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CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 1 Lecture 7 CS 1813 – Discrete Mathematics Equational Reasoning Back to the Future:

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Presentation on theme: "CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 1 Lecture 7 CS 1813 – Discrete Mathematics Equational Reasoning Back to the Future:"— Presentation transcript:

1 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 1 Lecture 7 CS 1813 – Discrete Mathematics Equational Reasoning Back to the Future: High-School Algebra

2 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 2 Some Laws of Algebra  a + 0 = a{+ identity}  (-a) + a = 0{+ complement}  a  1 = a{  identity}  a  0 = 0{  null}  a + b = b + a{+ commutative}  a + (b+c) = (a+b) + c{+ associative}  a  (b+c) = a  b + a  c {distributive law} Equations go both ways

3 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 3 Theorem (-1)  (-1) = 1 (-1)  (-1) = ((-1)  (-1)) + 0{+ id} = ((-1)  (-1)) + ((-1) + 1){+ comp} = (((-1)  (-1)) + (-1)) + 1{+ assoc} = (((-1)  (-1)) + (-1)  1) + 1{  id} = ((-1)  ((-1) + 1)) + 1{dist law} = ((-1)  0) + 1{+ comp} = 0 + 1{  null} = 1 + 0{+ comm} = 1{+ id} QED proof by equational reasoning

4 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 4 Laws of Boolean Algebra page 1 From Fig 2.1, Hall & O’Donnell, Discrete Math with a Computer, Springer, 2000

5 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 5 Laws of Boolean Algebra page 2 From Fig 2.1, Hall & O’Donnell, Discrete Math with a Computer, Springer, 2000

6 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 6 Theorem (a  False)  (b  True) = b equations {rule}substitution [formula in eqn / variable in rule] (p  False)  (q  True) names changed to clarify substitutions = False  (q  True) {  null}[p /a] = (q  True)  False {  comm} [False /a] [q  True /b] = q  True {  id}[q  True /a] = q {  id}[q /a] QED

7 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 7 Using Equational Proof Checker import Stdm th7 = (P `And` FALSE) `Or` ( Q `And` TRUE) `thmEq` Q pr7 = startProof ((P `And` FALSE) `Or` (Q `And` TRUE)) (FALSE `Or` (Q `And` TRUE), andNull) ((Q `And` TRUE) `Or` FALSE, orComm) (Q `And` TRUE, orID) (Q, andID) Notepad window Prelude> :cd DMf00 Prelude> :cd Lectures Prelude> :load lecture07.hs Reading file "lecture07.hs": Reading file "Stdm.lhs": Reading file "lecture07.hs": Hugs session for: C:\HUGS98\lib\Prelude.hs Stdm.lhs lecture07.hs Main> check_equation th7 pr7 The proof is correct Hugs Session Importing tools Green indicates cmd from user

8 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 8 Equations the Proof Checker Knows andNull orNull andID orID andIdempotent orIdempotent andComm orComm andAssoc orAssoc andDistOverOr orDistOverAnd deMorgansLawAnd deMorgansLawOr negTrue negFalse andCompl orCompl dblNeg currying implication contrapositive absurdity

9 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 9 Theorem (a  b)  b = b  absorption equations {rule}substitution [formula in eqn / variable in rule] (p  q)  q names changed to clarify substitutions = (p  q)  (q  True) {  id} [q /a] = (q  p)  (q  True) {  comm} [p /a] [q /b] = q  (p  True) {  dist over  }[q /a] [True /b] [p /c] = q  True {  null} [p /a] = q {  id}[q/a] QED

10 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 10 Theorem (a  b)  b = b  absorption equations {rule}substitution [formula in eqn / variable in rule] (p  q)  q names changed to clarify substitutions … exercise … = q

11 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 11 Consistent, But Not Minimal redundancy among laws of Boolean algebra equations {rule}substitution [formula in eqn / variable in rule] p  q = (  p)  q {imp} [p /a] [q /b] =  (  ((  p)  q)) {dbl neg} [(  p)  q /a] =  ((  (  p))  (  q)) {DeMorgan  } [  p /a] [q /b] =  (p  (  q)) {dbl neg} [p /a] = (  p)  (  (  q)) {DeMorgan  } [p /a] [  q /b] = (  (  q))  (  p) {  comm} [  p /a] [  (  (q)) /b] = (  q)  (  p) {imp} [  q /a] [  p /b] QED Deriving the contrapositive law Theorem (contrapositive): a  b =  b   a A proof using laws other than the contrapositive law This proof is sloppy! Where’s the shortcut?

12 CS 1813 Discrete Mathematics, Univ Oklahoma Copyright © 2000 by Rex Page 12 End of Lecture 7


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