T F F

• Satisfiable, but not a tautology

5.7.6 (P ⇒ Q) ⇔ (¬Q ⇒ ¬P )

P Q LHS=P ⇒ Q RHS=¬Q ⇒ ¬P LHS ⇔ RHS

T T T T TT F F F T

F T T T TF F T T T

• Tautology

• (This is just the contrapositive)

6 Natural Deduction: Inference Reasoning

6.1 Introduction

• Natural deduction allows us to reason about logical propositions without substituting

values for variables

• Logical Inference: reasoning formally about a set of statements. Requires:

– Object Language: to talk about the statements we’re reasoning about. e.g. Propo-sitional expressions

– Inference Rules: Rules to infer new facts from information given

– Meta Language: Defines the “form of argument” so that we can determinewhether arguments are actually valid

6.1.1 Definition: Sequent ()

• P 1, P 2, · · · , P n Q is called a sequent

• Means: if all propositions P 1, · · · , P n are known, then proposition Q can be inferredformally using the rules of natural deduction

• e.g. Difference between P Q ⇒ P and P Q ⇒ P

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– The version doesn’t say how we know that Q ⇒ P when we assume P . It justsays that it is True.

– The version says that there is a proof that shows Q ⇒ P when we assume P

6.1.2 Meta Variables

• Meta variables can stand for any WFF proposition

• e.g. we can have a be P ∧ Q

• Convention: meta variables will be written in lower case, propositional variables inupper case

6.1.3 Writing an Inference Rule

• Assumptions are written above a horizontal line

• Conclusion written below the line.

• (For latex simplicity, I’ll just use the sequent notation to talk about inference rules)

• Inference rule only works in one direction

6.2 Definitions of True, Not and Equivalence

6.2.1 Primitives

• Only primitive objects are constant False, and ∨, ∧, ⇒

• All other objects are built using these

• Below we give definitions for some primitives

– can be verified using truth tables

– But we’ll only manipulate them using rules of logical deduction

6.2.2 Definition: True

• True = False ⇒ False

• (Verify using truth table, but this particular definition is “trivial” because that’s howwe defined implication)

6.2.3 Definition: Negation

• ¬a = a ⇒ False

• Proof (via truth tables):

– Suppose a is True, then right side becomes True → False, which is False

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– Suppose a is False, then right side becomes False → False, which we defined tobe True

– So in both cases the definition is consistent

• Consistency (via logical inference only):

– If a is True, then ¬True is True ⇒ False which is (False ⇒ False) ⇒ False

– We have a rule called Reduction Ad Absurdum which allows us to infer False from(False ⇒ False) ⇒ False

– If a is False, then ¬False is $False ⇒ False$< which we defined to be True

6.2.4 Definition: Equivalence

• a ⇔ b = (a ⇒ b) ∧ (b ⇒ a)

• (Note that’s the definition of if and only if)

•

Proof by Truth Table:

a b a ⇒ b b ⇒ a LHS RHS

T T T T T TT F F T F FF T T F F FF F T T T T

• Hence, this formula is also correct

6.3 IMPORTANT: Note on Proofs of the Inference Rules

• I’ve “proven” each of the inference rules by making references to the Truth Table

• But it looks like that might be circular atleast based on the discussion in the sectionon “Inferring the Operator Truth Tables”

• Assume that you have a system with FALSE, and three operators OR, AND, IMPLIES

• Pretend that it is just a coincidence that they match the names of Boolean operators

6.4 And Introduction

6.4.1 Definition: a, b a ∧ b

• Proof:

– We know that a and b are True

– By definition of ∧, a ∧ b is True

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6.4.2 Usage Example 1

• Statement: (R ⇒ S ), ¬P (R ⇒ S ) ∧ ¬P

• Proof:

– Use and introduction

6.4.3 Usage Example 2

• Statement: P,Q,R (P ∧ Q) ∧ R

• Proof:

– If P and Q are True, P ∧ Q is True by and introduction

– Since, R is True, another usage of and introduction means (P ∧ Q) ∧ R is True aswell

• Note: This can be written in a tree diagram type form (Page 130)

6.4.4 Exercise 10:

• Statement: Prove P,Q,R P ∧ (Q ∧ R)

• Proof:

– If Q and R are True, Q ∧ R is true by and introduction

– If, P is true, another application of and introduction means P ∧ (Q ∧ R) is True aswell

• Note, we could have proven this simply by noting the associativity of ∧

6.4.5 Exercise 11

• Shape of x = A ∧ (B ∧ (C ∧ D))

– Right-heavy tree

– If there are 2n variables in x, then the height of the proof tree will also be 2n

– (At each level, we can reason about one of the variables in x)

• Shape of y = (A ∧ B) ∧ (C ∧ D))

– Balanced tree: height is logarithmic in number of variables

– (At each level, we can reason about half of the variables in y)

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6.5 And Elimination

6.5.1 Definition: (a ∧ b) a, (a ∧ b) b

• Proof:

– We know that a ∧ b is True

– But a ∧ b is True exactly when both a and b are True

– So, if the conjunction is True, they must individually be True as well

6.5.2 Usage Example

• Statement: P, Q ∧ R P ∧ Q

• Proof:

– P is True by assumption

– Q ∧ R means Q is True by and elimination

– Since both P and Q are true, P ∧ Q is True (by and introduction)

6.5.3 Usage Example

• Statement: (P ∧ Q) ∧ R R ∧ Q

• Proof:

– By and elimination, both P ∧ Q and R are True

– Another and elimination on P ∧ Q yields True for P and Q

– So, both Q and R are True, by and introduction, Q ∧ R is True

6.5.4 Usage Example

• Statement: P ∧ Q Q ∧ P

• Proof:

– By and elimination, both P and Q are True

– By and introduction, Q ∧ P is True

• This proves the commutativity of And (which we could have also done using TruthTables)

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6.6.3 Usage Example

• Statement: a, a ⇒ b, b ⇒ c c

• Proof:

– By Modus Ponens on a, a ⇒ b, we see that b is True

– By Modus Ponens on b, b ⇒ c, we see that c is True

6.6.4 Exercise 13

• Statement: P, P ⇒ Q, (P ∧ Q) ⇒ (R ∧ S ) S

• Proof:

– By Modus Ponens on P, P ⇒ Q, we see that Q is True

– By And Introduction, P ∧ Q is True

– By Modus Ponens on P ∧ Q, P ∧ Q ⇒ R ∧ S , we see that R ∧ S is True

– By And Elimination on R ∧ S , we see that both R and S are True

6.6.5 Exercise 14

• Statement: P ⇒ Q, R ⇒ S, P ∧ R S ∧ R

• Proof:

– By And Elimination on P ∧ R, we see that P and R are True

– By Modes Ponens on R, R ⇒ S , we see that S is True

– By And Introduction on R and S , we see that S ∧ R is True

6.7 Imply Introduction

6.7.1 Definition: (a b) (a ⇒ b)

• “In order to imply the logical implication a ⇒ b, you must have a proof of b using a asan assumption”

6.7.2 Example Usage

• Statement: (P ∧ Q) ⇒ Q

• Since the doesn’t have anything to the left, it is saying that this is a tautology

• What proof do we have for this?

– Consider the sequent P ∧ Q Q, which is just and elimination

– So, we have a proof for P ∧ Q Q, which means we can write P ∧ Q ⇒ Q byImplies introduction

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• Assumptions Made

– And Elimination assumed P ∧ Q is True.

– We can’t infer Q without this assumption.

– So, this appears in the sequent to the left of the in P ∧ Q Q– But the entire sequent P ∧ Q Q doesn’t have any further assumptions

∗ Whenever P ∧ Q is true, Q is true.

– So, this sequent can be put in as an assumption without adding any other assump-tions

– This is why when we do implies introduction, we get (P ∧ Q) ⇒ Q

6.7.3 Discharged assumptions

• Assumptions that are made temporarily to establish a sequent, and which is thenthrown away

• E.g. to establish the sequent P ∧ Q Q, we made an assumption P ∧ Q

• Once we got P ∧ Q Q, we don’t need to assumption for any other purpose

• Indicate discharged assumptions by boxes

6.7.4 Implication Chain Rule

• Statement: a ⇒ b, b ⇒ c a ⇒ c

• Proof:

– Since we are trying to prove a ⇒ c, we must use Imply Introduction somewhere– This means that we need to show a c

– Assume a is True, then by Modus Ponens on a ⇒ b, b is True

– Since b is True, by Modus Ponens on b ⇒ c, c is True

– So, we have a sequent a c, and we can discharge the assumption that a is True

– By Imply Introduction on a c, we get ac

6.7.5 Modus Tollens

• Statement: a ⇒ b, ¬b ¬a

• Proof:

– ¬a is an abbreviation for a ⇒ False

– This means that we need to use Imply Introduction

– Expand ¬a and ¬b to get: a ⇒ b, b ⇒ False a ⇒ False

– This is just implication chain rule

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6.7.6 Exercise 15

• Statement: P Q ⇒ P ∧ Q

• Proof:

– We are trying to prove an implication Q ⇒ P ∧Q, means we have to Imply Introduction

– To introduce imply, we need to prove the sequent Q P ∧ Q

– We know that P is true from what is given

– Assume that Q is true.

– By and introduction, we have P, Q P ∧ Q

– This sequent does not have any further assumptions, so we can discharge the as-sumption that B is True

– So, we have the sequent Q P ∧ Q, which means we have Q ⇒ P ∧ Q byintroduce imply

•

(Maybe incorrect)

6.7.7 Exercise 16

• Statement: P ∧ Q ⇒ Q ∧ P

• Proof:

– We need to use introduce imply on the sequent P ∧ Q Q ∧ P

– But that’s just commutativity of ∧, which we proved earlier, the sequent holdswithout any further assumptions

– So, we can introduce imply on the sequent to get the desired answer

6.8 Or Introduction

6.8.1 Definition: a a ∨ b and b a ∨ b

• Proof

– By the definition of , if one of its arguments is true, then the whole expressionbecomes true

6.8.2 Example Usage

• Statement: P ∧ Q P ∨ Q

• Proof

– By and elimination on P ∧ Q, we have P is True, Q is True

– By or introduction, we have P ∨ Q is True

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6.8.3 Exercise 17

• Statement: P ⇒ False ∨ P

• Proof

– To prove this implication, we need to prove the sequent P False ∨ P

– But that sequent is True by or introduction

– So, by introduce implication, we have P ⇒ False ∨ P

6.8.4 Exercise 18

• Statement: P, Q (P ∧ Q) ∨ (Q ∨ R)

• Proof:

– By and introduction, we have P ∧ Q is True

– By or introduction, we have Q ∨ R is True

– By or introduction, (P ∧ Q) ∨ (Q ∨ R) is True

– (Actually, knowing that P ∧ Q is True is sufficient because we can or introduce thatto get True)

6.9 Or Elimination

6.9.1 Definition (((a ∨ b), (a c), (b c)) c)

• This is Proof by Case Analysis

• Proof:

– Assumption is that a ∨ b is True

– By the definition of ∨ (from Truth Table), atleast one of a and b must be True

– Suppose a is True

∗ We have a c, i.e. a proof that c is True when a holds

– Otherwise, b is True

∗ We have b c, i.e. a proof that c is True when a holds

– So, c is True whenever the assumptions hold

• (Did i just prove Proof by Case Analysis by DOING proof of case analysis? Somethingis weird here. . . )

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6.9.2 Example Usage

• Statement: (P ∧ Q) ∨ (P ∧ R) P

• Proof:

– Suppose P ∧ Q is True, then by and elimination, P is True

– Suppose instead that P ∧ R is True, then by and elimination, P is True

– So, we have ((P ∧ Q) ∨ (P ∧ R)), (P ∧ Q) P, (P ∧ R) P

– Use or elimination to get that P is True

6.9.3 Example Usage

• Statement: (a ∧ b) ∨ (a ∧ c) (b ∨ c)

• Proof:

– Assume a ∧ b by and elimination, we get a ∧ b b

– Discharge the assumption a ∧ b because we have a sequent

– By or introduction, we have b b ∨ c

– Assume a ∧ c by and elimination, we get a ∧ c c

– Discharge the assumption a ∧ c

– By or introduction, we have c b ∨ c

– We can no do or elimination on (a ∧ b) ∨ (a ∧ c) (b ∨ c), ((a ∧ b) b) b ∨ c, ((a ∧ c)

c) b ∨ c

– This gives us b ∨ c

6.10 Identity

6.10.1 Definition a a

• It is necessary because we can’t prove P P without it.

6.10.2 Theorem: True

• By definition, True = False ⇒ False

• So, we need to introduce imply to prove this

• To introduce imply, we need to prove the sequent False False

• This is true by identity

• So, we have False ⇒ False or equivalently, True

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6.11 Contradiction

6.11.1 Definition: False a

• Anything can be inferred if the assumption is False

• (Remember the Truth table for P ⇒ Q gave True when P was False?)

• Expresses the fact that False is untrue purely using logical inference

– Can’t just say False is not True, because we defined True that way

– Also doesn’t rely on us making “intuitionistic” statements like “False is Wrong”

6.11.2 Example Usage

• Statement: P, ¬P Q

• Proof:

– ¬P is shorthand for P ⇒ False– So, from P, P ⇒ False by modus ponens, we have False

– False Q follows from contradiction


• Statement; a ∨ b, ¬a b

• Proof:

– ¬a is shorthand for a ⇒ False

– Performing or elimination on a ∨ b,

∗ Assume a is True, then we have a, a ⇒ False, which by Modus Ponens givesus False

∗ So, by contradiction, b is True

∗ (Discharge the assumption that a is True)

∗ Assume b is True, the by identity, b is True

∗ (Discharge the assumption that b is True)

∗ Combining, we have (a ∨ b, ¬a), (((a ∨ b) a), a ⇒ False) b, (((a ∨ b) b) b),we can now perform or elimination

6.12 Reduction Ad Absurdum

6.12.1 Definition: (¬a False) a

• If you can infer False from an assumption ¬a, then you must conclude that a is true

• This is the Proof by Contradiction strategy

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• Statement: ¬¬a a

• Proof:

– Strategy: Assume ¬a and show that it leads us to infer False (proof by contradic-tion)

– Assuming ¬a, we have ¬a, ¬¬a

– Expand out the abbreviations to get: a ⇒ False, (a ⇒ False) ⇒ False

– Since we have a ⇒ False and (a ⇒ False) ⇒ False, by Modus Ponens, we haveFalse

– So, by reductio ad absurdum, a Must be True if ¬¬a is True

6.13 Inferring the Operator Truth Tables

• Two fundamental questions:

– Are inference rules powerful enough?

– Will they ever allow us to make mistakes?

• profound and difficult questions, modern research has shown some astonishing results

• Easy version of first question: Are inference rules powerful enough to infer the truthtables of logical operators?

6.13.1 Example: Deriving the Truth Table value of True ∧ False

• Want to show that it is logically equivalent to False, and not logically equivalent to True

• To show equivalence with False, we need to show: False ⇒ (True ∧ False), True ∧

False ⇒ False

– False (True ∧ False) follows from contradiction

– By introduce implication, we have False ⇒ (True ∧ False)

– True ∧ False False

∗ Expand out True the expression is now False ∧ (False ⇒ False)

∗ By and elimination, we get False ∧ (False ⇒ False) False, which by identitygives us False False

∗ By introduce implication, we have True ∧ False ⇒ False

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6.13.2 Exercise: Calculating value of True ∧ True

• Proof that True ∧ True ⇒ True

– By and elimination, we have True ∧ True True

– By identity, we have True True

– By introduce implication, we have True ∧ True ⇒ True

• Proof that True ⇒ (True ∧ True)

– Need to prove the sequent True True ∧ True

– By and introduction, we have True True ∧ True

– By introduce implication, we have True ⇒ True ∧ True

6.13.3 Exercise: Calculating the value of True ∨ False

• Proof that True ∨ False ⇒ True

– We have True True from identity

– We have False True from contradiction

– So, we have (True ∨ False), (True True), (False True)

– By or elimination, we have (True ∨ False) True

– By introduce implication, we have True ∨ False ⇒ True

• Proof that True ⇒ (True ∨ False)

– From or introduction, we have True (True ∨ False)

– By introduce implication, we have True ⇒ (True ∨ False)

6.13.4 Extra Credit: What happens when we try to say True ∨ False is False?

• Need to show equivalence

• Proof that False ⇒ (True ∨ False)

– False (True ∨ False) by contradiction

– By introduce implication, we have False ⇒ (True ∨ False)

• Proof that (True ∨ False) ⇒ False

– Need to do an or elimination

– False False comes from identity

– True False needs to be shown

– Expand out True: False ⇒ False False needs to be shown

– Since False is not True, we can’t use Modus Ponens

– We have no applicable inference rule to use

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6.13.5 Exercise

• When we computed value of True ∧ False, we used and elimination to say that (False ⇒

False) ∧ False False

• But we could have said (False ⇒ False) False (False ⇒ False) which is True

• What would have happened then?

– We would have had to show that True False

– That is, False ⇒ False False

– Same situation as in extra credit: Have no valid inference law to use; so can’tdeduce.

7 Proof Checking by Computer

7.1 Proof Checkers and Theorem Provers

• Proof Checker: reads in a theorem and a proof. Determines if the proof is valid

• Theorem Prover: reads in a theorem and attempts to generate a proof

7.2 Example of Proof Checking

• See 06-proofs.hs

8 Boolean Algebra: Equational Reasoning

8.1 Introduction

8.1.1 Approaches to Propositional Logic

• Truth Tables: semantic approach

• Natural Deduction: syntactic approach

• Boolean Algebra: axiomatic approach

8.1.2 Equational Reasoning

• Show that two values are the same by building chains of equalities

– e.g. show that a = b and b = c, so a = c

• can substitute equals for equals

– if x = y, can substitute in x wherever y is used

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8.2 Laws of Boolean Algebra

8.2.1 Operations with Constants

• a ∧ False = False

• a ∧ True = True

• Identity for or: a ∨ True = True

• Identity for and: a ∨ False = a

• Exercise 25: Simplification

– (P ∧ False) ∨ (Q ∧ True)

– False ∨ (Q ∧ True)

– Q ∧ True

– Q

• Exercise 26

– Prove that: (P ∧ False) ∧ True) = False

– Direct Proof:

∗ (P ∧ False) ∧ True)

∗ False ∧ True

∗ False

8.2.2 Basic Properties of ∧ and ∨

• Disjunctive Implication: a ⇒ a ∨ b (Or Introduction)

• Conjunctive Implication: a ∧ b ⇒ a (And Elimination)

• Idempotence: a ∧ a = a

• Idempotence: a ∧ a = a

• Commutative: a ∧ b = b ∧ a

• Commutative: a ∨ b = b ∨ a

• Associativity: (a ∧ b) ∧ c = a ∧ (b ∧ c)

• Associativity: (a ∨ b) ∨ c = a ∧ (b ∨ c)

• Exercise 27

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– Prove: (P ∧ ((Q ∨ R) ∨ Q)) ∧ S = S ∧ ((R ∨ Q) ∧ P )

– Direct Proof starting from LHS

– (P ∧ ((Q ∨ R) ∨ Q)) ∧ S

– (P ∧ ((R ∨ Q) ∨ Q)) ∧ S (by commutativity of or)

– (P ∧ (R ∨ (Q ∨ Q)) ∧ S (by associativity of or)– (P ∧ (R ∨ Q)) ∧ S (by idempotence of or)

– ((R ∨ Q) ∧ P ) ∧ S (by commutivatiy of and)

– S ∧ ((R ∨ Q) ∧ P ) (by commutivatiy of and)

• Exercise 28

– Prove: P ∧ (Q ∧ (R ∧ S )) = ((P ∧ Q) ∧ R) ∧ S

– Direct proof starting from LHS

– P ∧ (Q ∧ (R ∧ S ))

– ((P ∧ Q) ∧ (R ∧ S )) (by associativity of and)

– ((P ∧ Q) ∧ R) ∧ S (by associativity of and)

8.2.3 Distributive and De Morgan’s Laws

• And distributes over Or: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)

• Or distributes over And: a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)

• De Morgan: ¬(a ∧ b) = ¬a ∨ ¬b

• De Morgan: ¬(a ∨ b) = ¬a ∧ ¬b

• Exercise 29

– Give an intuitive explanation of second De Morgan’s laws

– If none of (a, b) are true, it means that a is not true and b is not true

8.2.4 Laws on Negation

• ¬True = False

• ¬False = True

• a ∧ ¬a = False

• a ∨ ¬a = True

• ¬(¬a) = a

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8.2.5 Laws on Implication

• Modus Ponens: a ∧ (a ⇒ b) ⇒ b

• Modus Tollens: ¬b ∧ (a ⇒ b) ⇒ ¬a

• Disjunctive Syllogism: (a ∨ b) ∧ ¬ ⇒ b

• Implication Chain: (a ⇒ b) ∧ (b ⇒ c) ⇒ (a ⇒ c)

• Implication Combination: (a ⇒ b) ∧ (c ⇒ d) ⇒ ((a ∧ c) ⇒ (b ∧ d))

• Currying: (a ∧ b) ⇒ c = a ⇒ b ⇒ c

• Implication: a ⇒ b = ¬a ∧ b

• Contrapositive: a ⇒ b = ¬b ⇒ ¬a

• Absurdity: (a ⇒ b) ∧ (a ⇒ ¬b) = ¬a

•

Understanding the currying law

– Says there are two equivalent ways of establishing that a and b both hold

– One of them is to say a ∧ b

– The other is to have an implication on a and an implication on b

– If either of a or b is false,

∗ a ∧ b is false, so (a ∧ b) ⇒ c is vacuous

∗ Either a ∧ b or b ∧ c is vacuous (if a is False, then the first one is vacuous, if b

is false, the second one)

•

Understanding the Implication Law

– allows us to prove a ⇒ b

– Note that we don’t have an “Introduce Implication” rule like in natural deduction

– So we use this rule

8.2.6 Logical Equivalence

• a ⇔ b is the same as (a ⇒ b) ∧ (b ⇒ a)

• Subtle difference between ⇔ and =

– = is an operator in the meta language that says that two sides of an equation havethe same value

– ⇔ is in the object language itself and it gives a new proposition

• Exercise 30

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– Prove: (A ∨ B) ∧ BB

– Strategy: We’ll reduce the left side of the equivalence to be the right side

– (A ∨ B) ∧ B

– (A ∨ B) ∧ (B ∨ False) (by Identity of OR)

– (B ∨ A) ∧ (B ∨ False) (commutativity of AND)– B ∨ (A ∧ False) (distributivity of OR)

– B ∨ False

– B

• Exercise 31

– Prove: ((¬A ∧ B) ∨ (A ∧ ¬B)) ⇔ (A ∨ B) ∧ (¬(A ∧ B))

– Direct proof starting from “left side”

– (¬A ∧ B) ∨ (A ∧ ¬B)

– (¬A ∨ (A ∧ ¬B)) ∧ (B ∨ (A ∧ ¬B)) (by distributing the OR over the AND)

– ((¬A ∨ A) ∧ (¬A ∨ ¬B)) ∧ ((B ∨ A) ∧ (B ∨ ¬B)) (by distributing the OR over the AND)

– (True ∧ (¬A ∨ ¬B)) ∧ ((B ∨ A) ∧ True)

– (¬A ∨ ¬B) ∧ (B ∨ A)

– ¬(A ∧ B) ∧ (B ∨ A) (by Demorgan’s laws)

• Exercise 32

– Prove: ¬(A ∧ B) ≡ ¬A ∨ ¬B

– Direct proof starting from “right side”– ¬A ∨ ¬B

– ¬(¬(¬A ∨ ¬B)) (Double negation is just identity)

– ¬(A ∧ B) (by De Morgan’s other law)

• Exercise 33

– Prove: (A ∨ B) ∧ (¬A ∨ C ) ∧ (B ∨ C ) ⇔ (A ∨ B) ∧ (¬A ∨ C )

– Proof is by cases on A

– Case: A is True∗ LHS: (True ∨ B) ∧ ((¬True) ∨ C ) ∧ (B ∨ C )

∗ i.e. True ∧ C ∧ (B ∨ C )

∗ i.e. C ∧ (B ∨ C )

∗ i.e. C (proven in Review Exercise 45)

∗ RHS: (True ∨ B) ∧ ((¬True) ∧ C )

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∗ i.e. True ∧ C

∗ i.e. C

∗ So the equality holds in this case

– Case: A is False

∗ LHS: (False ∨ B) ∧ ((¬False) ∨ C ) ∧ (B ∨ C )

∗ i.e. B ∧ True ∧ (B ∨ C )

∗ i.e. B ∧ (B ∨ C )

∗ i.e. B

∗ RHS: (False ∨ B) ∧ ((¬False) ∨ C )

∗ i.e. B ∧ True

∗ i.e. B

∗ So, the equality holds in this case

– Since the equality holds in both cases, the statement is proven.

9 Logic in Computer Science

• Talks about applications of theorem provers

10 Meta Logic

• Making statements about the logical system itself

• Fundamental Meta-Logic operators:

– P 1, P 2, · · · , P n Q means there is a proof that shows Q from the assumptionsP 1, · · · , P n

– P 1, P 2, · · · , P n Q means Q must be tree if all assumptions are true, but saysnothing about the proof

10.1 Definition: Consistency of a Logical System

• A formal system is consistent if, for all well-formed formulas a and b, if a b then a b

• That is, each proposition that is provable is actually True

10.2 Definition: Completeness of a Logical System

• A formal system is complete if, for all well-formed formulas a and b, if a b then a b

• That is, every proposition that is true can be proved using the system

10.3 Theorem: Propositional Logic is consistent and complete

• No proof given

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10.4 Godel’s Theorem: A logical system powerful enough to express ordi-nary arithmetic must be either inconsistent or incomplete

• It is impossible to capture all of mathematics in a safe logical system

11 Further Reading• Forever Undecided: A Puzzle Guide to Godel by Raymond Smullyan

– “Don’t miss it”

• Godel, Escher, Bach: An Eternal Braid by Douglas Hofstadter

– “Another unmissable classic”

• How to Prove it: A Structured approach by Daniel Velleman

– Good source for hints on carrying out proofs with lots of example

• Computer-Aided Reasoning: An approach: by Matt Kaufmann, Panagiotis Mano-lios, and J Strother Moore

– Describes the ACL2 theorem prover

• Introduction to HOL: by M Gordon and T Melham

– Describes the HOL theorem prover

• Logic for Mathematics and Computer Science by Burris

– “more advanced treatement of mathematical logic”

• A Mathematical Introduction to Logic by Enderton

– “standard presentation of logic from a mathematical perspective”

• Type Theorey and Functional Programming by Thompson

– “detailed development of relationship between inference rules and rules used todefine type systems”

• Logic and Declarative Language by Downward

• Proofs and Types by Girard, Lafont and Taylor

– relationship between inference proofs and type systems at a research level

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Discrete Mathematics Using a Computer - 06 - Propositional Logic Notes

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