Predicate Logic: Semantics Alice Gao Lecture 13 CS 245 Logic and - - PowerPoint PPT Presentation

Predicate Logic: Semantics

Alice Gao

Lecture 13

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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Learning goals

By the end of this lecture, you should be able to:

▶ Defjne a valuation. ▶ Determine the value of a term given a valuation. ▶ Determine the truth value of a formula given a valuation. ▶ Give a valuation that makes a formula true or false. ▶ Determine and justify whether a formula is satisfjable and/or

valid.

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The Language of Predicate Logic

▶ Domain: a non-empty set of objects ▶ Individuals: concrete objects in the domain ▶ Functions: takes objects in the domain as arguments and

returns an object of the domain.

▶ Relations: takes objects in the domain as arguments and

returns true or false. They describe properties of objects or relationships between objects.

▶ Variables: placeholders for concrete objects in the domain ▶ Quantifjers: for how many objects in the domain is the

statement true?

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The semantics of a predicate formula

Given a well-formed formula of predicate logic, does the formula evaluate to 0 or 1 in some context? Example: What does (𝐺(𝑏) ∨ 𝐻(𝑏, 𝑐)) mean? The symbols 𝐺, 𝐻, 𝑏, and 𝑐 do not have intrinsic meanings. In propositional logic, we need a truth valuation to give a meaning to a formula. In predicate logic, we need a valuation to give a meaning to a term or a formula.

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Valuation

A valuation 𝑤 for our language ℒ consists of

• 1. A domain 𝐸,
• 2. A meaning for each individual symbol, e.g. 𝑏𝑤 ∈ 𝐸,
• 3. A meaning for each free variable symbol, e.g. 𝑣𝑤 ∈ 𝐸,
• 4. A meaning for each relation symbol, e.g. 𝐺 𝑤 ⊆ 𝐸𝑜,

≈𝑤= {⟨𝑦, 𝑦⟩}𝑦 ∈ 𝐸} ⊆ 𝐸2.

• 5. A meaning for each function symbol, e.g. 𝑔𝑤 ∶ 𝐸𝑛 → 𝐸.

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A function symbol must be interpreted as a total function

A function symbol 𝑔 must be interpreted as a function 𝑔𝑤 that is total on the domain 𝐸. 𝑔𝑤 ∶ 𝐸𝑛 → 𝐸

▶ Any 𝑛-tuple (𝑒1, ..., 𝑒𝑛) ∈ 𝐸𝑛 can be an input to 𝑔𝑤. ▶ For any legal 𝑛-tuple (𝑒1, ..., 𝑒𝑛) ∈ 𝐸𝑛,

𝑔𝑤(𝑒𝑤

1, ..., 𝑒𝑤 𝑛) ∈ 𝐸.

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CQ Which function is total?

Which of the following functions is total? (A) 𝑕(𝑦, 𝑧) = 𝑦 − 𝑧. 𝐸 = ℕ (natural numbers).

• Incorrect. 1 - 2 = -1 is not a natural number.

(B) 𝑔(𝑦) = √𝑦. 𝐸 = ℤ (integers).

• Incorrect. The square root of an integer may not be an integer

anymore. (C) 𝑔(𝑦) = 𝑦 + 1. 𝐸 = {1, 2, 3}.

• Incorrect. 3 + 1 = 4 not in domain.

(D) 𝑔(1) = 2, 𝑔(2) = 3 and 𝑔(3) = 3. 𝐸 = {1, 2, 3}. Correct (E) 𝑕(𝑦, 𝑧) = 𝑦 > 𝑧. 𝐸 = ℤ (integers).

• Incorrect. x > y produces true or false, not a domain element.

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Value of Terms

Defjnition (Value of Terms)

The value of terms of 𝑀 under valuation 𝑤 over domain 𝐸 is defjned by recursion:

• 1. 𝑏𝑤 ∈ 𝐸.
• 2. 𝑣𝑤 ∈ 𝐸.
• 3. 𝑔(𝑢1, … , 𝑢𝑜)𝑤 = 𝑔𝑤(𝑢𝑤

1, … , 𝑢𝑤 𝑜).

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The assignment override notation

𝑤(𝑣/𝛽) keeps all the mappings in 𝑤 intact EXCEPT reassigning 𝑣 to 𝛽 ∈ 𝐸. Consider a valuation: 𝑣𝑤

1 = 3, 𝑣𝑤 2 = 3, 𝑣𝑤 3 = 1. 𝐸 = {1, 2, 3}.

• 1. 𝑣𝑤(𝑣1/2)

1

= ?

• 2. 𝑣𝑤(𝑣1/2)

2

= ?

• 3. 𝑣𝑤(𝑣1/2)(𝑣2/1)

1

= ?

• 4. 𝑣𝑤(𝑣1/2)(𝑣2/1)

2

= ?

• 5. 𝑣𝑤(𝑣1/2)(𝑣2/1)

3

= ? 𝑣𝑤(𝑣1/2)

1

= 2, 𝑣𝑤(𝑣1/2)

2

= 3 𝑣𝑤(𝑣1/2)(𝑣2/1)

1

= 2, 𝑣𝑤(𝑣1/2)(𝑣2/1)

2

= 1, 𝑣𝑤(𝑣1/2)(𝑣2/1)

3

= 1

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True Value of Formulas

Defjnition (Truth Value of Formulas)

The truth value of formulas of 𝑀 under valuation 𝑤 over domain 𝐸 is defjned by recursion:

• 1. 𝐺(𝑢1, … , 𝑢𝑜)𝑤 = 1 ifg ⟨𝑢𝑤

1, … , 𝑢𝑤 𝑜⟩ ∈ 𝐺 𝑤.

• 2. (¬𝐵)𝑤 = 1 ifg 𝐵𝑤 = 0.
• 3. (𝐵 ∧ 𝐶)𝑤 = 1 ifg 𝐵𝑤 = 1 and 𝐶𝑤 = 1.
• 4. (𝐵 ∨ 𝐶)𝑤 = 1 ifg 𝐵𝑤 = 1 or 𝐶𝑤 = 1.
• 5. (𝐵 → 𝐶)𝑤 = 1 ifg 𝐵𝑤 = 0 or 𝐶𝑤 = 1.
• 6. (𝐵 ↔ 𝐶)𝑤 = 1 ifg 𝐵𝑤 = 𝐶𝑤.
• 7. (∀𝑦 𝐵(𝑦))𝑤 = 1 ifg for every 𝛽 ∈ 𝐸, 𝐵(𝑣)𝑤(𝑣/𝛽) = 1,

where 𝑣 does not occur in 𝐵(𝑦).

• 8. (∃𝑦 𝐵(𝑦))𝑤 = 1 ifg there exists 𝛽 ∈ 𝐸, 𝐵(𝑣)𝑤(𝑣/𝛽) = 1,

where 𝑣 does not occur in 𝐵(𝑦).

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Our predicate logic language

Our language of predicate logic: Individual symbols: 𝑏, 𝑐, 𝑑. Free variable symbols: 𝑣, 𝑤, 𝑥. Bound variable symbols: 𝑦, 𝑧, 𝑨. Function symbols: 𝑔 is a unary function. 𝑕 is a binary function. Relation symbols: 𝐺 is a unary relation. 𝐻 is a binary relation.

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An example of a valuation

Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = 1, 𝑐𝑤 = 2, 𝑑𝑤 = 3. ▶ Free variables: 𝑣𝑤 = 3, 𝑤𝑤 = 2, 𝑥𝑤 = 1. ▶ Functions:

𝑔𝑤: 𝑔𝑤(1) = 2, 𝑔𝑤(2) = 3, 𝑔𝑤(3) = 1. 𝑕𝑤: 𝑕𝑤(𝑦, 𝑧) = ((𝑦 + 𝑧) mod 3) + 1.

▶ Relations:

𝐺 𝑤: 𝐺 𝑤(𝑦) is true if and only if 𝑦 > 5. 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true if and only if 𝑦 > 𝑧.

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Another example of a valuation

Valuation 𝑤′:

▶ Domain: 𝐸 = {𝐵𝑚𝑗𝑑𝑓, 𝐶𝑝𝑐, 𝐷𝑏𝑢𝑓}. ▶ Individuals: 𝑏𝑤 = 𝐵𝑚𝑗𝑑𝑓, 𝑐𝑤 = 𝐶𝑝𝑐, 𝑑𝑤 = 𝐷𝑏𝑢𝑓. ▶ Free variables: 𝑣𝑤 = 𝐶𝑝𝑐, 𝑤𝑤 = 𝐵𝑚𝑗𝑑𝑓, 𝑥𝑤 = 𝐵𝑚𝑗𝑑𝑓. ▶ Functions:

𝑔𝑤: 𝑔𝑤(𝐵𝑚𝑗𝑑𝑓) = 𝐵𝑚𝑗𝑑𝑓, 𝑔𝑤(𝐶𝑝𝑐) = 𝐷𝑏𝑢𝑓, 𝑔𝑤(𝐷𝑏𝑢𝑓) = 𝐶𝑝𝑐. 𝑕𝑤: 𝑕𝑤(𝑦, 𝑧) = the person with the longer name. return 𝑦 if there is a tie.

▶ Relations:

𝐺 𝑤: 𝐺 𝑤(𝑦) is true ifg the person likes chocolates. (Alice and Cate like chocolates whereas Bob dislikes chocolates.) 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true ifg 𝑦 is older than or has the same age as 𝑧. (Alice is older than Cate, who is older than Bob.)

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Notation for functions and relations

Consider the domain 𝐸 = {1, 2, 3}. Functions:

▶ 𝑔𝑤 is the identify function. 𝑔𝑤(𝑦) = 𝑦. ▶ 𝑔𝑤(1) = 1, 𝑔𝑤(2) = 2 and 𝑔𝑤(3) = 3.

Relations:

▶ 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true if and only if 𝑦 > 𝑧. ▶ 𝐻𝑤 = {⟨2, 1⟩, ⟨3, 1⟩, ⟨3, 2⟩}

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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Evaluating terms and formulas w/o variables

Evaluate these terms and formulas under the valuation 𝑤. 𝑔(𝑔(𝑏)), (𝐺(𝑏) ∨ 𝐻(𝑏, 𝑐)). Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = 1, 𝑐𝑤 = 2, 𝑑𝑤 = 3. ▶ Free variables: 𝑣𝑤 = 3, 𝑤𝑤 = 2, 𝑥𝑤 = 1. ▶ Functions:

𝑔𝑤: 𝑔𝑤(1) = 2, 𝑔𝑤(2) = 3, 𝑔𝑤(3) = 1. 𝑕𝑤: 𝑕𝑤(𝑦, 𝑧) = ((𝑦 + 𝑧) mod 3) + 1.

▶ Relations:

𝐺 𝑤: 𝐺 𝑤(𝑦) is true if and only if 𝑦 > 5. 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true if and only if 𝑦 > 𝑧. 𝑔(𝑔(𝑏))𝑤 = ?, 𝑏𝑤 = 1, 𝑔𝑤(𝑏𝑤) = 2, 𝑔𝑤(𝑔𝑤(𝑏𝑤)) = 3 (𝐺(𝑏) ∨ 𝐻(𝑏, 𝑐))𝑤 = ?, 𝐺(𝑏)𝑤 = 0, 𝐻(𝑏, 𝑐)𝑤 = 0, (𝐺(𝑏) ∨ 𝐻(𝑏, 𝑐))𝑤 = 0

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Give a valuation that makes the formula true/false

Complete the valuation 𝑤 such that (A) 𝐻(𝑏, 𝑔(𝑔(𝑏)))𝑤 = 1 (B) 𝐻(𝑏, 𝑔(𝑔(𝑏)))𝑤 = 0 Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = ?, 𝑐𝑤 = ?, 𝑑𝑤 = ?. ▶ Functions: 𝑔𝑤 ∶ ?, 𝑕𝑤 ∶ ? ▶ Relations: 𝑄 𝑤 ∶ ?, 𝐻𝑤 ∶ ?

𝑏𝑤 = 1 𝑔𝑤 is the identity function. To make the formula true, make sure ⟨1, 1⟩ ∈ 𝐻. To make the formula false, make sure ⟨1, 1⟩ ∉ 𝐻. If 𝐻𝑤 = ∅, the formula is false. If 𝐻𝑤 contains all possible tuples, the formula is true.

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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A valuation for interpreting free variables

Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = 1, 𝑐𝑤 = 2, 𝑑𝑤 = 3. ▶ Free variables: 𝑣𝑤 = 3, 𝑤𝑤 = 2, 𝑥𝑤 = 1. ▶ Functions:

𝑔𝑤: 𝑔𝑤(1) = 2, 𝑔𝑤(2) = 3, 𝑔𝑤(3) = 1. 𝑕𝑤: 𝑕𝑤(𝑦, 𝑧) = ((𝑦 + 𝑧) mod 3) + 1.

▶ Relations:

𝐺 𝑤: 𝐺 𝑤(𝑦) is true if and only if 𝑦 > 5. 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true if and only if 𝑦 > 𝑧.

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Evaluating terms & formulas w/o bound variables

Evaluate these terms and formulas under the valuation 𝑤. 𝑕(𝑣, 𝑔(𝑐)), 𝐻(𝑏, 𝑔(𝑔(𝑣))). Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = 1, 𝑐𝑤 = 2, 𝑑𝑤 = 3. ▶ Free variables: 𝑣𝑤 = 3, 𝑤𝑤 = 2, 𝑥𝑤 = 1. ▶ Functions:

𝑔𝑤: 𝑔𝑤(1) = 2, 𝑔𝑤(2) = 3, 𝑔𝑤(3) = 1. 𝑕𝑤: 𝑕𝑤(𝑦, 𝑧) = ((𝑦 + 𝑧) mod 3) + 1.

▶ Relations:

𝐺 𝑤: 𝐺 𝑤(𝑦) is true if and only if 𝑦 > 5. 𝐻𝑤: 𝐻𝑤(𝑦, 𝑧) is true if and only if 𝑦 > 𝑧. 𝑕(𝑣, 𝑔(𝑐))𝑤 = 1. 𝐻(𝑏, 𝑔(𝑔(𝑣)))𝑤 = 0.

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Give a valuation that makes the formula true/false

Complete a valuation such that (A) 𝐻(𝑏, 𝑔(𝑔(𝑣)))𝑤 = 1 (B) 𝐻(𝑏, 𝑔(𝑔(𝑣)))𝑤 = 0 Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Individuals: 𝑏𝑤 = ?, 𝑐𝑤 = ?, 𝑑𝑤 = ?. ▶ Free variables: 𝑣𝑤 = ?, 𝑣𝑤 = ? 𝑣𝑤 = ?. ▶ Functions: 𝑔𝑤 ∶ ?, 𝑕𝑤 ∶ ? ▶ Relations: 𝑄 𝑤 ∶ ?, 𝐻𝑤 ∶ ?

𝑣𝑤 = 1. 𝑔𝑤 is the identity function. 𝑏𝑤 = 1. To make 𝐻 true, let ⟨1, 1⟩ ∈ 𝐻𝑤. To make 𝐻 true, let ⟨1, 1⟩ ∉ 𝐻𝑤.

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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Evaluate quantifjed formulas under a valuation

Evaluate these formulas under the valuation 𝑤. (A) (∀𝑦 (∃𝑧 𝐻(𝑦, 𝑧))) (B) (∃𝑦 (∀𝑧 𝐻(𝑦, 𝑧))) Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ Relations: 𝐻𝑤 = {⟨1, 2⟩, ⟨3, 1⟩, ⟨2, 3⟩}.

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(A) (∀𝑦 (∃𝑧 𝐻(𝑦, 𝑧))) (B) (∃𝑦 (∀𝑧 𝐻(𝑦, 𝑧))) ⟨𝑣𝑤(𝑣/1)(𝑥/2), 𝑥𝑤(𝑣/1)(𝑥/2)⟩ = ⟨1, 2⟩ ∈ 𝐻𝑤, ⟨𝑣𝑤(𝑣/3)(𝑥/1), 𝑥𝑤(𝑣/3)(𝑥/1)⟩ = ⟨3, 1⟩ ∈ 𝐻𝑤, ⟨𝑣𝑤(𝑣/2)(𝑥/3), 𝑥𝑤(𝑣/2)(𝑥/3)⟩ = ⟨2, 3⟩ ∈ 𝐻𝑤, Therefore, (∀𝑦 (∃𝑧 𝐻(𝑦, 𝑧)))𝑤 = 1. ⟨𝑣𝑤(𝑣/1)(𝑥/1), 𝑥𝑤(𝑣/1)(𝑥/1)⟩ = ⟨1, 1⟩ ∉ 𝐻𝑤, ⟨𝑣𝑤(𝑣/3)(𝑥/2), 𝑥𝑤(𝑣/3)(𝑥/2)⟩ = ⟨3, 2⟩ ∉ 𝐻𝑤, ⟨𝑣𝑤(𝑣/2)(𝑥/1), 𝑥𝑤(𝑣/2)(𝑥/1)⟩ = ⟨2, 1⟩ ∉ 𝐻𝑤, Therefore, (∃𝑦 (∀𝑧 𝐻(𝑦, 𝑧)))𝑤 = 0.

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Give a valuation that makes the formula true/false

Complete the valuation 𝑤 to make the following formula true/false. (When satisfying the formula, try making 𝐻𝑤 as small as possible.) (A) (∀𝑧 (∃𝑦 𝐻(𝑦, 𝑧))) (B) (∃𝑧 (∀𝑦 𝐻(𝑦, 𝑧))) Valuation 𝑤:

▶ Domain: 𝐸 = {1, 2, 3}. ▶ ...

(A) 𝐻𝑤 = {⟨1, 1⟩, ⟨1, 2⟩, ⟨1, 3⟩} makes the formula true. (B) 𝐻𝑤 = {⟨1, 1⟩, ⟨2, 1⟩, ⟨3, 1⟩} makes the formula true.

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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Difgerence between Individual and Free Variable Symbols

Let our domain be the set of people. Let the predicate 𝑀(𝑣) be true if 𝑣 likes chocolates. Let 𝑏 be an individual symbol referring to Alice.

▶ 𝑀(𝑏)

This formula only contains individual symbols. Since 𝑏 refers to Alice, the truth value of this formula is already determined (It’s true because Alice likes chocolates =).

▶ 𝑀(𝑣)

This formula only contains free variable symbols. We do not know the truth value of this formula because 𝑣 can refer to any person in the domain. We need to assign 𝑣 to a particular person because we can determine whether this formula is true or false.

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Difgerence between Free and Bound Variables

Let our domain be the set of integers.

▶ 𝑣 + 𝑣 = 𝑤

The variables are free. We do not know the truth value of this formula until we assign the free variables to elements of the domain.

▶ ∀𝑦 ∀𝑧 (𝑦 + 𝑦 = 𝑧)

The variables are bound. We know the truth value of this formula because the meanings of the variables are given by the quantifjers.

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Outline

The Learning Goals Evaluating Terms and Formulas w/o Variables Evaluating Terms and Formulas w/o Bound Variables Evaluating Quantifjed Formulas A few clarifjcations Satisfjable and Valid Revisiting the Learning Goals

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Satisfjable and Valid

A formula 𝐵 is satisfjable: there exists a valuation 𝑤, 𝐵𝑤 = 1. A formula 𝐵 is valid: for every valuation 𝑤, 𝐵𝑤 = 1. Most predicate formulas are satisfjable but not valid because we have a great deal of freedom to choose the valuation.

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Proving that a formula is satisfjable or not

Is the following formula satisfjable? If it’s satisfjable, give a valuation that satisfjes it. If it’s not satisfjable, give a proof. (∃𝑦 𝐺(𝑦)) → (∀𝑦 𝐺(𝑦)) Answer: This formula is satisfjable. Proof: Consider the valuation: 𝐸 = {1, 2}, and 𝐺 𝑤 = ∅. Since 𝐺 𝑤 = ∅, 𝐺(𝑣)𝑤(𝑣/𝛽) = 0 for every 𝛽 ∈ 𝐸. Thus, (∃𝑦 𝐺(𝑦))𝑤 = 0 and ((∃𝑦 𝐺(𝑦)) → (∀𝑦 𝐺(𝑦)))𝑤 = 1. Here is another valuation that works: 𝐸 = {1, 2} and 𝐺 𝑤 = {1, 2}.

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Proving that a formula is valid/not valid

Is the following formula valid? If it’s valid, give a proof. If it’s not valid, give a counterexample. (∀𝑦 𝐺(𝑦)) → (∃𝑦 𝐺(𝑦))

▶ Determine whether the formula is valid or not. ▶ How do I prove that a formula is NOT valid?

Find a valuation under which the formula is false. In this case, fjnd a valuation under which ∀𝑦 𝐺(𝑦) is true and ∃𝑦 𝐺(𝑦) is false.

▶ How do I prove that a formula is valid?

Consider any valuation. Prove that the formula must be true. In this case, consider any valuation 𝑤 under which ∀𝑦 𝐺(𝑦) is

• true. Show that ∃𝑦 𝐺(𝑦) is also true under 𝑤.

CS 245 Logic and Computation Fall 2019 33 / 36

Proving that a formula is valid or not

Is the following formula valid? If it’s valid, give a proof. If it’s not valid, give a counterexample. (∀𝑦 𝐺(𝑦)) → (∃𝑦 𝐺(𝑦)) Answer: This formula is valid. Proof: We prove this by contradiction. Assume that there is a valuation 𝑤 such that (∀𝑦𝐺(𝑦))𝑤 = 1 and (∃𝑦𝐺(𝑦))𝑤 = 0. Form 𝐺(𝑣) from 𝐺(𝑦), 𝑣 not occurring in 𝐺(𝑦). Since (∀𝑦𝐺(𝑦))𝑤 = 1, then for every 𝛽 ∈ 𝐸, 𝐺(𝑣)𝑤(𝑣/𝛽) = 1. (1) Since (∃𝑦𝐺(𝑦))𝑤 = 0, then there exists 𝛽 ∈ 𝐸, 𝐺(𝑣)𝑤(𝑣/𝛽) = 0, which contradicts (1). QED

CS 245 Logic and Computation Fall 2019 34 / 36

Proving that a formula is valid or not

Is the following formula valid? If it’s valid, give a proof. If it’s not valid, give a counterexample. (∃𝑦 𝐺(𝑦)) → (∀𝑦 𝐺(𝑦)) Answer: This formula is not valid. Proof: Consider the valuation below.

▶ 𝐸 = {1, 2} ▶ 𝐺 𝑤 = {1}

Since 1 ∈ 𝐺 𝑤, 𝐺(𝑣)𝑤(𝑣/1) = 1. Thus, (∃𝑦 𝐺(𝑦))𝑤 = 1. Since 2 ∉ 𝐺 𝑤, 𝐺(𝑣)𝑤(𝑣/2) = 0. Thus. (∀𝑦 𝐺(𝑦))𝑤 = 0.

CS 245 Logic and Computation Fall 2019 35 / 36

Revisiting the learning goals

By the end of this lecture, you should be able to:

▶ Defjne a valuation. ▶ Determine the value of a term given a valuation. ▶ Determine the truth value of a formula given a valuation. ▶ Give a valuation that makes a formula true or false. ▶ Determine and justify whether a formula is satisfjable and/or

valid.

CS 245 Logic and Computation Fall 2019 36 / 36