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Transcript 
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
Equality of propositions ( if and only if )
=
True
False
True
True
False
False
False
True
Calculational Logic
Equality of propositions ( if and only if )
=
True
False
True
True
False
False
False
True
Equality of numbers ( is )
=
0
1
2
3
4
0
True
False
False
False
False
1
False
True
False
False
False
2
False
False
True
False
False
3
False
False
False
True
False
4
False
False
False
False
True
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
(y × z is positive =
= 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
(y × z is positive =
= 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
ab

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
ab

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
ab

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 ab and c0,
a < b  ac < bc  0 < c
Formal Proof
0.
if a > 0 and b > c > 0 then a + b > a + c > 0
1.
if a > b > 0 then a > b > 0
2.
224 > 9 > 0
3.
224 > 9 > 0
( 1 and 2 )
4.
414 > 3 > 0
( 3 and arithmetic )
5.
57 + 414 > 57 + 3 > 0
( 0 and 4 )
6.
(57 + 414) > (57 + 3) > 0
( 1 and 5 )
7.
1 + 214 > 215 > 0
( 6 and arithmetic )
8.
8 + 1 + 214 > 8 + 215 > 0
( 0 and 7 )
9.
(8 + 1 + 214) > (8 + 215) > 0
( 1 and 8 )
10. 2 + 7 > 3 + 5 > 0
( 9 and arithmetic )
Formal Proof
0.
if a > 0 and b > c > 0 then a + b > a + c > 0
1.
if a > b > 0 then a > b > 0
2.
224 > 9 > 0
3.
224 > 9 > 0
( 1 and 2 )
4.
414 > 3 > 0
( 3 and arithmetic )
5.
57 + 414 > 57 + 3 > 0
( 0 and 4 )
6.
(57 + 414) > (57 + 3) > 0
( 1 and 5 )
7.
1 + 214 > 215 > 0
( 6 and arithmetic )
8.
8 + 1 +214 > 8 + 215 >0
( 0 and 7 )
9.
(8 + 1 + 214) (8 + 215) > 0
( 1 and 8 )
10.
2 + 7 > 3 + 5 > 0
( 9 and arithmetic )
224 > 9 > 0
2 + 7 > 3 + 5
Goal-Directed Construction
=
2 + 7 ? 3 + 5
{ property of squaring
}
(2 + 7)2 ? (3 + 5)2
=
{ arithmetic
}
9 + 214 ? 8 + 215
=
{ property of arithmetic
}
1 + 214 ? 215
=
{ property of squaring
}
(1 + 214)2 ? (215)2
=
{ arithmetic
}
57 + 414 ? 60
=
{ property of arithmetic
}
414 ? 3

{ property of squaring, and arithmetic }
? is “ > ”.
Goal-Directed Construction
2 + 7
  ? 3
 + 5
=
{ property
of squaring }
 
 
(2 + 7) 2 ?  (3 + 5)2
=


2+ 7 ?
=
{
3+ 5
property of squaring
}
( 2 + 7)2 ? ( 3 + 5)2
=
{
arithmetic
}
9 + 2 14 ? 8 + 2 15
=
{
property of arithmetic
}
1 + 2 14 ? 2 15
=
{
property of squaring
}
(1 + 214)2 ? (215)2
=
{
arithmetic
}
57 + 414 ? 60

=
{
property of arithmetic
}

{ property
of squaring, and arithmetic }
? is “ > ”.
4 14 ? 3

{
property of squaring, and arithmetic
? is “ > ”.
}
Complicating Proof
0+0 = 0
0(0+0) = 00
00 + 00 = 00
(00 + 00) + x = 00 + x
00 + (00 + x) = 00 + x
00 + 0 = 0
00 = 0
Rule (i) with a = 0
multiplying both sides by 0
“distributive rule”:
a(b+c) = ab + ac
adding x to both sides, where x
is the solution of 00 + x = 0
guaranteed by rule (ii)
“associative rule”:
a + (b+c) = (a+b) + c
using the property of x
using rule (i) with a = 00
Complicating Proof
0+0 = 0
0+0 = 0
0(0+0) = 00
00 + 00 = 00
(00 + 00) + x = 00 + x
00 + (00 + x) = 00 + x
00 + 0 = 0
00 = 0
00 = 0
Rule (i) with a = 0
multiplying both sides by 0
“distributive rule”:
a(b+c) = ab + ac
adding x to both sides, where x
is the solution of 00 + x = 0
guaranteed by rule (ii)
“associative rule”:
a + (b+c) = (a+b) + c
using the property of x
using rule (i) with a = 00
Goal-Directed Proof
=
=
=

00 = 0
{ a+0 = a }
00 + 0 = 0
{ assume 00 + x = 0, Leibniz }
00 + (00 + x) = 00 + x
{ associativity of + }
(00 + 00) + x = 00 + x
{ Leibniz }
00 + 00 = 00
First four steps are the crucial ones ...
Goal-Directed Proof
00 + 00 = 00
=
{ distributivity }
0(0 + 0) = 00

{ Leibniz }
0+0 = 0
=
{ a+0 = a }
true .
… the rest is easy.
Construction versus
Verification
1 + 2 + … + n = n ( n + 1) / 2
12 + 22 + … + n2 = n ( n + 1) ( 2n + 1) / 6
13 + 23 + … + n3 = ?
127 + 227 + … + n27 = ?
Verification versus
Construction
1 + 2 + … + n = n ( n + 1) / 2
12 + 22 + … + n2 = n ( n + 1) ( 2n + 1) / 6
13 + 23 + … + n3 = ?
127 + 227 + … + n27 = ?
Verification versus
Construction
1 + 2 + … + n = n ( n + 1) / 2
12 + 22 + … + n2 = n ( n + 1) ( 2n + 1) / 6
13 + 23 + … + n3 = ?
127 + 227 + … + n27 = ?
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 in n 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 movement of 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
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