3' Typographical Number Theory TNT

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3. Typographical Number Theory (TNT)

• One last step to develop a formal system with
  enough structure to explore the limits of what
  can be computed
• TNT is a system which encodes statements about
  the natural numbers (0,1,2,3,4,5,). That is, no
  negative numbers, fractions, irrationals,
  imaginaries, etc.
• Based on Hofstadter, Chapter 8

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Typical statements in Number Theory

• (1) 5 is prime
• (2) 2 is not a square
• (3) 1729 is a sum of two cubes
• (4) No sum of two positive cubes is itself a
  cube
• (5) There are infinitely many prime numbers
• (6) 6 is even.

We want a formal system in which these can be
written and even proved or disproved
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Symbols of TNT

• Most of those of propositional calculus carry
  over with the same interpretations
• ? ? ? ?
• ? ?
• Some new ones as well
• S 0 .
• a b c d e
• ? ? ( )

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Rules of Well-Formedness

• NUMERALS
• 0 is a numeral.
• A numeral preceded by S is also a numeral where
  S stands for the successor of.
• For example,
• zero 0
• one S0
• two SS0
• three SSS0 etc.
•  

5

• VARIABLES
• a is a variable.
• b, c, d, and e can also be variables
• A variable followed by a prime is also a
  variable a b c etc.
•  

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• TERMS
• All numerals and variables are terms.
• A term preceded by S is also a term.
• If s and t are terms, then so are
• (s t) and (s.t).
• The mathematical symbols have their usual
  interpretations
• Plus
• Times .
• Groupings ( )
• Equals
•  

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ATOMS If s and t are terms, then s t is an
atom.These will be strings which are statements
of equality such as SS0 S0 or (S0 . S0)
S0. If an atom contains a variable u, then u
is free in it

• Closed formulas contain no free variables. They
  are sentences of TNT.
• Open formulas contain at least one free variable.
  They are sentences without subjects and are
  called predicates.
• (b S0) SS0 is an open formula and b is
  a free variable.

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Quantifiers

• Open formula can be closed using quantifiers
• If u is a variable, and x is a well-formed
  formula in which u is free, then the following
  strings are well-formed formulas
• ?u x, ?u x
• For example,
• ?b (b S0) SS0
• ?b (b S0) SS0
• These mean, there exists a number b and for
  all numbers b.

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Well formed rules from before.
Negation A well-formed formula preceded by a
bar is well-formed Compound statements If x
and y are well-formed formulas, and provided
that no variable which is free in one is
quantified in the other, then the following are
all well-formed formulas ? x ? y?, ?x ? y?,
? x? y?  

•  

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Translating our sentences into TNT

• Lets start with the easy ones
• (6) 6 is an even number.
• There exists a number e such that 2 times e
  equals 6.
• ?e (SS0 . e) SSSSSS0
• 2 is not a square.
• There does not exist a number b, such that b
  times b equals 2.
• ??b (b . b) SS0 or ?b?(b . b) SS0

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• 1729 is a sum of two cubes
• There exist numbers b and c such that b times b
  times b, plus c times c times c equals 1729.
• ?b ?c SSSSSS........SSSSSSS0 (((b . b) . b)
  ((c . c) . c))
• ??________________?
• 1729 of them

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• Alternatively
• ?b ?c (((SSSSSSSSSS0 . SSSSSSSSSS0) .
  (SSSSSSSSSS0) ((SSSSSSSSS0 . SSSSSSSSS0) .
  SSSSSSSSS0)) (((b . b) .b) ((c .c) .c))
•   or
• ?b ?c (((SSSSSSSSSSSS0 . SSSSSSSSSSSS0) .
  SSSSSSSSSSSS0) ((S0 . S0) . S0)) (((b .b) .b)
  ((c . c) . c ))

1729 93 103 123 13
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• (4) No sum of two positive cubes is itself a cube
• For all numbers b and c greater than 0, there
  is no number a such that a times a times a equals
  b times b times b plus c times c times c.
• Suppose we want 7 is not the sum of two positive
  cubes, we negate 7 is the sum of two positive
  cubes 
• ??b ?c SSSSSSS0 (((Sb . Sb) . Sb) ((Sc .
  Sc) . Sc))
• Now replace SSSSSSS0 by ((a .a) . a), which is a
  cubed 
• ??b ?c ((a . a) .a) (((Sb .Sb) . Sb)
  ((Sc.Sc) . Sc)) 
• This formula is open, since a is free, hence we
  need 
• ?a ??b ?c ((a . a) .a) (((Sb .Sb) . Sb)
  ((Sc.Sc) . Sc))
•  

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• 5 is prime.
• There do not exist numbers a and b, both
  greater than 1, such that 5 equals a times b.
• ??a ?b SSSSS0 (SSa . SSb)
• (5) There are infinitely many prime numbers.
• For each number a there exists a number b
  greater than a, with the property that there do
  not exist numbers c and d, both greater than 1,
  such that b equals c times d.
• If we wanted a plus e plus 1 is prime we would
  write 
• ??c ?d (a Se) (SSc . SSd) 
• Now to get a number greater than a and prime 
• ?e ??c ?d (a Se) (SSc . SSd)
•   This is independent of our choice of a 
• ?a ?e ??c ?d (a Se) (SSc . SSd)

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Five Axioms of TNT

• AXIOM 1 ?a ?Sa 0
• AXIOM 2 ?a (a 0) a
• AXIOM 3 ?a ?b (a Sb) S(a b)
• AXIOM 4 ?a (a . 0) 0
• AXIOM 5 ?a ?b (a . Sb) ((a . b) a)

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Rules of TNT

• Rules of Equality
• SYMMETRY
• If r s is a theorem, then so is s r.
• TRANSITIVITY
• If r s and s t are theorems, then
  r t is a theorem.
•  Rules of Successorship
• ADD S
• If r t is a theorem, then so is Sr St.
• DROP S
• If Sr St is a theorem, then so is r t.

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• Rule of Interchange
• Suppose u is a variable. Then the strings
  ?u? and ??u are interchangeable anywhere
  inside any theorem.
• For all u, it is not the case that
• There does not exist a u such that
•  
• Example
• AXIOM 1 ?a ?Sa 0
• ? ?a Sa 0

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•  Rule of Existence
• Suppose a term (which may contain variables as
  long as they are free) appears once, or several
  times, in a theorem. Then any (or several, or
  all) of the appearances of the term may be
  replaced by a variable which otherwise does not
  occur in the theorem, and the corresponding
  existential quantifier must be placed in front.
• if weve seen something is true once, then we
  know a solution exists
• AXIOM 1 ?a ?Sa 0
• ?b?a ?Sa b

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Some more complicated rules

• Rule of Specification
• Suppose u is a variable which occurs inside
  the string x. If the string ?ux is a
  theorem, then so is x, and so are any strings
  made from x by replacing u, wherever it occurs,
  by one and the same term.
• If its true every time, its true for a specific
  instance
• ?a ?Sa 0
• ?S0 0
• ?Sa 0 Theorems can have
  ?S(bS0) 0 free variables!
• Restriction the term which replaces u must not
  contain any variable that is quantified in x.

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• Rule of Generalization
• Suppose x is a theorem in which u, a variable,
  occurs free. Then ?u x is a theorem.
• ?Sa 0
• ?a ?Sa 0
• Theorems about free variables are implicitly
  about all possible instances.
• Restriction no generalization is allowed in a
  fantasy on any variable which appeared free in
  the fantasys premise

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Sample derivation

• (1) ?a ?b (a Sb) S(a b) axiom 3
• (2) ?b (S0 Sb) S(S0 b) specification
• (S0 for a)
• (3) (S0 S0) S(S0 0) specification
• (0 for b)
• (4) ?a (a 0) a axiom 2
• (5) (S0 0) S0 specification
• (S0 for a)
• (6) S(S0 0) SS0 add S
• (7) (S0 S0) SS0 transitivity
  (lines 3,6)

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Are all true statements theorems?

• (0 0) 0
• (0 S0) S0
• (0 SS0) SS0
• (0 SSS0) SSS0
• (0 SSSS0) SSSS0
• etc.
• But what about ?a (0 a) a ?
• This is not a theorem of the system we have!
• This is called ?-incompleteness

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• A system is ?-incomplete if all the strings in a
  pyramidal family are theorems, but the
  universally quantified summarizing string is not
  a theorem.
• Note that both
• ?a (0 a) a
• and ??a (0 a) a
• are nontheorems of the TNT we have described.
• They are undecidable.
• Can we fix it?
• Only by adding another rule, the rule of induction

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First, some notation

• Let Xa stand for a well-formed formula in
  which a is free.
• Then XSa/a will stand for the string but with
  every occurrence of a replaced by Sa.
• Similarly X0/a will stand for the formula but
  with every occurrence of a replaced by 0.

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This allows us to state the

• Rule of Induction
• Suppose u is a variable, and Xu is a
  well-formed formula in which u occurs free.
• If both ?u ?Xu ? XSu/u? and X0/u are
  theorems, then ?u Xu is also a theorem.
•  

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Can we now prove the earlier true formula?

• (1) ?a ?b (a Sb) S(a b) axiom 3
• (2) ?b (0 Sb) S(0 b) specification
• (3) (0 Sb) S(0 b) specification
• (4) push
• (5) (0 b) b premise
• (6) S(0 b) Sb add S
• (7) (0 Sb) S(0 b) carry over
  line 3
• (8) (0 Sb) Sb transitivity
• (9) pop

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• So then
• (10) ?(0 b) b ? (0 Sb) Sb ? fantasy
  rule
• (11) ?b ?(0 b) b ? (0 Sb) Sb?
  generalization
• (12) (0 0) 0 specification of
  axiom 2
• By induction, we can deduce
•   ?b (0 b) b
• Thus, we have resolved the ?-incompleteness
  problem!
• Is this new version of TNT now complete?