Logic in Action Chapter 9: Proofs http://www.logicinaction.org/ ( - - PowerPoint PPT Presentation

Logic in Action

Chapter 9: Proofs

http://www.logicinaction.org/

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Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method.

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Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method. It is a refutation method.

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Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method. It is a refutation method. It does not follow the way humans reason.

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Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method. It is a refutation method. It does not follow the way humans reason. Issues with the presented derivation systems.

(http://www.logicinaction.org/) 2 / 24

Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method. It is a refutation method. It does not follow the way humans reason. Issues with the presented derivation systems. Proofs are not very natural (e.g., try to prove ϕ → ¬¬ϕ).

(http://www.logicinaction.org/) 2 / 24

Natural Deduction for Propositional Logic

Systems revised so far

Issues with the tableau method. It is a refutation method. It does not follow the way humans reason. Issues with the presented derivation systems. Proofs are not very natural (e.g., try to prove ϕ → ¬¬ϕ). They do not facilitate conditional reasoning.

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Natural Deduction for Propositional Logic

The deduction property

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Natural Deduction for Propositional Logic

The deduction property Σ, ϕ | = ψ if and only if Σ | = ϕ → ψ

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Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ.

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Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps.

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Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps. But if we can make assumptions . . .

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Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps. But if we can make assumptions . . . 1 ϕ

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Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps. But if we can make assumptions . . . 1 ϕ 2 ϕ

repetition 1 (http://www.logicinaction.org/) 4 / 24

Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps. But if we can make assumptions . . . 1 ϕ 2 ϕ

repetition 1

3 ϕ → ϕ

deduction 1-2 (http://www.logicinaction.org/) 4 / 24

Natural Deduction for Propositional Logic

What if we can make assumptions?

Consider a proof for ϕ → ϕ. Using the derivation system presented in Chapter 2, the proof takes several steps. But if we can make assumptions . . . 1 ϕ 2 ϕ

repetition 1

3 ϕ → ϕ

deduction 1-2

This is the main idea for the deduction rule.

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Natural Deduction for Propositional Logic

The deduction rule

Suppose you want to prove ϕ → ψ.

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Natural Deduction for Propositional Logic

The deduction rule

Suppose you want to prove ϕ → ψ. Assume ϕ. ϕ

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Natural Deduction for Propositional Logic

The deduction rule

Suppose you want to prove ϕ → ψ. Assume ϕ. If after further steps ϕ . . .

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Natural Deduction for Propositional Logic

The deduction rule

Suppose you want to prove ϕ → ψ. Assume ϕ. If after further steps you can prove ψ, ϕ . . . ψ

(http://www.logicinaction.org/) 5 / 24

Natural Deduction for Propositional Logic

The deduction rule

Suppose you want to prove ϕ → ψ. Assume ϕ. If after further steps you can prove ψ, then you actually have ϕ → ψ. ϕ . . . ψ ϕ → ψ

deduction (http://www.logicinaction.org/) 5 / 24

Natural Deduction for Propositional Logic

Recall

The three axioms for propositional logic

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Natural Deduction for Propositional Logic

Recall

The three axioms for propositional logic

1 ϕ → (ψ → ϕ) (http://www.logicinaction.org/) 6 / 24

Natural Deduction for Propositional Logic

Recall

The three axioms for propositional logic

1 ϕ → (ψ → ϕ) 2 (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) (http://www.logicinaction.org/) 6 / 24

Natural Deduction for Propositional Logic

Recall

The three axioms for propositional logic

1 ϕ → (ψ → ϕ) 2 (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) 3 (¬ϕ → ¬ψ) → (ψ → ϕ) (http://www.logicinaction.org/) 6 / 24

Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction:

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Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction: 1 ϕ

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Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction: 1 ϕ 2 ψ

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Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction: 1 ϕ 2 ψ 3 ϕ

repetition 1 (http://www.logicinaction.org/) 7 / 24

Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction: 1 ϕ 2 ψ 3 ϕ

repetition 1

4 ψ → ϕ

deduction 2-3 (http://www.logicinaction.org/) 7 / 24

Natural Deduction for Propositional Logic

Proving the axioms (1)

The axiom ϕ → (ψ → ϕ) can be proved from deduction: 1 ϕ 2 ψ 3 ϕ

repetition 1

4 ψ → ϕ

deduction 2-3

5 ϕ → (ψ → ϕ)

deduction 1-4 (http://www.logicinaction.org/) 7 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction:

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Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ)

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Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ

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Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ

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Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2

5 ψ → χ

modus ponens 3,1 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2

5 ψ → χ

modus ponens 3,1

6 χ

modus ponens 4,5 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2

5 ψ → χ

modus ponens 3,1

6 χ

modus ponens 4,5

7 ϕ → χ

deduction 3-6 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2

5 ψ → χ

modus ponens 3,1

6 χ

modus ponens 4,5

7 ϕ → χ

deduction 3-6

8 (ϕ → ψ) → (ϕ → χ)

deduction 2-7 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

Proving the axioms (2)

The axiom (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)) can be proved from modus ponens and deduction: 1 ϕ → (ψ → χ) 2 ϕ → ψ 3 ϕ 4 ψ

modus ponens 3,2

5 ψ → χ

modus ponens 3,1

6 χ

modus ponens 4,5

7 ϕ → χ

deduction 3-6

8 (ϕ → ψ) → (ϕ → χ)

deduction 2-7

9 (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ))

deduction 1-8 (http://www.logicinaction.org/) 8 / 24

Natural Deduction for Propositional Logic

We need more

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) cannot be proved from modus ponens and deduction.

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Natural Deduction for Propositional Logic

We need more

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) cannot be proved from modus ponens and deduction. We need a way to deal with negations.

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ.

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ. Assume ¬ϕ. ¬ϕ

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ. Assume ¬ϕ. If after further steps ¬ϕ . . .

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ. Assume ¬ϕ. If after further steps you can prove a contradiction ⊥, ¬ϕ . . . ⊥

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ. Assume ¬ϕ. If after further steps you can prove a contradiction ⊥, then ¬ϕ cannot be true ¬ϕ . . . ⊥

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Natural Deduction for Propositional Logic

The refutation rule

Suppose you want to prove ϕ. Assume ¬ϕ. If after further steps you can prove a contradiction ⊥, then ¬ϕ cannot be true so you actually have ϕ. ¬ϕ . . . ⊥ ϕ

refutation (http://www.logicinaction.org/) 10 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation:

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Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ

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Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ

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Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ

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Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1 (http://www.logicinaction.org/) 11 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1

5 ⊥

modus ponens 2,4 (http://www.logicinaction.org/) 11 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1

5 ⊥

modus ponens 2,4

6 ϕ

refutation 3-5 (http://www.logicinaction.org/) 11 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1

5 ⊥

modus ponens 2,4

6 ϕ

refutation 3-5

7 ψ → ϕ

deduction 2-6 (http://www.logicinaction.org/) 11 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1

5 ⊥

modus ponens 2,4

6 ϕ

refutation 3-5

7 ψ → ϕ

deduction 2-6

8 (¬ϕ → ¬ψ) → (ψ → ϕ)

deduction 1-7 (http://www.logicinaction.org/) 11 / 24

Natural Deduction for Propositional Logic

Proving the axioms (3)

The axiom (¬ϕ → ¬ψ) → (ψ → ϕ) can be proved from modus ponens, deduction and refutation: 1 ¬ϕ → ¬ψ 2 ψ 3 ¬ϕ 4 ¬ψ

modus ponens 3,1

5 ⊥

modus ponens 2,4

6 ϕ

refutation 3-5

7 ψ → ϕ

deduction 2-6

8 (¬ϕ → ¬ψ) → (ψ → ϕ)

deduction 1-7

For step 5, note that ¬ψ can be seen as an abbreviation of ψ → ⊥.

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Natural Deduction for Propositional Logic

So . . .

ϕ , ϕ → ψ ψ

modus ponens

ϕ . . . ψ ϕ → ψ

deduction

¬ϕ . . . ⊥ ϕ

refutation (http://www.logicinaction.org/) 12 / 24

Natural Deduction for Propositional Logic

So . . .

ϕ , ϕ → ψ ψ

modus ponens

ϕ . . . ψ ϕ → ψ

deduction

¬ϕ . . . ⊥ ϕ

refutation

The modus ponens, deduction and refutation rules are a complete system for propositional logic.

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Natural Deduction for Propositional Logic

To facilitate things . . .

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Natural Deduction for Propositional Logic

To facilitate things . . .

Natural deduction introduces rules to manipulate all the connectives in an easy way.

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Natural Deduction for Propositional Logic

For implication →

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Natural Deduction for Propositional Logic

For implication →

ϕ , ϕ → ψ ψ

modus ponens (http://www.logicinaction.org/) 14 / 24

Natural Deduction for Propositional Logic

For implication →

ϕ , ϕ → ψ ψ

modus ponens

E→

(http://www.logicinaction.org/) 14 / 24

Natural Deduction for Propositional Logic

For implication →

ϕ , ϕ → ψ ψ

modus ponens

ϕ . . . ψ ϕ → ψ

deduction

E→

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Natural Deduction for Propositional Logic

For implication →

ϕ , ϕ → ψ ψ

modus ponens

ϕ . . . ψ ϕ → ψ

deduction

E→ I→

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Natural Deduction for Propositional Logic

For negation ¬

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Natural Deduction for Propositional Logic

For negation ¬

¬ϕ , ϕ ⊥

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Natural Deduction for Propositional Logic

For negation ¬

¬ϕ , ϕ ⊥

E¬

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Natural Deduction for Propositional Logic

For negation ¬

¬ϕ , ϕ ⊥ ¬ϕ . . . ⊥ ϕ

refutation

E¬

(http://www.logicinaction.org/) 15 / 24

Natural Deduction for Propositional Logic

For negation ¬

¬ϕ , ϕ ⊥ ¬ϕ . . . ⊥ ϕ

refutation

E¬ I¬

(http://www.logicinaction.org/) 15 / 24

Natural Deduction for Propositional Logic

For conjunction ∧

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Natural Deduction for Propositional Logic

For conjunction ∧

ϕ ∧ ψ ϕ ϕ ∧ ψ ψ

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Natural Deduction for Propositional Logic

For conjunction ∧

ϕ ∧ ψ ϕ ϕ ∧ ψ ψ

E∧

(http://www.logicinaction.org/) 16 / 24

Natural Deduction for Propositional Logic

For conjunction ∧

ϕ ∧ ψ ϕ ϕ ∧ ψ ψ ϕ , ψ ϕ ∧ ψ

E∧

(http://www.logicinaction.org/) 16 / 24

Natural Deduction for Propositional Logic

For conjunction ∧

ϕ ∧ ψ ϕ ϕ ∧ ψ ψ ϕ , ψ ϕ ∧ ψ

E∧ I∧

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Natural Deduction for Propositional Logic

For disjunction ∨

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Natural Deduction for Propositional Logic

For disjunction ∨

ϕ ∨ ψ , ϕ . . . χ , ψ . . . χ χ

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Natural Deduction for Propositional Logic

For disjunction ∨

ϕ ∨ ψ , ϕ . . . χ , ψ . . . χ χ

E∨

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Natural Deduction for Propositional Logic

For disjunction ∨

ϕ ∨ ψ , ϕ . . . χ , ψ . . . χ χ ϕ ϕ ∨ ψ ψ ϕ ∨ ψ

E∨

(http://www.logicinaction.org/) 17 / 24

Natural Deduction for Propositional Logic

For disjunction ∨

ϕ ∨ ψ , ϕ . . . χ , ψ . . . χ χ ϕ ϕ ∨ ψ ψ ϕ ∨ ψ

E∨ I∨

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Natural Deduction for Predicate Logic

For predicate logic

In order to present introduction and elimination rules for both ∀ and ∃, we need to recall two notions.

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Natural Deduction for Predicate Logic

For predicate logic

In order to present introduction and elimination rules for both ∀ and ∃, we need to recall two notions. Bounded variable.

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Natural Deduction for Predicate Logic

For predicate logic

In order to present introduction and elimination rules for both ∀ and ∃, we need to recall two notions. Bounded variable. Substitution of a variable for a term in a formula.

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Natural Deduction for Predicate Logic

Bounded variable

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Natural Deduction for Predicate Logic

Bounded variable

Scope of a quantifier. In a formula of the form ∀xϕ (∃xϕ), the subformula ϕ is said to be the scope of the quantifier ∀ (∃).

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Natural Deduction for Predicate Logic

Bounded variable

Scope of a quantifier. In a formula of the form ∀xϕ (∃xϕ), the subformula ϕ is said to be the scope of the quantifier ∀ (∃). Binding a variable. In a formula of the form ∀xϕ (∃xϕ), the quantifier ∀ (∃) binds any occurrence of x in ϕ that is not bounded by another quantifier inside ϕ.

(http://www.logicinaction.org/) 19 / 24

Natural Deduction for Predicate Logic

Bounded variable

Scope of a quantifier. In a formula of the form ∀xϕ (∃xϕ), the subformula ϕ is said to be the scope of the quantifier ∀ (∃). Binding a variable. In a formula of the form ∀xϕ (∃xϕ), the quantifier ∀ (∃) binds any occurrence of x in ϕ that is not bounded by another quantifier inside ϕ. Bounded variable. An occurrence of a variable x is bounded in a formula ϕ if there is a quantifier in ϕ that binds it.

(http://www.logicinaction.org/) 19 / 24

Natural Deduction for Predicate Logic

Substitution (1)

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Natural Deduction for Predicate Logic

Substitution (1)

Substitution inside a term. Replacing the occurrences of the variable y for the term t inside the term s produces the term denoted by (s)y

t (http://www.logicinaction.org/) 20 / 24

Natural Deduction for Predicate Logic

Substitution (1)

Substitution inside a term. Replacing the occurrences of the variable y for the term t inside the term s produces the term denoted by (s)y

t

Formally, For a constant: (c)y

t := c

For a variable: 8 < : (x)y

t := x

for x different from y (y)y

t := t (http://www.logicinaction.org/) 20 / 24

Natural Deduction for Predicate Logic

Substitution (1)

Substitution inside a term. Replacing the occurrences of the variable y for the term t inside the term s produces the term denoted by (s)y

t

Formally, For a constant: (c)y

t := c

For a variable: 8 < : (x)y

t := x

for x different from y (y)y

t := t

Examples: (a)x

c := a

(x)y

a := x

(z)z

y := y (http://www.logicinaction.org/) 20 / 24

Natural Deduction for Predicate Logic

Substitution (2)

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Natural Deduction for Predicate Logic

Substitution (2)

Substitution inside a formula. Replacing the free occurrences of the variable y for the term t inside the formula ϕ produces the formula denoted by (ϕ)y

t (http://www.logicinaction.org/) 21 / 24

Natural Deduction for Predicate Logic

Substitution (2)

Substitution inside a formula. Replacing the free occurrences of the variable y for the term t inside the formula ϕ produces the formula denoted by (ϕ)y

t

Formally, (P t1 · · · tn)y

t := P (t1)y t · · · (tn)y t

(¬ϕ)y

t := ¬(ϕ)y t

(ϕ ∧ ψ)y

t := (ϕ)y t ∧ (ψ)y t

(ϕ ∨ ψ)y

t := (ϕ)y t ∨ (ψ)y t

(ϕ → ψ)y

t := (ϕ)y t → (ψ)y t

(ϕ ↔ ψ)y

t := (ϕ)y t ↔ (ψ)y t

8 < : (∀xϕ)y

t := ∀x(ϕ)y t

(∀yϕ)y

t := ∀yϕ

8 < : (∃xϕ)y

t := ∃x(ϕ)y t

(∃yϕ)y

t := ∃yϕ (http://www.logicinaction.org/) 21 / 24

Natural Deduction for Predicate Logic

For the universal quantifier ∀

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Natural Deduction for Predicate Logic

For the universal quantifier ∀

∀x ϕ (ϕ)x

t

provided that no variable in t

• ccurs bounded in ϕ

(http://www.logicinaction.org/) 22 / 24

Natural Deduction for Predicate Logic

For the universal quantifier ∀

∀x ϕ (ϕ)x

t

provided that no variable in t

• ccurs bounded in ϕ

E∀

(http://www.logicinaction.org/) 22 / 24

Natural Deduction for Predicate Logic

For the universal quantifier ∀

∀x ϕ (ϕ)x

t u

. . . (ϕ)x

u

∀x ϕ provided that no variable in t

• ccurs bounded in ϕ

for u a special symbol not used anywhere else in the proof

E∀

(http://www.logicinaction.org/) 22 / 24

Natural Deduction for Predicate Logic

For the universal quantifier ∀

∀x ϕ (ϕ)x

t u

. . . (ϕ)x

u

∀x ϕ provided that no variable in t

• ccurs bounded in ϕ

for u a special symbol not used anywhere else in the proof

E∀ I∀

(http://www.logicinaction.org/) 22 / 24

Natural Deduction for Predicate Logic

For the existential quantifier ∃

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Natural Deduction for Predicate Logic

For the existential quantifier ∃

∃x ϕ ,

u

(ϕ)x

u

. . . ψ ψ for u a special symbol not used anywhere in the proof

(http://www.logicinaction.org/) 23 / 24

Natural Deduction for Predicate Logic

For the existential quantifier ∃

∃x ϕ ,

u

(ϕ)x

u

. . . ψ ψ for u a special symbol not used anywhere in the proof

E∃

(http://www.logicinaction.org/) 23 / 24

Natural Deduction for Predicate Logic

For the existential quantifier ∃

∃x ϕ ,

u

(ϕ)x

u

. . . ψ ψ (ϕ)x

t

∃x ϕ for u a special symbol not used anywhere in the proof provided that no variable in t

• ccurs bounded in ϕ

E∃

(http://www.logicinaction.org/) 23 / 24

Natural Deduction for Predicate Logic

For the existential quantifier ∃

∃x ϕ ,

u

(ϕ)x

u

. . . ψ ψ (ϕ)x

t

∃x ϕ for u a special symbol not used anywhere in the proof provided that no variable in t

• ccurs bounded in ϕ

E∃ I∃

(http://www.logicinaction.org/) 23 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1]

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1]

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that t2 contains only variables that

• ccurr freely in ϕ, and

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1]

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that t2 contains only variables that

• ccurr freely in ϕ, and

t1 contains only variables that do not get bounded after replacement.

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1]

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that t2 contains only variables that

• ccurr freely in ϕ, and

t1 contains only variables that do not get bounded after replacement.

E=

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1] t = t

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that t2 contains only variables that

• ccurr freely in ϕ, and

t1 contains only variables that do not get bounded after replacement.

for any term t.

E=

(http://www.logicinaction.org/) 24 / 24

Natural Deduction for Predicate Logic

For the identity symbol =

t1 = t2 , ϕ ϕ[t1/t2] t1 = t2 , ϕ ϕ[t2/t1] t = t

where ϕ[t1/t2] is the result of replac- ing, in ϕ, some ocurrences of t2 by t1, provided that t2 contains only variables that

• ccurr freely in ϕ, and

t1 contains only variables that do not get bounded after replacement.

for any term t.

E= I=

(http://www.logicinaction.org/) 24 / 24