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

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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 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 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. 1

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. 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. 2

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. 3

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], 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 x p(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. 4

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, 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 5 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. 6

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. 7

Predicate Logic: Sematics Part 1

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