Using Definite Knowledge: Axiomatizing the Natural Numbers

1
Using Definite KnowledgeAxiomatizing the
Natural Numbers

• Notes for Ch.3 of Poole et al.
• CSCE 580
• Marco Valtorta

2
Constants, function, intended interpretation

• The domain the natural numbers and a special
  symbol S
• Two constants one and zero
• One function
• one --- denotes the number 1
• zero---denotes S it does not denote a number!
• b(N,zero)
• denotes 2 x N, where N denotes the number N
• b(N,one)
• denotes 2 x N 1

3
Mnemonic

• A number represented in b notation can be read as
  a binary number by replacing one with 1 and zero
  with 0. Example
• b(b(b(one, zero),one), one)
• 1 0 1 1
  (decimal 11)
• b(b(b(one, one), one), one)
• 1 1 1 1
  (decimal 15)
• B(b(b(b(one,one),zero),one),zero)
• 1 1 0 1 0
  (decimal 26)

4
number

• numb(N) is true if N denotes (has the form of)
  a number.
• For short, we may say, if N is a number.
• Use numb, because number/1 is predefined in CILOG
• numb(one).
• numb(b(N,zero)) ? numb(N).
• numb(b(N,one)) ? numb(N).

5
successor

• succ(N,M) is true if M denotes the successor of
  (the number after) N
• succ(one, b(one, zero)).
• succ(b(N, zero), b(N, one)) ?numb(N) numb(M).
• succ(b(N,one), b(M,zero)) lt- succ(N,M).
• Adding 1 to 1 results in 0 with a carry of 1.
  The carry may propagate.
• Check it!

6
Plus

• plus(X,Y,Z) is true if the number denoted by Z is
  the sum of the numbers denoted by X and Y (i.e.,
  Z X Y, or simply Z X Y).
• plus(one, Y, Z) lt- succ(Y, Z).
• plus(X, Y, Z) lt- succ(XMinus1, X)
  succ(Y, YPlus1) plus(XMinus1, YPlus1, Z).
• X Y Z iff (X-1) (Y1) Z

7
Times

• X Y Z iff (X-1) Y Y Z
• times(one, Y, Y).
• times(X,Y,Z) lt- succ(XMinus1, X)
  times(XMinus1,Y,Z1) plus(Z1,Y,Z).

8
is

• Is does expression evaluation
• We use is1 because is is predefined in CILOG.
• is1(E,E) lt- numb(E).
• is1(V, XY) lt- is1(V1,X) is1(V2,Y)
  plus(V1,V2,V).
• is1(V, XY) lt- is1(V1,X) is1(V2,Y)
  times(V1,V2,V).
• is1(V, X-Y) lt- is1(V1,X) is1(V2,Y)
  plus(V2,V,V1).

9
Less than

• lt(one, b(N,zero)) lt- numb(N).
• lt(one, b(N,one)) lt- numb(N).
• lt(N,M) lt- succ(NMinus1,N) succ(MMinus1,M)
  lt(NMinus1, MMinus1).

10
Not Equal

• neq(N,M) lt- lt(N,M).
• neq(N,M) lt- lt(M,N).

11
Complexity of plus

• Each time the second clause of plus is applied,
  the first argument is decremented by one
• plus takes linear time in the value of the first
  argument.
• This is exponential time in the size of the
  binary representation of a number!
• A more efficient plus (linear in the size of the
  representation) can be obtained by simulating
  addition by hand (with carry). We call it bplus.

12
Efficient plus

• bplus(one, X, S) ? succ(X,S).
• bplus(b(N,D),one,S) ? succ(b(N,D),S).
• bplus(b(R,zero),b(S,zero),b(RS,zero)) ?
  bplus(R,S,RS).
• bplus(b(R,zero),b(S,one),b(RS,one)) ?
  bplus(R,S,RS).
• bplus(b(R,one),b(S,zero),b(RS,one)) ?
  bplus(R,S,RS).
• bplus(b(R,one),b(S,zero),b(SRS,zero)) ?
  bplus(R,S,RS) succ(RS,SRS).

13
Complexity of times

• Each time the second clause of times is applied,
  the first argument is decremented by one and plus
  is called
• Plus takes quadratic time in the value of the
  first argument.
• This is doubly exponential in the size of the
  binary representation!
• A more efficient (quadratic in the size of the
  representation) solution is obtained by
  simulating hand multiplication. We call this
  btimes.

14
Efficient times

• btimes(one,X,X).
• btimes(b(R,zero), S, b(RS,zero) ?
  btimes(R,S,RS).
• btimes(b(R,one), S, RSS) ? btimes(R,S,RS)
  bplus(S,b(RS,zero), RSS).
• When the third clause is applied, one number is
  stripped at the end of the multiplicand. This
  appears to be linear-time, but it is not, because
  the addend in bplus grows in size.