Logic as a Tool Chapter 3: Understanding First-order Logic 3.3 - - PowerPoint PPT Presentation

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Logic as a Tool Chapter 3: Understanding First-order Logic 3.3 Basic grammar and use of first-order languages

Valentin Goranko Stockholm University October 2016

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Translation from natural languages to first-order logic: examples in the structure of real numbers R

There is a real number greater than 2 and less than 3.” ∃x(x > 2 ∧ x < 3). There is an integer greater than 2 and less than 3.” ∃x(I(x) ∧ x > 2 ∧ x < 3). where I(x) is interpreted as ‘x is an integer. There is no real number the square of which equals −1.” How about ∃x(¬x2 = −1)? No! The sentence actually says “It is not true that there is a real number the square of which equals −1.” The correct translation is ¬∃x(x2 = −1).

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Translation from natural languages to first-order logic: examples in the structure of humans H

Translate to first-order logic “Every man loves a woman.” ∀x∃yL(x, y)? No! This means ‘Everybody loves somebody.’. We must restrict the quantification of x to men, and of y respectively to women. For that purpose we transform the sentence to: “For every human, if he is a man, then there is a human who is a woman and the man loves that woman.” Now the translation into LH is immediate: ∀x(M(x) → ∃y(W(y) ∧ L(x, y))). Now, translate “Every mother has a child whom she loves.” ∀x(∃y(x = m(y)) → ∃z(C(z, x) ∧ L(x, z))).

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Restricted quantification

To quantify only over those elements of the domain that satisfy a given (definable) property P, we use restricted quantification.

• For existential restricted quantification we use the template:

∃x(P(x) ∧ . . .)

• For universal restricted quantification we use the template:

∀x(P(x) → . . .) For instance: ∃x(x > 0 ∧ x2 + x < 5) interpreted in R, says that there exists a real number x which is positive and which satisfies x2 + x < 5. Likewise, ∀x(x > 0 → x2 + x < 5) interpreted in R says that all real numbers x which are positive satisfy x2 + x < 5.

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Free and bound variables intuitively

Two essentially different ways in which we use individual variables in first-order formulae:

• 1. Free variables: used to denote unknown or unspecified objects, as in

(5 < x) ∨ (x2 + x − 2 = 0).

• 2. Bound variables: used to quantify, as in

∃x((5 < x) ∨ (x2 + x − 2 = 0)) and ∀x((5 < x) ∨ (x2 + x − 2 = 0)).

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Scope of a quantifier. Bound and free occurrences of variables.

Scope of (an occurrence of) a quantifier Q in a given formula A is the unique subformula QxB beginning with that occurrence of the quantifier Q in A. Some examples of scopes of quantifiers: ∀x((x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3)))). ∀x((x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3)))). ∀x((x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3)))). An occurrence of the quantifier Q followed by a variable x in a formula A is said to bind the variable x in A. An occurrence of a variable x in a formula A is bound if it is in the scope

• f an occurrence of a quantifier Q binding x in A.

Otherwise, that occurrence of x is free in A.

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Which quantifiers binds which occurrences of variables?

A bound occurrence of a variable x in a given formula A is bound by the innermost occurrence of a quantifier Qx in A, in the scope of which it

• ccurs.

For example, the first occurrence of ∀ in the following formula binds the first 3 occurrences of x: ∀x((x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3)))), whereas the occurrence of ∃ binds the last 2 occurrences of x: ∀x((x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3)))).

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Bound and free variables in a formula

A variable x is bound in the formula A if it has at least one bound

• ccurrence in A. Then x is said to be a bound variable of A.

A variable x is free in the formula A if it has at least one free occurrence in A. Then x is said to be a free variable of A. For instance, in the formula A = (x > 5) → ∀y(y < 5 → (y < x ∧ ∃x(x < 3))). the first two occurrences of x are free, while all other occurrences of variables are bound. Thus, the only free variable in A is x, while both x and y are bound in A. Note that the same variable can be both free and bound in the same formula, e.g. x in the formula x > 0 ∧ ∃x(5 < x).

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Open and closed formulae

NB: The truth of a formula A in a given structure S only depends on the assignment of values to the free variables occurring in A. That is, if FV (A) is the set of free variables in A and v1, v2 are variable assignments in S such that v1 |FV (A)= v2 |FV (A), then S, v1 | = A iff S, v2 | = A. An open formula is a formula with no bound variables, i.e. not containing any occurrences of quantifiers. A closed formula, or a sentence is a formula with no free variables. Thus, the truth of a sentence in a structure does not depend on the variable assignment.

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Using bound and free variables in a formula

Free variables have their own values in a given formula (determined by a variable assignment), while bound variables only play a dummy role and can be replaced (with care!) by one another. For instance, the sentence ∃x(5 < x ∧ x2 + x − 2 = 0) means exactly the same as ∃y(5 < y ∧ y2 + y − 2 = 0), in the sense that both formulae always have the same truth value. Likewise, ∀x(5 < x ∨ x2 + x − 2 = 0) means the same as ∀y(5 < y ∨ y2 + y − 2 = 0). On the other hand, the meaning of 5 < x ∧ x2 + x − 2 = 0 is essentially different from the meaning of 5 < y ∧ y2 + y − 2 = 0. Both formulae state the same, but about different individuals.

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Reusing variables as free and bound in a formula

The same variable can occur both free and bound in a formula: x > 5 → ∀x(2x > x). However, the free occurrence of x has nothing to do with the bound

• ccurrences of x:

x > 5 → ∀x(2x > x). Thus, the formula above has the same meaning as x > 5 → ∀y(2y > y), but not the same meaning as y > 5 → ∀x(2x > x).

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Binding a variable by different quantifiers in a formula

Different occurrences of the same variable can be bound by different quantifiers: ∃x(x > 5) ∨ ∀x(2x > x). Again, the occurrences of x, bound by the first quantifier, have nothing to do with those bound by the second one. For instance, the two x’s claimed to exist in the formula ∃x(x > 5) ∧ ∃x(x < 3). need not (and, in fact, cannot) be the same. Thus, the formula above has the same meaning as each of ∃y(y > 5) ∧ ∃x(x < 3), ∃x(x > 5) ∧ ∃z(z < 3), ∃y(y > 5) ∧ ∃z(z < 3).

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Nested bindings of a variable in a formula

Different bindings of the same variable can be nested, e.g.: ∀x(x > 5 → ∃x(x < 3)). Again, the occurrences of x in the subformula ∃x(x < 3) are bound by ∃ and not related to the first two occurrences of x, bound by ∀: ∀x(x > 5 → ∃x(x < 3)). Thus, the formula above has the same meaning as each of ∀x(x > 5 → ∃y(y < 3)), ∀z(z > 5 → ∃x(x < 3)), ∀z(z > 5 → ∃y(y < 3)).

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Renaming of a bound variable in a formula

Using the same variable for different purposes in a formula can be confusing, and is often unwanted, so we may want to eliminate it. Renaming of the variable x in a formula A is the substitution of all

• ccurrences of x bound by the same occurrence of a quantifier in A with

another variable, not occurring in A. E.g., a possible renaming of (x > 5) ∧ ∀x(x > 5 → ¬∃x(x < y)) is the formula (x > 5) ∧ ∀x(x > 5 → ¬∃z(z < y)) However, neither of the following formulae is a correct renaming: (z > 5) ∧ ∀x((x > 5) → ¬∃x(x < y)), (x > 5) ∧ ∀z((z > 5) → ¬∃z(z < y)), (x > 5) ∧ ∀x(x > 5 → ¬∃y(y < y)). Claim: The result of renaming a variable in a formula A always has the same truth value as A.

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Clean formulae

A formula A is clean if no variable occurs both free and bound in A and every two occurrences of quantifiers bind different variables. Thus, ∃x(x > 5) ∧ ∃y(y < z) is clean, while ∃x(x > 5) ∧ ∃y(y < x) and ∃x(x > 5) ∧ ∃x(y < x) are not. Claim: Every formula can be transformed into a clean formula by means of several consecutive renamings of variables. E.g., (x > 5) ∧ ∀x((x > 5) → ¬∃x(x < y)) can be transformed into a clean formula as follows: (x > 5) ∧ ∀x1((x1 > 5) → ¬∃x(x < y)), (x > 5) ∧ ∀x1((x1 > 5) → ¬∃x2(x2 < y)).

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Substitution of a term for a variable in a formula

Unform substitution of a term t for a variable x in a formula A means that all free occurrences x in A are simultaneously replaced by t. The result of the substitution is denoted A[t/x]. Example: given the formula A = ∀x(P(x, y) → (¬Q(y) ∨ ∃yP(x, y))) we have A[f (y, z)/y] = ∀x(P(x, f (y, z)) → (¬Q(f (y, z)) ∨ ∃yP(x, y))), while A[f (y, z)/x] = A because x does not occur free in A. Intuitively, A[t/x] is supposed to say about the individual denoted by t the same as what A says about the individual denoted by x. Question: is that always the case? Is a substitution of a term for a formula always ’safe’?

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Capture of a variable in substitution

The formula A = ∃y(x < y) is true in N for any value of x. However, A[(y + 1)/x] = ∃y(y + 1 < y), which is false in N. Therefore, the formula A[(y + 1)/x] does not say about the term y + 1 the same as what A says about x. What went wrong? The occurrence of y in the term y + 1 got captured by the quantifier ∃y, because we mixed the free and the bound uses of y. Capture: new occurrences of a variable y in the scope of a quantifier Qy introduced as a result of substitution of a term t containing y for another variable x in a formula.

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Terms free for substitution for a variable in a formula

A term t is free for (substitution for) a variable x in a formula A, if no variable in t is captured by a quantifier when t is substituted for x in A. Examples: The term f (x, y) is free for substitution for y in the formula A = ∀x(P(x, z) ∧ ∃yQ(y)) → P(y, z), resulting in A[f (x, y)/y] = ∀x(P(x, z) ∧ ∃yQ(y)) → P(f (x, y), z), but it is not free for substitution for z in A, resulting in A[f (x, y)/z] = ∀x(P(x, f (x, y)) ∧ ∃yQ(y)) → P(y, f (x, y)), because a capture occurs: A[f (x, y)/z] = ∀x(P(x, f (x, y)) ∧ ∃yQ(y)) → P(y, f (x, y)). Note that every ground term (not containing variables), in particular every constant symbol, is always free for substitution.

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Renamings and substitutions in a formula

NB: renaming and substitution are different operations: renaming always acts on bound variables, while substitution always acts on free variables. Also, renamings preserve the formula up to logical equivalence, while substitutions do not. On the other hand, a suitable renaming of a formula can prepare it for a substitution, by rendering the term to be substituted free for such substitution in the renamed formula. For instance, the term f (x, y) is not free for substitution for y in A = ∀x(P(x, y) ∧ ∃yQ(y)), but it becomes free for such substitution after renaming of A to A′ = ∀x′(P(x′, y) ∧ ∃yQ(y)).