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Explore the principles of calculational logic, including inequalities and formal proofs, as a powerful alternative to verbal reasoning. Discover the beauty of mathematical logic first introduced by J. Robert Oppenheimer. Join us as we delve into the world of logical constructions and verifications. Learn about Towers of Hanoi and inductive reasoning, guided by the quote, "The purpose of logic is not to mimic verbal reasoning but to provide a calculational alternative". Presented by Roland Backhouse in Tallinn, Estonia on 19th November 2002.
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The Art of Effective Reasoning Roland Backhouse Tallinn, Estonia, 19th November 2002
Quote of Oppenheimer “The hallmark of a science is the avoidance of error” J. Robert Oppenheimer
Logic for all there exists (anno 1966)
Calculational Logic For all non-zero y and z y ×z is positive = (y is positive = z is positive)
Calculational Logic For all non-zero y and z y × z is positive = (y is positive = z is positive) (y ×z is positive = y is positive) = z is positive
Calculational Logic For all non-zero y and z y × z is positive = (y is positive = z is positive) (y ×z is positive = y is positive) = z is positive “The purpose of logic is not to mimic verbal reasoning but to provide a calculational alternative.” Edsger W. Dijkstra
Inequalities (Trichotomy) If a b then either a < b or b < a (Translation) If a < b then a+c < b+c If a < b and c > 0 then ac < bc
Inequalities (Trichotomy) If a b then either a < b or b < a ab a < b b < a (Translation) If a < b then a+c < b+c If a < b and c > 0 then ac < bc
Inequalities (Trichotomy) If a b then either a < b or b < a ab a < b b < a (Translation) If a < b then a+c < b+c a < b a+c < b+c If a < b and c > 0 then ac < bc
Inequalities (Trichotomy) If a b then either a < b or b < a ab a < b b < a (Translation) If a < b then a+c < b+c a < b a+c < b+c If a < b and c > 0 then ac < bc For all a,b,c, where ab and c0, a < b ac < bc 0 < c
Formal Proof 224 > 9 > 0 2 + 7 > 3 + 5
Goal-Directed Construction 2 + 7 ?3 + 5 = { property of squaring } (2 + 7)2? (3 + 5)2 = { property of squaring, and arithmetic } ? is “ > ”.
Complicating Proof 0 + 0 = 0 Rule (i) with a = 0 0(0+0) = 00 multiplying both sides by 0 00 + 00 = 00 “distributive rule”: a(b+c) = ab + ac (00 + 00) + x = 00 + x adding x to both sides, where x is the solution of 00 + x = 0 guaranteed by rule (ii) 00 + (00 + x) = 00 + x “associative rule”: a + (b+c) = (a+b) + c 00 + 0 = 0 using the property of x 00 = 0 using rule (i) with a = 00
Complicating Proof 0 + 0 = 0 0 + 0 = 0 Rule (i) with a = 0 0(0+0) = 00 multiplying both sides by 0 00 + 00 = 00 “distributive rule”: a(b+c) = ab + ac (00 + 00) + x = 00 + x adding x to both sides, where x is the solution of 00 + x = 0 guaranteed by rule (ii) 00 + (00 + x) = 00 + x “associative rule”: a + (b+c) = (a+b) + c 00 + 0 = 0 using the property of x 00 = 0 using rule (i) with a = 00 0 0 = 0
Goal-Directed Proof 00 = 0 = { a + 0 = a } 00 + 0 = 0 = { assume 00 + x = 0, Leibniz } 00 + (00 + x) = 00 + x = { associativity of + } (00 + 00) + x = 00 + x { Leibniz } 00 + 00 = 00 First four steps are the crucial ones ...
Goal-Directed Proof 00 + 00 = 00 = { distributivity } 0(0 + 0) = 00 { Leibniz } 0 + 0 = 0 = { a + 0 = a } true . … the rest is easy.
Inductive Construction 1° + 2° + + n° n 1¹ + 2¹ + + n¹ n(n+1)/2 1² + 2² + + n² n(n+1)(2n+1)/6 Conjecture: 1k + 2k + + nk is a polynomial inn of degree k+1.
Let S.n = 1 + 2 + + n P.n = (S.n = a + bn + cn² ) Then, n:: P.n = { induction } P.0 n:: P.n P.(n+1) = { P.0 = (S.0 = a) } 0 = a n:: P.n P.(n+1) = { substitution of equals for equals } 0 = a n:: P.n P.(n+1) [a:= 0]
P.(n+1) [a:= 0] = { definition } S.(n+1) = b(n+1) + c(n+1)² = { S.(n+1) = S.n + (n+1) } S.n + n + 1 = b(n+1) + c(n+1)² = { assume S.n = bn + cn² } bn + cn² + n + 1 = b(n+1) + c(n+1)² = 0 = (b + c - 1) + (2c - 1)n Hence n:: P.n a = 0 b = ½ c = ½
Towers of Hanoi An atomic move is a pair k,d where k is the disk number and d is the direction (clockwise or anticlockwise) the disk is to be moved. H.(n,d) is a sequence of atomic moves that results in moving the top n disks from one pole to the next pole in a direction d. H.(0,d) = [] H.(n+1,d) = H.(n,d) ; [n,d] ; H.(n,d)
Invariant The value of even.n d remains constant for every call of H. Proof: Compare H.(n+1,d) = H.(n,d) ; [n,d] ; H.(n,d) with even.(n+1) d = { contraposition: p q p q } (even.(n+1)) d = { property of even } even.n d .
Invariant The value of even.n d remains constant for every call of H. Hence: • The direction of movement of individual disks is constant. • Alternate disks move in alternate directions. • The direction of movementof disk 0 (the smallest disk) is given by even.0 d0 even.n dn where dk donates the direction of movement of disk k.
Invariant Equivalently, d0 dn even.n . That is, the smallest disk is moved in the same direction as the largest disk exactly when the number given to the largest disk is even.
Summary • Calculational Logic • Goal-directed • Construction, not Verification