Logic as a Tool Chapter 4: Deductive Reasoning in First-Order Logic - - PowerPoint PPT Presentation

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Logic as a Tool Chapter 4: Deductive Reasoning in First-Order Logic 4.3 Natural Deduction for First-Order Logic

Valentin Goranko Stockholm University October 2016

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Natural Deduction

◮ ND: System for structured deduction from a set of assumptions,

based on rules, specific to the logical connectives.

◮ For each logical connective:

introduction rule(s) and elimination rule(s)

◮ Introduction (opening) and cancelation (closing, discharge) of

assumptions. Assumptions can be re-used many times before canceled.

◮ Cancelation of assumptions: only when the rules allow it, but not an

• bligation.

◮ All open assumptions at the end of the derivation must be declared.

The less assumptions, the stronger the derivation.

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Quantifier rules for Natural Deduction

Introduction rules: Elimination rules: (∀I)∗ A[c/x] ∀xA(x) (∃I)∗∗ A[t/x] ∃xA(x) (∀E)∗∗ ∀xA(x) A[t/x] (∃E)∗∗∗ ∃xA(x) [A[c/x]]1 . . . C C 1

∗where c is a constant symbol, not occurring in A(x), nor in any open

assumption used in the derivation of A[c/x].

∗∗for any term t free for x in A. ∗∗∗where c is a constant symbol, not occurring in A(x), nor in C or in

any open assumption in the derivation of C, except for A[c/x].

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First-order Natural Deduction: Warning: a common mistake

How about replacing the rule (∃E) with the following, simpler one? (∃E ′) ∃xA(x) A[c/x] where c is a new constant symbol. Though simple and looking natural, this rule is not valid! Using it, one can derive the invalid implication ∃xA(x) → A[c/x].

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Notational convention

When A = A(x), i.e., x occurs free in A, and t is free for substitution for x in A, we sometimes write A(t) instead of A[t/x].

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First-order Natural Deduction: Example 1

⊢ND ∀x∀yP(x, y) → ∀y∀xP(x, y) : (∀E) [∀x∀yP(x, y)]1 (∀E) ∀yP(c1, y)

(∀I)

P(c1, c2)

(∀I)

∀xP(x, c2)

(→I)

∀y∀xP(x, y) ∀x∀yP(x, y) → ∀y∀xP(x, y)

1

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First-order Natural Deduction: Example 2

∀x(P(x) ∧ Q(x)) ⊢ND ∀xP(x) ∧ ∀xQ(x) : (∧I) (∀I)

(∧E)

(∀E) ∀x(P(x) ∧ Q(x))

P(c) ∧ Q(c) P(c) ∀xP(x) (∀I)

(∧E)

(∀E) ∀x(P(x) ∧ Q(x))

P(c) ∧ Q(c) Q(c) ∀xQ(x) ∀xP(x) ∧ ∀xQ(x)

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First-order Natural Deduction: Example 3

¬∃x¬A(x) ⊢ND ∀xA(x) : ¬∃x¬A(x) [¬A(c)]1 ∃x¬A(x) ⊥ A(c) 1 ∀xA(x)

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First-order Natural Deduction: Example 4

¬∀x¬A(x) ⊢ND ∃xA(x) : ¬∀x¬A(x) [¬∃xA(x)]2 [A(c)]1 ∃xA(x) ⊥ ¬A(c)

1

∀x¬A(x) ⊥ ∃xA(x) 2

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First-order Natural Deduction: Example 5

Suppose x is not free in P. Then ⊢ND ∀x(P → Q(x)) → (P → ∀xQ(x)): (→ I) (→ I)

(∀I)

(→E) (∀E) [∀x(P → Q(x))]1

P → Q(c)

,

[P]2 Q(c) ∀xQ(x) P → ∀xQ(x) 2 ∀x(P → Q(x)) → (P → ∀xQ(x)) 1

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First-order Natural Deduction: Example 6

Suppose x is not free in P. Then ∀x(Q(x) → P), ∃xQ(x) ⊢ND P : (∃E) ∃xQ(x) (→ E)

(∀E) ∀x(Q(x) → P)

Q(c) → P

,

[Q(c)]1 P P 1

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First-order Natural Deduction: a challenge!

Derive in ND the formula: ∃x(P(x) → ∀yP(y))