Semantic Entailment and Natural Deduction Alice Gao Lecture 6, - - PowerPoint PPT Presentation

Semantic Entailment and Natural Deduction

Alice Gao

Lecture 6, September 26, 2017

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Learning goals

Semantic entailment

• Defjne semantic entailment.
• Explain subtleties of semantic entailment.
• Determine whether a semantic entailment holds by using truth tables,

valuation trees, and/or logical identities.

• Prove semantic entailment using truth tables and/or valuation trees.

Natural deduction in propositional logic

• Describe rules of inference for natural deduction.
• Prove a conclusion from given premises using natural deduction

inference rules.

• Describe strategies for applying each inference rule when proving a

conclusion formula using natural deduction.

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A review of the conditional

Consider the formulas 𝑞1 ∧ 𝑞2 ∧ 𝑞3 and 𝑑. The following two statements are equivalent:

• for any truth valuation 𝑢, if (𝑞1 ∧ 𝑞2 ∧ 𝑞3) is true, then 𝑑 is true.
• (𝑞1 ∧ 𝑞2 ∧ 𝑞3) → 𝑑 is a tautology.

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Subtleties about the conditional

Consider the formulas 𝑞1 ∧ 𝑞2 ∧ 𝑞3 and 𝑑. How many of the following statements are true?

• a. If 𝑞1 is false, then (𝑞1 ∧ 𝑞2 ∧ 𝑞3) → 𝑑 is true.
• b. If 𝑞1 = 𝑦 and 𝑞2 = (¬𝑦), then (𝑞1 ∧ 𝑞2 ∧ 𝑞3) → 𝑑 is false.
• c. If 𝑑 is a tautology, then (𝑞1 ∧ 𝑞2 ∧ 𝑞3) → 𝑑 is true.
• d. Two of (a), (b), and (c) are true.
• e. All of (a), (b), and (c) are true.

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Proving arguments valid

Recall that logic is the science of reasoning. One important goal of logic is to infer that a conclusion is true based on a set of premises. A logical argument: Premise 1 Premise 2 ... Premise n ——— Conclusion A common problem is to prove that an argument is valid, that is the set of premises semantically entails the conclusion.

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Formalizing argument validity: Semantic Entailment

Let Σ = {𝑞1, 𝑞2, ..., 𝑞𝑜} be a set of premises and let 𝛽 be the conclusion that we want to derive. Σ semantically entails 𝛽, denoted Σ ⊨ 𝛽, if and only if

• Whenever all the premises in Σ are true, then the conclusion 𝛽 is true.
• For any truth valuation 𝑢, if every premise in Σ is true under 𝑢, then

the conclusion 𝛽 is true under 𝑢.

• For any truth valuation 𝑢, if 𝑢 satisfjes Σ (denoted Σ𝑢 = T), then 𝑢

satisfjes 𝛽 (𝛽𝑢 = T).

• (𝑞1 ∧ 𝑞2 ∧ ... ∧ 𝑞𝑜) → 𝛽 is a tautology.

If Σ semantically entails 𝛽, then we say that the argument (with the premises in Σ and the conclusion 𝛽) is valid. What does Σ𝑢 = T (𝑢 satisfjes Σ) mean? See the next slide.

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What does Σ𝑢 = T mean?

Σ𝑢 = T (𝑢 satisfjes Σ) means ...

• Every formula in Σ is true under the valuation 𝑢.
• If a formula 𝛾 is in Σ, then 𝛾 is true under 𝑢.

If Σ is the empty set ∅, then any valuation satisfjes Σ. Why? The defjnition of “𝑢 satisfjes Σ” says

• If a formula 𝛾 is in Σ, then 𝛾 is true under 𝑢.

There is no formula in ∅, so the premise of the above statement is false, which means the statement is vacuously true. Thus, any valuation satisfjes the empty set ∅.

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Subtleties about entailment

Consider a set of formulas Σ and the formula 𝛽. How many of the following statements are true?

• a. If 𝑞1 in Σ is false, then Σ ⊨ 𝛽 is false.
• b. If Σ = {𝑦, (¬𝑦)}, then Σ ⊨ 𝛽 is true.
• c. If ∅ ⊨ 𝛽 is true, then 𝛽 is a tautology (∅ is the empty set).
• d. Two of (a), (b), and (c) are true.
• e. All of (a), (b), and (c) are true.

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Proving or disproving entailment

Proving that Σ entails 𝛽, denoted Σ ⊨ 𝛽:

• Using a truth table: Consider all rows of the truth table in which all of

the formulas in Σ are true. Verify that 𝛽 is true in all of these rows.

• Direct proof: For every truth valuation under which all of the premises

are true, show that the conclusion is also true under this valuation.

• Proof by contradiction: Assume that the entailment does not hold,

which means that there is a truth valuation under which all of the premises are true and the conclusion is false. Derive a contradiction. Proving that Σ does not entail 𝛽, denoted Σ ⊭ 𝛽:

• Find one truth valuation 𝑢 under which all of the premises in Σ are

true and the conclusion 𝛽 is false.

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Proving entailment using a truth table

Let Σ = {(¬(𝑞 ∧ 𝑟)), (𝑞 → 𝑟)}, 𝑦 = (¬𝑞), and 𝑧 = (𝑞 ↔ 𝑟). Based on the truth table, which of the following statements is true?

• a. Σ ⊨ 𝑦 and Σ ⊨ 𝑧.
• b. Σ ⊨ 𝑦 and Σ ⊭ 𝑧.
• c. Σ ⊭ 𝑦 and Σ ⊨ 𝑧.
• d. Σ ⊭ 𝑦 and Σ ⊭ 𝑧.

𝑞 𝑟 (¬(𝑞 ∧ 𝑟)) (𝑞 → 𝑟) 𝑦 = (¬𝑞) 𝑧 = (𝑞 ↔ 𝑟) 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Proving entailment

What is {(¬(𝑞 ∧ 𝑟)), (𝑞 ∧ 𝑟)} ⊨ (𝑞 ↔ 𝑟)?

• a. True
• b. False

𝑞 𝑟 (¬(𝑞 ∧ 𝑟)) (𝑞 ∧ 𝑟) (𝑞 ↔ 𝑟) 1 1 1 1 1 1 1 1 1 1

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Equivalence and Entailment

Equivalence can be expressed using the notion of entailment.

• Lemma. 𝛽 ≡ 𝛾 if and only if both {𝛽} ⊨ 𝛾 and {𝛾} ⊨ 𝛽.

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Proofs in Propositional Logic: Natural Deduction

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Solution to the previous puzzle

A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet three inhabitants: Alice, Rex and Bob.

• 1. Alice says, “Rex is a knave.” This means Alice and Rex are difgerent.
• 2. Rex says, “it’s false that Bob is a knave (or Bob is a knight).” This

means Rex and Bob are the same.

• 3. Bob claims, “I am a knight or Alice is a knight.” Bob is a knight, or

Bob and Alice are both knaves. Based on 1 and 2, Alice and Bob are difgerent, so they cannot both be knaves (2nd option in 3). Thus, the only possibility left is Alice is a knave, and Rex and Bob are knights.

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Labyrinth Puzzle

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Learning goals

Natural deduction in propositional logic

• Describe rules of inference for natural deduction.
• Prove a conclusion from given premises using natural deduction

inference rules.

• Describe strategies for applying each inference rule when proving a

conclusion formula using natural deduction.

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The Natural Deduction Proof System

We will consider a proof system called Natural Deduction.

• It closely follows how people (mathematicians, at least) normally make

formal arguments.

• It extends easily to more-powerful forms of logic.

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Why would you want to study natural deduction proofs?

• It is impressive to be able to write proofs with nested boxes and

mysterious symbols as justifjcations.

• Be able to prove or disprove that Superman exists (on Tuesday).
• Be able to prove or disprove that the onnagata are correct to insist

that males should play female characters in Japanese kabuki theatres.

• To realize that writing proofs and problem solving in general is both a

creative and a scientifjc endeavour.

• To develop problem solving strategies that can be used in many other

situations.

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A proof is syntactic

First, we think about proofs in a purely syntactic way. A proof

• starts with a set of premises,
• transforms the premises based on a set of inference rules (by pattern

matching),

• and reaches a conclusion.

We write Σ ⊢ND 𝜒

• r simply

Σ ⊢ 𝜒 if we can fjnd such a proof that starts with a set of premises Σ and ends with the conclusion 𝜒.

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Goal is to show semantic entailment

Next, we think about connecting proofs to semantic entailment. We will answer these questions:

• (Soundness) Does every proof establish a semantic entailment?

If I can fjnd a proof from Σ to 𝜒, can I conclude that Σ semantically entails 𝜒? Does Σ ⊢ 𝜒 imply Σ ⊨ 𝜒?

• (Completeness) For every semantic entailment, can I fjnd a proof for

it? If I know that Σ semantically entails 𝜒, can I fjnd a proof from Σ to 𝜒? Does Σ ⊨ 𝜒 imply Σ ⊢ 𝜒?

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Refmexivity / Premise

If you want to write down a previous formula in the proof again, you can do it by refmexivity. Name ⊢-notation inference notation Refmexivity,

• r Premise

Σ, 𝛽 ⊢ 𝛽 𝛽 𝛽 The notation on the right: Given the formulas above the line, we can infer the formula below the line. The version in the center reminds us of the role of assumptions in Natural

• Deduction. Other rules will make more use of it.

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An example using refmexivity

Here is a proof of {𝑞, 𝑟} ⊢ 𝑞. 1. 𝑞 Premise 2. 𝑟 Premise 3. 𝑞 Refmexivity: 1 Alternatively, we could simply write 1. 𝑞 Premise and be done.

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For each symbol, the rules come in pairs.

• An “introduction rule” adds the symbol to the formula.
• An “elimination rule” removes the symbol from the formula.

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Rules for Conjunction

Name ⊢-notation inference notation ∧-introduction (∧i) If Σ ⊢ 𝛽 and Σ ⊢ 𝛾, then Σ ⊢ (𝛽 ∧ 𝛾) 𝛽 𝛾 (𝛽 ∧ 𝛾) Name ⊢-notation inference notation ∧-elimination (∧e) If Σ ⊢ (𝛽 ∧ 𝛾), then Σ ⊢ 𝛽 and Σ ⊢ 𝛾 (𝛽 ∧ 𝛾) 𝛽 (𝛽 ∧ 𝛾) 𝛾

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Example: Conjunction Rules

Example. Show that {(𝑞 ∧ 𝑟)} ⊢ (𝑟 ∧ 𝑞). 1. (𝑞 ∧ 𝑟) Premise 2. 𝑟 ∧e: 1 3. 𝑞 ∧e: 1 4. (𝑟 ∧ 𝑞) ∧i: 2, 3

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Example: Conjunction Rules (2)

Example. Show that {(𝑞 ∧ 𝑟), 𝑠} ⊢ (𝑟 ∧ 𝑠). 1. (𝑞 ∧ 𝑟) Premise 2. 𝑠 Premise 3. 𝑟 ∧e: 1 4. (𝑟 ∧ 𝑠) ∧i: 3, 2

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Rules for Implication: →e

Name ⊢-notation inference notation →-elimination (→e) (modus ponens) If Σ ⊢ (𝛽 → 𝛾) and Σ ⊢ 𝛽, then Σ ⊢ 𝛾 (𝛽 → 𝛾) 𝛽 𝛾 In words: If you assume 𝛽 is true and 𝛽 implies 𝛾, then you may conclude 𝛾.

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Rules for Implication: →i

Name ⊢-notation inference notation →-introduction (→i) If Σ, 𝛽 ⊢ 𝛾, then Σ ⊢ (𝛽 → 𝛾) 𝛽 . . . . 𝛾 (𝛽 → 𝛾) The “box” denotes a sub-proof. In the sub-proof, we starts by assuming that 𝛽 is true (a premise of the sub-proof), and we conclude that 𝛾 is true. Nothing inside the sub-proof may come out. Outside of the sub-proof, we could only use the sub-proof as a whole.

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Example: Rule →i and sub-proofs

• Example. Give a proof of {(𝑞 → 𝑟), (𝑟 → 𝑠)} ⊢ (𝑞 → 𝑠).

To start, we write down the premises at the beginning, and the conclusion at the end. 1. (𝑞 → 𝑟) Premise 2. (𝑟 → 𝑠) Premise 3. 𝑞 Assumption 4. 𝑟 →e: 1, 3 5. 𝑠 →e: 2, 4 6. (𝑞 → 𝑠) ??? What next? The goal “(𝑞 → 𝑠)” contains →. Let’s try rule →i…. Inside the sub-proof, we can use rule →e. Done!

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Rules of Disjunction: ∨i and ∨e

Name ⊢-notation inference notation ∨-introduction (∨i) If Σ ⊢ 𝛽, then Σ ⊢ 𝛽 ∨ 𝛾 and Σ ⊢ 𝛾 ∨ 𝛽 𝛽 𝛽 ∨ 𝛾 𝛽 𝛾 ∨ 𝛽 ∨-elimination (∨e) If Σ, 𝛽1 ⊢ 𝛾 and Σ, 𝛽2 ⊢ 𝛾, then Σ, 𝛽1 ∨ 𝛽2 ⊢ 𝛾 𝛽1 ∨ 𝛽2 𝛽1 . . . . 𝛾 𝛽2 . . . . 𝛾 𝛾 ∨e is also known as “proof by cases”.

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Example: Or-Introduction and -Elimination

Example: Show that {𝑞 ∨ 𝑟} ⊢ (𝑞 → 𝑟) ∨ (𝑟 → 𝑞).

1. 𝑞 ∨ 𝑟 Premise 2. 𝑞 Assumption 3. 𝑟 Assumption 4. 𝑞 Refmexivity: 2 5. 𝑟 → 𝑞 →i: 3–4 6. (𝑞 → 𝑟) ∨ (𝑟 → 𝑞) ∨i: 5 7. 𝑟 Assumption 8. 𝑞 Assumption 9. 𝑟 Refmexivity: 7 10. 𝑞 → 𝑟 →i: 8–9 11. (𝑞 → 𝑟) ∨ (𝑟 → 𝑞) ∨i: 10 12. (𝑞 → 𝑟) ∨ (𝑟 → 𝑞) ∨e: 1, 2–6, 7–11

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Negation

We shall treat negation by considering contradictions. We shall use the notation ⟂ to represent any contradiction. It may appear in proofs as if it were a formula. The elimination rule for negation: Name ⊢-notation inference notation ⟂-introduction, or ¬-elimination (¬e) Σ, 𝛽, (¬𝛽) ⊢ ⟂ 𝛽 (¬𝛽) ⟂ If we have both 𝛽 and (¬𝛽), then we have a contradiction.

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Negation Introduction (¬i)

If an assumption 𝛽 leads to a contradiction, then derive (¬𝛽). Name ⊢-notation inference notation ¬-introduction (¬i) If Σ, 𝛽 ⊢ ⟂, then Σ ⊢ (¬𝛽) 𝛽 . . . . ⟂ (¬𝛽)

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Example: Negation

• Example. Show that {𝛽 → (¬𝛽)} ⊢ (¬𝛽).

1. 𝛽 → (¬𝛽) Premise 2. 𝛽 Assumption 3. (¬𝛽) →e: 1, 2 4. ⟂ ¬e: 2, 3 5. (¬𝛽) ¬i: 2–4

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The Last Two Basic Rules

Double-Negation Elimination: Name ⊢-notation inference notation ¬¬-elimination (¬¬e) If Σ ⊢ (¬(¬𝛽)), then Σ ⊢ 𝛽 (¬(¬𝛽)) 𝛽 Contradiction Elimination: Name ⊢-notation inference notation ⟂-elimination (⟂e) If Σ ⊢ ⟂, then Σ ⊢ 𝛽 ⟂ 𝛽

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A Redundant Rule

The rule of ⟂-elimination is not actually needed. Suppose a proof has 27. ⟂ ⟨some rule⟩ 28. 𝛽 ⟂e: 27. We can replace these by 27. ⟂ ⟨some rule⟩ 28. (¬𝛽) Assumption 29. ⟂ Refmexivity: 27 30. (¬(¬𝛽)) ¬i: 28–29 31. 𝛽 ¬¬e: 30. Thus any proof that uses ⟂e can be modifjed into a proof that does not.

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Example: “Modus tollens”

The principle of modus tollens: {𝑞 → 𝑟, (¬𝑟)} ⊢ (¬𝑞). 1. 𝑞 → 𝑟 Premise 2. (¬𝑟) Premise 3. 𝑞 Assumption 4. 𝑟 →e: 3, 1 5. ⟂ ¬e: 2, 4 6. (¬𝑞) ?? Modus tollens is sometimes taken as a “derived rule”: 𝛽 → 𝛾 (¬𝛾) (¬𝛽)

MT

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Derived Rules

Whenever we have a proof of the form Γ ⊢ 𝛽, we can consider it as a derived rule: Γ 𝛽 If we use this in a proof, it can be replaced by the original proof of Γ ⊢ 𝛽. The result is a proof using only the basic rules. Using derived rules does not expand the things that can be proved. But they can make it easier to fjnd a proof.

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Strategies for natural deduction proofs

• 1. Work forward from the premises. Can you apply an elimination rule?
• 2. Work backwards from the conclusion. What introduction rule do you

need to use at the end?

• 3. Stare at the formula. Notice its structure. Use it to guide your proof.
• 4. If a direct proof doesn’t work, try a proof by contradiction.

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Further Examples of Natural Deduction

Example. Show that {𝑞 → 𝑟} ⊢ (𝑠 ∨ 𝑞) → (𝑠 ∨ 𝑟). In the sub-proof, try ∨-elimination on the assumption (step 2).

1. 𝑞 → 𝑟 Premise 2. 𝑠 ∨ 𝑞 Assumption 3. 𝑠 Assumption 4. 𝑠 ∨ 𝑟 ∨i: 3 5. 𝑞 Assumption 6. 𝑟 →e: 5, 1 7. 𝑠 ∨ 𝑟 ∨i: 6 8. 𝑠 ∨ 𝑟 ∨e: 2, 3–4, 5–7 9. (𝑠 ∨ 𝑞) → (𝑠 ∨ 𝑟) →i: 2–8

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Life’s Not Always So Easy…

Example. Show that ⊢ ((𝑞 → 𝑟) → 𝑞) → 𝑞. 1. (𝑞 → 𝑟) → 𝑞 Assumption 2. No elimination applies. 3. 4. ????? 5. 𝑞 No connective. 6. ((𝑞 → 𝑟) → 𝑞) → 𝑞 Try →i… Time to try something ingenious….

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Some Common Derived Rules

Proof by contradiction (reductio ad absurdum): if Σ, (¬𝛽) ⊢⟂, then Σ ⊢ 𝛽. The “Law of Excluded Middle” (tertiam non datur): ⊢ 𝛽 ∨ (¬𝛽). Double-Negation Introduction: if Σ ⊢ 𝛽 then Σ ⊢ (¬(¬𝛽)). You can try to prove these yourself, as exercises. (Hint: in the fjrst two, the last step uses rule ¬¬e: (¬(¬𝛽)) ⊢ 𝛽.) Or see pages 24–26 of Huth and Ryan.

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Soundness and Completeness

• f Natural Deduction

for Propositional Logic

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Soundness and Completeness of Natural Deduction

We want to prove that Natural Deduction is both sound and complete. Soundness of Natural Deduction means that the conclusion of a proof is always a logical consequence of the premises. That is, If Σ ⊢ND 𝛽, then Σ ⊨ 𝛽 . Completeness of Natural Deduction means that all logical consequences in propositional logic are provable in Natural

• Deduction. That is,

If Σ ⊨ 𝛽, then Σ ⊢ND 𝛽 .

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Proof of Soundness

To prove soundness, we use induction on the length of the proof: For all deductions Σ ⊢ 𝛽 which have a proof of length 𝑜 or less, it is the case that Σ ⊨ 𝛽. That property, however, is not quite good enough to carry out the

• induction. We actually use the following property of a natural number 𝑜.

Suppose that a formula 𝛽 appears at line 𝑜 of a partial deduction, which may have one or more open sub-proofs. Let Σ be the set of premises used and Γ be the set of assumptions of

• pen sub-proofs. Then Σ ∪ Γ ⊨ 𝛽.

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Basis of the Induction

Base case. The shortest deductions have length 1, and thus are either 1. 𝛽 Premise.

• r

1. 𝛽 Assumption. 2. We have either 𝛽 ∈ Σ (in the fjrst case), or 𝛽 ∈ Γ (in the second case). Thus Σ ∪ Γ ⊨ 𝛽, as required.

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Proof of Soundness: Inductive Step

Inductive step. Hypothesis: the property holds for each 𝑜 < 𝑙; that is, If some formula 𝛽 appears at line 𝑙 or earlier of some partial deduction, with premises Σ and un-closed assumptions Γ, then Σ ∪ Γ ⊨ 𝛽. To prove: if 𝛽′ appears at line 𝑙 + 1, then Σ ∪ Γ′ ⊨ 𝛽′ (where Γ′ = Γ ∪ 𝛽′ when 𝛽′ is an assumption, and Γ′ = Γ otherwise). The case that 𝛽′ is an assumption is trivial. Otherwise, formula 𝛽′ must have a justifjcation by some rule. We shall consider each possible rule.

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Inductive Step, Case I

Case I: 𝛽′ was justifjed by ∧i. We must have 𝛽′ = 𝛽1 ∧ 𝛽2, where each of 𝛽1 and 𝛽2 appear earlier in the proof, at steps 𝑛1 and 𝑛2, respectively. Also, any sub-proof open at step 𝑛1 or 𝑛2 is still open at step 𝑙 + 1. Thus the induction hypothesis applies to both; that is, Σ ∪ Γ ⊨ 𝛽1 and Σ ∪ Γ ⊨ 𝛽2. By the defjnition of ⊨, this yields Σ ∪ Γ ⊨ 𝛽′, as required.

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Inductive Step, Case II

Case II: 𝛽′ was justifjed by →i. Rule →i requires that 𝛽′ = 𝛽1 → 𝛽2 and there is a closed sub-proof with assumption 𝛽1 and conclusion 𝛽2, ending by step 𝑙. Also, any sub-proof open before the assumption of 𝛽1 is still open at step 𝑙 + 1. The induction hypothesis thus implies Σ ∪ (Γ ∪ {𝛽1}) ⊨ 𝛽2. Hence Σ ∪ Γ ⊨ 𝛽1 → 𝛽2, as required.

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Inductive Step, Cases III fg.

Case III: 𝛽′ was justifjed by ¬e. This requires that 𝛽′ be the pseudo-formula ⟂, and that the proof contain formulas 𝛽 and (¬𝛽) for some 𝛽, each using at most 𝑙 steps. By the induction hypothesis, both Σ ⊨ 𝛽 and Σ ⊨ (¬𝛽). Thus Σ is contradictory, and Σ ⊨ 𝛽′ for any 𝛽′. Cases IV–XIII: The other cases follow by similar reasoning. This completes the inductive step, and the proof of soundness.

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Completeness of Natural Deduction

We now turn to completeness. Recall that completeness means the following. Let Σ be a set of formulas and 𝜒 be a formula. If Σ ⊨ 𝜒, then Σ ⊢ 𝜒 . That is, every consequence has a proof. How can we prove this?

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Proof of Completeness: Getting started

We shall assume that the set Σ of hypotheses is fjnite. The theorem is also true for infjnite sets of hypotheses, but that requires a completely difgerent proof.

Suppose that Σ ⊨ 𝜒, where Σ = {𝜏1, 𝜏2, … , 𝜏𝑛}. Thus the formula (𝜏1 ∧ 𝜏2 ∧ … ∧ 𝜏𝑛) → 𝜒 is a tautology.

• Lemma. Every tautology is provable in Natural Deduction.

Once we prove the Lemma, the result follows. Given a proof of (𝜏1 ∧ 𝜏2 ∧ … ∧ 𝜏𝑛) → 𝜒, one can use ∧i and →e to complete a proof of Σ ⊢ 𝜒.

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Tautologies Have Proofs

For a tautology, every line of its truth table ends with T. We can mimic the construction of a truth table using inferences in Natural Deduction.

• Claim. Let 𝜒 have 𝑙 variables 𝑞1, … , 𝑞𝑙. Let 𝑤 be a valuation,

and defjne ℓ1, ℓ2, … , ℓ𝑙 as ℓ𝑗 = ⎧ { ⎨ { ⎩ 𝑞𝑗 if 𝑤(𝑞𝑗) = T ¬𝑞𝑗 if 𝑤(𝑞𝑗) = F. If 𝜒𝑤 = T, then {ℓ1, … ℓ𝑙} ⊢ 𝜒, and if 𝜒𝑤 = F, then {ℓ1, … ℓ𝑙} ⊢ (¬𝜒). To prove the claim, use structural induction on formulas (which is induction on the column number of the truth table). Once the claim is proven, we can prove a tautology as follows….

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Outline of the Proof of a Tautology

1. 𝑞1 ∨ (¬𝑞1) L.E.M. 2. 𝑞2 ∨ (¬𝑞2) L.E.M. ⋮ ⋮ 𝑙. 𝑞𝑙 ∨ (¬𝑞𝑙) L.E.M. 𝑙 + 1. 𝑞1 assumption 𝑞2 assumption ⋮ 𝜒 (¬𝑞2) assumption ⋮ 𝜒 𝑛. 𝜒 ∨e: 2, … 𝑛 + 1. (¬𝑞1) assumption ⋮ ⋮ 𝜒 ∨e: 𝑛 + 1, … 𝑜. 𝜒 ∨e: 1, 𝑛 − (𝑙 + 1), 𝑜 − (𝑛 + 1) Once each variable is assumed true or false, the previous claim provides a proof.

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Proving the Claim

Hypothesis: the following hold for formulas 𝛽 and 𝛾: If {ℓ1, … , ℓ𝑙} ⊨ 𝛽, then {ℓ1, … , ℓ𝑙} ⊢ 𝛽; If {ℓ1, … , ℓ𝑙} ⊭ 𝛽, then {ℓ1, … , ℓ𝑙} ⊢ (¬𝛽); If {ℓ1, … , ℓ𝑙} ⊨ 𝛾, then {ℓ1, … , ℓ𝑙} ⊢ 𝛾; and If {ℓ1, … , ℓ𝑙} ⊭ 𝛾, then {ℓ1, … , ℓ𝑙} ⊢ (¬𝛾). If {ℓ1, … , ℓ𝑙} ⊨ (𝛽 ∧ 𝛾), put the two proofs of 𝛽 and 𝛾 together, and then infer (𝛽 ∧ 𝛾), by ∧i. If {ℓ1, … , ℓ𝑙} ⊭ (𝛽 → 𝛾) (i.e., {ℓ1, … , ℓ𝑙} ⊨ 𝛽 and {ℓ1, … , ℓ𝑙} ⊭ 𝛾),

• Prove 𝛽 and (¬𝛾).
• Assume (𝛽 → 𝛾); from it, conclude 𝛾 (→e) and then ⟂ (¬e).
• From the sub-proof, conclude (¬(𝛽 → 𝛾)), by ¬i.

The other cases are similar.

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