Lecture 10 CS 1813 Discrete Mathematics. Quantify What? Reasoning with Predicates

Transcription

1 Lecture 10 CS 1813 Discrete Mathematics Quantify What? Reasoning with Predicates 1

2 More Examples with Forall the Universal Quantifier L predicate about qsort L(n) length(qsort[a 1, a 2,, a n ] ) = n Universe of discourse: U = N = {0, 1, 2, } n.l(n) Do you expect n.l(n) to be True? I another predicate about qsort I(n, k) (qsort[a 1, a 2,, a n ] = [b 1, b 2,, b n ] ) (b k b k+1 ) Universe of discourse: U = {(n, k) N N 0 < k < n} (n, k).i(n, k) A predicate calculus formula because each I(n, k) is a proposition (a non-atomic one in this case) Do you expect (n, k).i(n, k) to be True? Alternative formulation: n > 1. 0 < k < n. I(n, k) I(n, k) is a proposition So, 0 < k < n. I(n, k) is a WFF in predicate calculus So, n > 1. 0 < k < n. I(n, k) is a WFF in predicate calc 2

3 Still More Examples with Forall the Universal Quantifier Predicates about (++) C(n, m) length( [a 1, a 2,, a n ] ++ [b 1, b 2,, b m ] ) = n + m Universe of discourse U = N N = {(0,0), (1,0), (0,1), (2,0), } (n, k).c(n, k) Do you expect (n, k).c(n, k) to be True? A more common way to express this idea Universe of discourse: U = N n. k. C(n, k) Nested formula, same universe of discourse on each level A(xs, ys, zs) xs ++ (ys ++ zs) = (xs ++ ys) ++ zs Universe of discourse: U = {xs xs :: [a] }, a Haskell types xs. ys. zs. A(xs, ys, zs) Note three levels of nesting Do you expect this formula to be True? 3

4 the Existential Quantifier, There Exists x.p(x) This formula is a WFF of predicate calculus whenever P(x) is a WFF of predicate calculus True if there is at least one x in the universe of discourse for which the proposition P(x) is True False if x. P(x) is True Equivalent to forming the Logical Or of all P(x) s Example E predicate about maximum E(n, k) maximum[s 1, s 2,, s n ] = s k k.e(23, k) Universe of discourse: U = {1, 2,, 23} k.e(23, k) means E(23,1) E(23,2) E(23,23) Do you think k.e(23, k) is True? Note: When U is finite, quantifiers not required Clumsy to write big formulas without quantifiers, though Without quantifiers, reasoning can be more complex, too 4

5 Another Example with There Exists the Existential Quantifier R predicate about qsort R(n, j, k) (qsort[a 1, a 2,, a n ] = [b 1, b 2,, b n ] ) (a j =b k ) Universe of discourse Triples of non-zero natural numbers where no number in the triple exceeds the first number in the triple U = {(n, j, k) N N N 0 < k n, 0 < j n} k.r(1009, 503, k) Universe of discourse: U = {1, 2, 3,, 1009} Do you expect k.r(1009, 503, k) to be True? Forall and There Exists, in combination n > 0. 0 < j n. 0 < k n. R(n, j, k) Universe of discourse: U = N = {0, 1, 2, } Must use nesting in this case (because of mixture of and ) The universe of discourse is actually different for n than for j and k in this formula, but the constraints spell this out 5

6 Free Variables and Bound Variables Variables in WFFs of predicate calculus Denoted by lower-case letters Examples of predicate calculus WFFs with variables Which variables are free? F(p, q) G(q, r) ( x.f(x)) (G(y) H(y)) ( x.f(x, y) G(y)) (H(z) K(x)) variables: p, q, r variables: x, y variables: x, y, z Free variables and bound variables Let e stand for a WFF of predicate calculus Bound variable x. e x. e x is bound in the formula x. e x is bound in the formula x. e Free variables are variables that are not bound 6

7 Arbitrary Variables A variable is arbitrary in a proof if it does not occur free in any undischarged assumption of that proof Examples x arbitrary? x. F(x) { E} F(x) G(x, y) { I} y. G(x, y) x arbitrary? Yes No x arbitrary? discharged P(x) Q { E L } P(x) Yes { I} P(x) Q P(x) { I} x. P(x) Q P(x) 7

8 Inference Rules of Predicate Calculus Renaming Variables F(x) {x arbitrary, y not in F(x)} {R} F(y) x. F(x) {y not in F(x)} { R} y. F(y) x. F(x) {y not in F(x)} { R} y. F(y) Introducing/Eliminating Quantifiers F(x) {x arbitrary} { I} x. F(x) F(x) { I} x. F(x) Which rules trigger discharges? x. F(x) {universe is not empty} { E} F(x) x. F(x) F(x) A {x not free in A} { E} A...plus the inference rules of propositional calculus 8

9 An Easy Proof, Just to Warm Up using { I} and { E} Theorem ( commutes) x. y. F(x,y) y. x. F(x,y) proof x. y.f(x,y) { E} y.f(x,y) { E} F(x,y) {note: x arbitrary} { I} x. F(x,y) {note: y arbitrary} { I} y. x. F(x,y) x.f(x) { E} F(x) plays role of F(x) in { E} rule F(x) {x arb} { I} x.f(x) 9

10 Existential Elimination something like { E} Theorem 31 proof x. P(x), x. P(x) Q(x) x. Q(x) x. P(x) Q(x) { E} P plays role of F P(x) P(x) Q(x) { E} in { E} rule Q(x) { I} x. P(x) x. Q(x) { E} x. Q(x) {x not free in x.q(x)} discharge x. F(x) F(x) A {x not free in A} { E} A x U. Q(x)plays role of A in { E} rule 10

11 Bad Theorem An Incorrect Proof x. P(x), x. P(x) Q(x) x. Q(x) Purported proof x. P(x) Q(x) { E} P(x) P(x) Q(x) Problem is here { E} Q(x) { I} x. P(x) x. Q(x) { E} x. Q(x) y is free in this assumption So, { I} rule is not properly cited. F(x) {x arbitrary} { I} x. F(x) 11

12 Existential Qualifier Can Move In but it can t move out like a roach motel Theorem 32 y. F(y) { E} F(y) F(x) { I} x. F(x) x. y.f(x, y) y. x. F(x, y) F(, y) F(y) {y arbitrary} { I} y. F(y) x.f(x, ) F(x, ) y. F(x, y) { E} F(x, y) { I} x. F(x, y) {y arb} { I} x. y.f(x, y) y. x. F(x, y) { E} y.f(, y) y. x.f(x, y) x. F(x) F(x) A {x not free in A} { E} A 12

13 End of Lecture 13