Quizzes CS 1813 – Discrete Mathematics Lecture 7 - CS 1813 Discrete Math, University of Oklahoma 12/2/2018 Quizzes CS 1813 – Discrete Mathematics CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page

CS 1813 Discrete Mathematics, Univ Oklahoma But First … a quiz Prove the following sequent a  (b) |– (a  b) Proof {I} triggers discharge a  (b False) {EL} a a  b a  (b False) {E} b { ER} b False {E} False {I} (a  b)  False CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page

CS 1813 Discrete Mathematics, Univ Oklahoma But First … a quiz Prove the following sequent |– (a  b)  (b  c) Proof {I} triggers discharge a  b { ER} b {IL} b  c {I} a  b  b  c CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page

CS 1813 Discrete Mathematics, Univ Oklahoma But First … a quiz Prove the following Boolean equation (a  b) = a  (b) Proof (a  b) = ((a)  b) {implication} = ((a))  (b) {DeMorgan } = a  (b) {double neg} QED CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page

CS 1813 Discrete Mathematics, Univ Oklahoma Quiz map f [ ] = [ ] (map).[] map f (x: xs) = (f x): (map f xs) (map).: Prove k  {1, 2, …, (length xs)}. (map f xs)k = f(xsk) ysk denotes the kth element of ys for any sequence ys = [y1, y2, … yn], [y1, y2, … yn]k = yk Assume that xs has a finite number of elements Proof k  {1, 2, …, 0}. (map f xs)k = xsk (Isaac’s rule, since {1, …, 0} = ) k  {1, 2, …, (length(x: xs))}. (map f (x: xs))k = f((x: xs)k) = k  {1, 2, …, 1+(lengthxs)}. ((f x): (map f xs))k = f((x: xs)k) True for k=1 ((f x): (map f (x: xs)))1 = (f x) = f((x: xs)1) True for 2  k  1+(length xs) induction hypothesis, since (y: ys)k = ysk-1 when k  2 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page