CSE 311 Lecture 08: Inference Rules and Proofs for ... · Proving implications with the direct proof rule Direct Proof Rule The premise means “Given , we can prove .” So the direct

CSE 311 Lecture 08: Inference Rules and Proofsfor Predicate LogicEmina Torlak and Kevin Zatloukal

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture08.html?print-pdf#/

TopicsPropositional logic proofs

A brief review of .A quick look at predicate logic proofs

Inference rules for quantifiers and a “hello” world example.An in-depth look at predicate logic proofs

Understanding rules for quantifiers through more advanced examples.

Lecture 07

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture07.html


Propositional logic proofsA brief review of .Lecture 07

http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture07.html

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Inference rules for propositional logicTwo rules per binary connective: to eliminate and introduce it.

Intro

Elim

Intro

Elim

Direct Proof Rule

Modus Ponens

Direct Proof Rule is special: not like the other rules.

∧ A; B∴ A ∧ B

∧ A ∧ B∴ A, B

∨ A∴ A ∨ B, B ∨ A

∨ A ∨ B; ¬A∴ B

A ⟹ B∴ A → B

A; A → B∴ B

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Proving implications with the direct proof rule

Direct Proof Rule

The premise means “Given , we can prove .”

So the direct proof rule says that if we have such a proof, then we canconclude that is true.

A ⟹ B∴ A → B

A ⟹ B A B

A → B

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Direct Proof Rule



Example: prove .

1.1. Assumption1.2.1.3.

2. Direct Proof Rule

A ⟹ B∴ A → B

A ⟹ B A B

A → B(p ∧ q) → (p ∨ q)

p ∧ qpp ∨ q

(p ∧ q) → (p ∨ q)

Indent the proof subroutine.Write the assumption and the goal.

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Direct Proof Rule



Example: prove .

1.1. Assumption1.2. Elim : 1.11.3. Intro : 1.2


A ⟹ B∴ A → B

A ⟹ B A B

A → B(p ∧ q) → (p ∨ q)

p ∧ qp ∧p ∨ q ∨

(p ∧ q) → (p ∨ q)

Indent the proof subroutine.Write the assumption and the goal.Fill in the steps.

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Why does the direct proof rule work?Inference rules let us derive facts that are implied by the existing facts.

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So, for every rule , is a tautology ( ).P

∴ QP → Q P → Q ≡ '

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So, for every rule , is a tautology ( ).

The proof shows that is a tautology ( ),since it just a series of implications that we know are tautologies.

P∴ Q

P → Q P → Q ≡ '

P ⟹ Q P → Q P → Q ≡ '

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So the Direct Proof Rule says that we can add to

our set of facts, if we can show that is a tautology.

P∴ Q

P → Q P → Q ≡ '

P ⟹ Q P → Q P → Q ≡ '

A ⟹ B∴ A → B

A → B

A → B

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So the Direct Proof Rule says that we can add to

our set of facts, if we can show that is a tautology.

One way to show that is by writing a subproof, using allthe facts we have inferred up to that point.

P∴ Q

P → Q P → Q ≡ '

P ⟹ Q P → Q P → Q ≡ '

A ⟹ B∴ A → B

A → B

A → BA → B ≡ '

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An example proofProve .((p → q) ∧ (q → r)) → (p → r)

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An example proofProve .


1.5.

2. DirectProofRule

((p → q) ∧ (q → r)) → (p → r)(p → q) ∧ (q → r)

p → r((p → q) ∧ (q → r)) → (p → r)

Write the premise and the conclusion.

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1.1. Assumption1.2. Elim : 1.11.3. Elim : 1.1

1.5.

2. DirectProofRule

((p → q) ∧ (q → r)) → (p → r)(p → q) ∧ (q → r)p → q ∧q → r ∧

p → r((p → q) ∧ (q → r)) → (p → r)

Write the premise and the conclusion.Work forwards and backwards.

We’ll need parts of 1.1 so Elim toget 1.2, 1.3.

∧

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1.4.1. Assumption1.4.2.1.4.3.

1.5. Direct Proof Rule

2. DirectProofRule

((p → q) ∧ (q → r)) → (p → r)(p → q) ∧ (q → r)p → q ∧q → r ∧

p

rp → r

((p → q) ∧ (q → r)) → (p → r)


We’ll need parts of 1.1 so Elim toget 1.2, 1.3.We can use DPR to get 1.5.

∧

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3.


2. DirectProofRule

((p → q) ∧ (q → r)) → (p → r)(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)


We’ll need parts of 1.1 so Elim toget 1.2, 1.3.We can use DPR to get 1.5.Using MP on 1.2, 1.4.1 gives us 1.4.2.

∧

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule

((p → q) ∧ (q → r)) → (p → r)(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)


We’ll need parts of 1.1 so Elim toget 1.2, 1.3.We can use DPR to get 1.5.Using MP on 1.2, 1.4.1 gives us 1.4.2.Using MP on 1.3, 1.4.2 gives us 1.4.3.

∧

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Which facts can be used in a subproof?


1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule

A line in a (sub)proof can use a factat line if the set of assumptions andgivens that is derived from containsall the assumptions and givens that is derived from.

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule

A line in a (sub)proof can use a factat line if the set of assumptions andgivens that is derived from containsall the assumptions and givens that is derived from.So, 1.4.2 can use 1.2 because they arederived from the assumptions {1.1,1.4.1} and {1.1}, respectively.

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

8




1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule

A line in a (sub)proof can use a factat line if the set of assumptions andgivens that is derived from containsall the assumptions and givens that is derived from.So, 1.4.2 can use 1.2 because they arederived from the assumptions {1.1,1.4.1} and {1.1}, respectively.Can 1.5 use 1.4.3?

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule


No. Because 1.5 is derived from{1.1} and 1.4.3 from {1.1, 1.4.1}.

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule



Can 2 use 1.2?

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

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1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2


2. DirectProofRule



Can 2 use 1.2?No. Because 2 is derived from {}and 1.2 from {1.1}.

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r((p → q) ∧ (q → r)) → (p → r)

ki

ki

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Which facts can be used in a subproof? A mnemonic{

{


{

1.4.1. Assumption1.4.2. MP: 1.2, 1.4.11.4.3. MP: 1.3, 1.4.2

}

1.5. Direct Proof Rule}

2. Direct ProofRule

}

(p → q) ∧ (q → r)p → q ∧q → r ∧

pqr

p → r

((p → q) ∧ (q → r)) → (p → r)

This is just like Java’s scoping rules.

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A general proof strategy

1. Look at the rules for introducing connectives to see how you wouldbuild up the formula you want to prove from pieces of what is given.

2. Use the rules for eliminating connectives to break down the givenformulas so that you get the pieces need for 1

3. Write the proof beginning with what you figured out for 2 followed by 1.

Intro

Elim

Intro

Elim

Direct Proof Rule

Modus Ponens

∧ A; B∴ A ∧ B

∧ A ∧ B∴ A, B

∨ A∴ A ∨ B, B ∨ A

∨ A ∨ B; ¬A∴ B

A ⟹ B∴ A → B

A; A → B∴ B

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A quick look at predicate logic proofsInference rules for quantifiers and a “hello” world example.

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Inference rules for quantifiers

Elim

Intro

Intro

Elim

∀ ∀x. P(x)∴ P(a) for any a

∀ P(a); a is arbitrary∴ ∀x. P(x)

The name stands for anarbitrary value in the domain. Noother name in depends on .

a

P a

∃ P(c) for some c∴ ∃x. P(x)

∃ ∃x. P(x)∴ P(c) for a specific c

The name is fresh and stands for avalue in the domain where is true.List all dependencies for .

cP(c)

c

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Predicate logic proofs can use …

Predicate logic inference rulesApplied to whole formulas only.

Predicate logic equivalencesEven on subformulas.

Propositional logic inference rulesApplied to whole formulas only.

Propositional logic equivalencesEven on subformulas.

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A “hello world” proofProve .

Elim

Intro

Intro

Elim

(∀x. P(x)) → (∃x. P(x))





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Elim

Intro

Intro

Elim

(∀x. P(x)) → (∃x. P(x))∀x. P(x)

∃x. P(x)(∀x. P(x)) → (∃x. P(x))

Given , so use Direct Proof Rule.→





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1.1. Assumption1.2.1.3. Intro : 1.2


Elim

Intro

Intro

Elim

(∀x. P(x)) → (∃x. P(x))∀x. P(x)P(c)∃x. P(x) ∃

(∀x. P(x)) → (∃x. P(x))

Given , so use Direct Proof Rule.We can use Intro to get 1.3, but

need for some .

→∃

P(c) c





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Elim

Intro

Intro

Elim

(∀x. P(x)) → (∃x. P(x))∀x. P(x)P(c) ∀∃x. P(x) ∃

(∀x. P(x)) → (∃x. P(x))


need for some .We have from Elim on 1.1.

→∃

P(c) cP(c) ∀





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Elim

Intro

Intro

Elim

Working forwards and backwards:In applying Intro rule, we didn’tknow what expression we mightbe able to prove for, so weworked forwards to figure outwhat might work.

(∀x. P(x)) → (∃x. P(x))∀x. P(x)P(c) ∀∃x. P(x) ∃

(∀x. P(x)) → (∃x. P(x))


need for some .We have from Elim on 1.1.

→∃

P(c) cP(c) ∀





∃

P(c)

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An in-depth look at predicate logic proofsUnderstanding rules for quantifiers through more advanced examples.

15


Advanced proofs: considering domain semanticsSo far, we have treated the predicate definitions as black boxes, and thedomain of discourse as a set of objects with no additional properties.

In practice, we want to prove theorems for specific domains, and usethe properties of those domains in our proofs.

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For example, the set of integers is equipped with the operators .+, ⋅, =

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For example, the set of integers is equipped with the operators .

We can use these operators in our predicates (below) and proofs (next):

+, ⋅, =

Domain of discourseIntegers

Predicate definitionsEven(x) ::= ∃y. x = 2 ⋅ y

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A not so odd exampleProve that there is an even number: .

Elim

Intro

Intro

Elim

∃x. Even(x)







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1.2.3.4.

Elim

Intro

Intro

Elim

∃x. Even(x)

∃x. Even(x)







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1.2.3.4. Intro : 3

Elim

Intro

Intro

Elim

∃x. Even(x)

Even(2)∃x. Even(x) ∃







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1.2.3. Definition of Even: 24. Intro : 3

Elim

Intro

Intro

Elim

∃x. Even(x)

∃y. 2 = 2 ⋅ yEven(2)∃x. Even(x) ∃







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1. Arithmetic2. Intro : 13. Definition of Even: 24. Intro : 3

Elim

Intro

Intro

Elim

∃x. Even(x)2 = 2 ⋅ 1∃y. 2 = 2 ⋅ y ∃Even(2)∃x. Even(x) ∃







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A prime exampleProve that there is an even prime number: .

Elim

Intro

Intro

Elim

We use a black-box definition of Prime because the proof won’t need to break it down further.

∃x. Even(x) ∧ Prime(x)


Predicate definitions

“ is prime”Even(x) ::= ∃y. x = 2 ⋅ yPrime(x) ::= x





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1.2.3.4.5.6.

Elim

Intro

Intro

Elim











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1.2.3.4.5.6. Intro : 5

Elim

Intro

Intro

Elim



Even(2) ∧ Prime(2)∃x. Even(x) ∧ Prime(x) ∃








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1.2.3.4.5. Intro : 3, 46. Intro : 5

Elim

Intro

Intro

Elim



Even(2)Prime(2)Even(2) ∧ Prime(2) ∧∃x. Even(x) ∧ Prime(x) ∃








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1.2.3.4. Property of integer 25. Intro : 3, 46. Intro : 5

Elim

Intro

Intro

Elim



Even(2)Prime(2)Even(2) ∧ Prime(2) ∧∃x. Even(x) ∧ Prime(x) ∃








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1. Arithmetic2. Intro : 13. Definition of Even: 24. Property of integer 25. Intro : 3, 46. Intro : 5

Elim

Intro

Intro

Elim


∃x. Even(x) ∧ Prime(x)2 = 2 ⋅ 1∃y. 2 = 2 ⋅ y ∃Even(2)Prime(2)Even(2) ∧ Prime(2) ∧∃x. Even(x) ∧ Prime(x) ∃








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An equal exampleProve that follows from .

Elim

Intro

Intro

Elim

∀y. ∃z. y = z ∀x. x = x






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1. Given2.3.4.

Elim

Intro

Intro

Elim

∀y. ∃z. y = z ∀x. x = x∀x. x = x

∀y. ∃z. y = z






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1. Given2. Elim : 1, is arbitrary3.4.

Elim

Intro

Intro

Elim

∀y. ∃z. y = z ∀x. x = x∀x. x = xa = a ∀ a

∀y. ∃z. y = z






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1. Given2. Elim : 1, is arbitrary3. Intro : 24.

Elim

Intro

Intro

Elim

∀y. ∃z. y = z ∀x. x = x∀x. x = xa = a ∀ a∃z. a = z ∃∀y. ∃z. y = z






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1. Given2. Elim : 1, is arbitrary3. Intro : 24. Intro : 3

Elim

Intro

Intro

Elim

∀y. ∃z. y = z ∀x. x = x∀x. x = xa = a ∀ a∃z. a = z ∃∀y. ∃z. y = z ∀






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A square exampleProve that the square of every even number is even: .

Elim

Intro

Intro

Elim

∀x. Even(x) → Even( )x2







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4.

Elim

Intro

Intro

Elim









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1. Let be an arbitrary integer.

3.4. Intro : 1, 3

Elim

Intro

Intro

Elim


a

Even(a) → Even( )a2

∀x. Even(x) → Even( )x2 ∀

Use Intro on 1 and 2.∀







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2.1. Assumption2.2.2.3.2.4.2.5.2.6.

3. Direct Proof Rule4. Intro : 1, 3

Elim

Intro

Intro

Elim


aEven(a)

Even( )a2



Use Intro on 1 and 2. so use DRP to get 3.

∀→







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2.1. Assumption2.2. Definition of Even: 2.12.3.2.4.2.5.2.6. Definition of Even: 2.5


Elim

Intro

Intro

Elim


aEven(a)∃y. a = 2y

∃y. = 2ya2

Even( )a2




Use definition of Even to breakdown 2.1 and 2.6.

∀→







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2.1. Assumption2.2. Definition of Even: 2.12.3. Elim : 2.2, depends on 2.4.2.5.2.6. Definition of Even: 2.5


Elim

Intro

Intro

Elim


aEven(a)∃y. a = 2ya = 2b ∃ b a

∃y. = 2ya2

Even( )a2





Use Elim on 2.2.

∀→

∃







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2.1. Assumption2.2. Definition of Even: 2.12.3. Elim : 2.2, depends on 2.4. Algebra2.5.2.6. Definition of Even: 2.5


Elim

Intro

Intro

Elim



= 4 = 2(2 )a2 b2 b2

∃y. = 2ya2

Even( )a2





Use Elim on 2.2.Use algebra on 2.3 to match the

body of 2.5.

∀→

∃







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2.1. Assumption2.2. Definition of Even: 2.12.3. Elim : 2.2, depends on 2.4. Algebra2.5. Intro : 2.42.6. Definition of Even: 2.5


Elim

Intro

Intro

Elim



= 4 = 2(2 )a2 b2 b2

∃y. = 2ya2 ∃Even( )a2





Use Elim on 2.2.Use algebra on 2.3 to match the

body of 2.5.Use Intro on 2.4 to get 2.5.

∀→

∃

∃







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Why list dependencies? To avoid incorrect proofs.Over the integer domain: is True but is False.

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x



The name stands for an arbitrary value in thedomain. No other name in depends on .

aP a



The name is fresh and stands for a value in the domainwhere is true. List all dependencies for .

cP(c) c

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1. Given2. 3.4.5.6.

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x

∀x. ∃y. y ≥ x

∃y. ∀x. y ≥ x

Example: an incorrect proof.




aP a




cP(c) c

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1. Given2. Let be an arbitrary integer. 3.4.5.6. Intro : 5

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x

∀x. ∃y. y ≥ xa

∃y. ∀x. y ≥ x ∃





aP a




cP(c) c

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1. Given2. Let be an arbitrary integer. 3. Elim : 14.5.6. Intro : 5

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x

∀x. ∃y. y ≥ xa

∃y. y ≥ a ∀

∃y. ∀x. y ≥ x ∃





aP a




cP(c) c

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1. Given2. Let be an arbitrary integer. 3. Elim : 14. Elim : 3, depends on 5.6. Intro : 5

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x

∀x. ∃y. y ≥ xa

∃y. y ≥ a ∀b ≥ a ∃ b a

∃y. ∀x. y ≥ x ∃





aP a




cP(c) c

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1. Given2. Let be an arbitrary integer. 3. Elim : 14. Elim : 3, depends on 5. Intro : 2, 46. Intro : 5

Elim

Intro

Intro

Elim

∀x. ∃y. y ≥ x ∃y. ∀x. y ≥ x

∀x. ∃y. y ≥ xa

∃y. y ≥ a ∀b ≥ a ∃ b a∀x. b ≥ x ∀∃y. ∀x. y ≥ x ∃


Can’t get rid of sinceanother name, , in thesame formula depends on it!

ab




aP a




cP(c) c

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SummaryPredicate logic proofs extend propositional logic proofs.

Can use all rules and equivalences for propositional logic.Plus inference rules for quantifiers and equivalences for predicate logic.

When applying Intro to , make sure that is arbitrary, and

no other name depends on .When applying Elim to , make sure that

in is fresh, andall the dependencies for are listed.

∀ P(a)a

a∃ ∃x. P(x)

c P(c)c

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Documents

CSE 311 Lecture 08: Inference Rules and Proofs for ... · Proving implications with the direct proof rule Direct Proof Rule The premise means “Given , we can prove .” So the direct