Section 8.3 Higher-Order Logic

1. Section 8.3 Higher-Order Logic A 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 xp(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 Order The 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 xp(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. Semantics A 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 wff W = 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 pp(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 pp(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 pp(x) is valid. Proof: 1. x p ¬ p(x) P[for IP], T 2. p ¬ p(c) 1, EI 3. ¬ p(c) 2, UI 4. ¬ ¬ p(c) 2, UI 5. False 5, T QED 1–6, IP. Alternative Proof: 1. x p ¬ p(x) P[for IP], T 2. p ¬ p(c) 1, EI 3. ¬ True(c) 2, UI 4. False 3, T QED 1–4, IP. Quiz. Give a formal proof that p xp(x) is valid. Proof: 1. p x ¬ p(x) P[for IP], T 2. x ¬ True(x) 1, UI 3. ¬ True(c) 2, EI 4. 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 ≠ yP [for CP] 2. l(x)  l(y) Ax1, UI, UI, 1, MP, EI 3. l(x)  l(y)  M(x)  M(y)  l = M Ax2, UI, UI, 1, MP,UI, UI 4. M(x)  M(y) P [for CP] 5. l(x)  l(y)  M(x)  M(y) 2, 4, Conj 6. l = M 3, 5, MP 7. M(x)  M(y)  l = M 4–6, CP 8. M (M(x)  M(y)  l = M) 7, UG 9. l(x)  l(y)  M (M(x)  M(y)  l = M) 2, 8, Conj 10. L (L(x)  L(y)  M (M(x)  M(y)  M = L)) 9, EG 11. (x ≠ y)  L (L(x)  L(y)  M (M(x)  M(y)  M = L)) 1–3, 7–10, CP 12. 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 wff B (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 wff B (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.