1

Section 8.3 Higher-Order LogicA logic is higher-order if it allows predicate names or function names to be quantified or to be arguments of a predicate.Example. The sentence, “There is a function with a fixed point.” can be formalized as

f x p(ƒ(x), x), where p denotes equality.Example. Given the sentence, “ Every binary relation that is irreflexive and transitive is antisymmetric.” The sentence can be formalized as

p (x ¬ p(x, x) x y z (p(x, y) p(y, z) p(x, z)) x y (p(x, y) ¬ p(y, x))).We can define higher-order logic in terms of sets because predicates and functions are sets.Example (predicates). P(x) is true iff x P.

Q(x, y) is true iff (x, y) Q.Example (functions). ƒ(x) = y iff ƒ(x, y) is true iff (x, y) ƒ.So a higher-order logic allows sets to be quantified or to be elements of other sets.Classifying Logics by OrderThe order of a predicate is 1 if its arguments are terms. Otherwise the order is n + 1 where n is the maximum order of the arguments that are not terms. The order of a function is always 1 since it’s arguments are always terms.Examples. In the wff p(x) q(x, p) the order of p is one and the order of q is two.In the wff p(x) q(p) r(q) the orders of p, q, and r are 1, 2, and 3, respectively. Note: We agree that a set cannot be defined in terms of itself. So we don’t allow something like p(q) q(p) in the same wff since it would mean q p and p q.

2

The order of a quantifier is 1 if it quantifies a variable and 2 if it quantifies a function. Otherwise the order is n + 1 where n is the order of the predicate being quantified. The order of wff is the maximum of the orders of its predicates and quantifiers.An nth-order logic is a logic with wffs having order n or less. Example. The wff f x p(ƒ(x), x) has order 2.Example/Quiz. What is the order of the following wff?

B (H(B) R W (B(R) R(W)))Answer: The order is 3. As sets, we have W R B H, where W is a variable.SemanticsA wff has meaning only with respect to an interpretation. Choose a domain D ≠ . • Assign constants and free variables to elements of D.• Assign free functions to functions over D.• Assign free predicates to relations with respect to D. Example. Given the wff

x p (p(x) S(p)).The predicate p has order 1 and the predicate S has order 2. So p(x) S(p) can be written as x p and p S, or x p S. So for any domain D we have

x D, p power(D), and S power(power(D)). So p D and S power(D). We’ll examine some different interpretations of the wff.

3

Example. Given the wffW = x p (p(x) S(p)).

We’ll look at three sample interpretations.(1) Let I be the interpretation with domain D = {a, b} and S = {{a}, {b}}.Since x varies over {a, b}, the wff W wrt I can be written as

W = p (p(a) S(p)) p (p(b) S(p)) True True (Let p = {a} for the first part. So a p and p S;

Let p = {b} for the second part. So b p and p S.) True.

(2) Let I be the interpretation with domain D = {a, b} and S = {, {b}}.Since x varies over {a, b}, the wff W wrt I can be written as

W = p (p(a) S(p)) p (p(b) S(p)) False True (For each p S it follows that p(a) is false. i.e., a p.) False.

(3) Let I be the interpretation with domain D = {a, b} and S = .Then S(p) is false for all p D. So W is false wrt I.Example. Given the wff

W = x p (p(x) S(p)).W is unsatisfiable because for any interpretation with domain D, p must vary over all predicates, including p = , which means p(x) is false for all x D.

4

Quiz. Do a truth analysis for the wff x p p(x).Answer: The wff is valid because for any domain D, let p(x) be true for all x D. From a set viewpoint, this means choose p = D. Alternatively, for each x, choose p = {x}.Quiz. Do a truth analysis for the following wff x p p(x).Answer: The wff is unsatisfiable because for any domain D, let p(x) be false for all x D. From a set viewpoint, this means choose p = .Formal Proofs. The quantifier inference rules extend to higher-order logics. Example. Formal proof that x p p(x) is valid.

Proof: 1. x p ¬ p(x) P [for IP], T2. p ¬ p(c) 1, EI3. ¬ p(c) 2, UI4. ¬ ¬ p(c) 2, UI5. False 5, T

QED 1–6, IP.

Quiz. Give a formal proof that p x p(x) is valid.Proof: 1. p x ¬ p(x) P [for IP], T

2. x ¬ True(x) 1, UI3. ¬ True(c) 2, EI4. False 3, T

QED 1–4, IP.

Alternative Proof:1. x p ¬ p(x) P [for IP], T2. p ¬ p(c) 1, EI3. ¬ True(c) 2, UI4. False 3, T

QED 1–4, IP.

5

Hilbert’s Euclidean Geometry. The axioms (See Text Page 499) are as follows.Ax1: On any two distinct points there is a line.

x y ((x ≠ y) L (L(x) L(y)).Ax2: On any two distinct points there is at most one line.

x y ((x ≠ y) L M (L(x) L(y) M(x) M(y) L = M)).Ax3: On every line there are at least two points.

L x y ((x ≠ y) L(x) L(y)).Ax4: There are at least three points which are not on the same line.

x y z ((x ≠ y) (x ≠ z) (y ≠ z) L (L(x) L(y) ¬ L(z))).Example/Quiz. Ax1 and Ax2 imply that on any two distinct points there is a unique line.

x y ((x ≠ y) L (L(x) L(y) M (M(x) M(y) M = L))).1. x ≠ y P [for CP]2. l(x) l(y) Ax1, UI, UI, 1, MP, EI3. l(x) l(y) M(x) M(y) l = M Ax2, UI, UI, 1, MP,UI, UI4. M(x) M(y) P [for CP]5. l(x) l(y) M(x) M(y) 2, 4, Conj6. l = M 3, 5, MP7. M(x) M(y) l = M 4–6, CP8. M (M(x) M(y) l = M) 7, UG9. l(x) l(y) M (M(x) M(y) l = M) 2, 8, Conj10. L (L(x) L(y) M (M(x) M(y) M = L)) 9, EG11. (x ≠ y) L (L(x) L(y) M (M(x) M(y) M = L)) 1–3, 7–10, CP12. x y ((x ≠ y) L (L(x) L(y) M (M(x) M(y) M = L))) 11, UG, UG

QED.

6

Example. Suppose someone interprets the third order wffB (H(B) R w (B(R) R(w)))

as, “Every house has a room with a window.”From a formal viewpoint, is this an interpretation I with domain D?The wff uses w as an individual variable and the predicates have the following meanings in terms of sets: H(B) means B H; B(R) means R B; and R(w) means w R. A possible formal Interpretation:Let D be the set of all windows. Then we have the following statements. • w varies over D. So w varies over windows.• R varies over subsets of D. So R(w) means w is a window in R. So R varies over sets of

windows. So we’ll let a room be a set of windows.• B varies over subsets of subsets of D. So B(R) means R is a set of windows in B. So B

varies over sets of sets of windows. So B varies over sets of rooms. So we’ll let a building be a set of rooms.

• H is a free predicate of order 3 so it must be defined as a subset of a subset of a subset of D. So H(B) means B is a set of a set of windows. So H(B) means B is a set of rooms. So H(B) means B is a building. We’ll let H(B) mean that B is a building with 10 or less rooms, which we’ll call a house.

So the wff might be described by:Every building (a set of rooms) that is a house (a building with 10 or less rooms) has a room (a set of windows) with a window.

7

Example/Quiz. In the previous example we considered the third order wffB (H(B) R w (B(R) R(w)))

together with the interpretation, “Every house has a room with a window.”What if we want to think of house, room, and window as predicates to describe things (I.e, the domain D is the set of all things. For example, house(x) means x is a house, room(x) means x is a room, and window(x) means x is a window. How would we formalize the sentence using these predicates?A solution:

X (house(X) Y (room(Y) X(Y) Z (window(Z) Y(Z)))).A less intuitive solution: Replace house by H, room by R, and window by W.

X (H(X) Y (R(Y) X(Y) Z (W(Z) Y(Z)))).Quiz. What is the order of the wff?Answer: 3.