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Incompleteness theorem for first order number theory

Incompleteness theorem for first order number theory. Plan. Provide axioms and proof rules for number theory: N = {0, 1, 2, …} with +, *, <, constants 0 and 1. Then we can prove everything that is correct about number theory (including e.g., Fermat’s conjecture, which was open for 300 years).

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Incompleteness theorem for first order number theory

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  1. Incompleteness theorem for first order number theory

  2. Plan • Provide axioms and proof rules for number theory: N = {0, 1, 2, …} with +, *, <, constants 0 and 1. • Then we can prove everything that is correct about number theory (including e.g., Fermat’s conjecture, which was open for 300 years). • Problem: plan fails. We’ll prove that this cannot be done.

  3. Some first order sentences on number theory… to warm up “When dividing x by y, q is the quotient and r the remainder”:INTDIV(x,y,q,r) = x = q*y+r /\ r<y “y divides x”:DIV(y,x) = q INTDIV(x,y,q,0) “x is even (odd)”:EVEN(x)=DIV(2,x), ODD(X)=EVEN(X) “x is prime”:PRIME(x)=x2 /\ y (DIV(y,x)(y=1\/y=x))

  4. More… • POWERp(x)=y ((DIV(y,x)/\PRIME(y))y=p)

  5. Gadget: treat number as string of bits • BIT(x,y) – the bit in position y (a power of two!!) of x is 1. • y = 1 0 0 0 0 0 0 • x= 1 1 0 1 1 0 1 1 0 1 0 1 0 1 0 1 • How to obtain this?Divide x by y. Take the quotient. Check if it is odd. • Instead of bits, we will use representation in some prime number base.

  6. First proof • The set of provable is recursive enumerable: we can enumerate proofs until we obtain the correct one. • We will show that the set of correct sentences in the theory of naturals is not r.e. • We will show a reduction from non halting problem: {M#x|M does not halts on input x}. • Construct a sentence in natural number theory that is true iff M does not halts on x.

  7. Continuing construction • VALCOMPM,x(y): y is a valid halting computation of M on x. That is: y is of the form a0, a1, … , aN, where a0 is a start configuration of M on x, ai follows from ai according to M. aN is a halting configuration.Sequence is represented as a number. • M not halting on x: y VALCOMPM,x(y)

  8. How to encode successive configurations? d b b c d b b e  d b e c d b b e q r Every position contains tape letter, and optionally also a symbol for the current state (if the head is in that position). Corresponding triples of (maximal) tape length distance correspond to transition relation:(q, b)=(r,e,R) then b b c corresponds to b e c qr • We assume there is a set C of matching triples.

  9. Formulization • v represents at most d symbols.LENGTH(v,d)=POWERp(d)/\v<d • Digit v at position y (power of p) is b:DIGIT(v,y,b)=ua( v=a+b*y+u*p*y/\a<y/\b<p) • 3 digits at position y are b, c, d:3DIGITS(v,y,b,c,d)= ua(v=a+b*y+c*p*y+ d*p*p*y+u*p*p*p*y/\a<y/\b<p/\ c<p/\d<p)

  10. More formulizations • Matching between triples of positions y and z:MATCH(v,y,z)=\/(a,b,c,d,e,f)C 3DIGIT(v,y,a,b,c)/\3DIGIT(v,z,d,e,f) • v represents successive configurations of M of length g, of distance c, up to length h. MOVE(v,g,h)=y(powerp(y)/\y*p*p*g<h) MATCH(v,y,y*g)

  11. Some more • v starts with the start configuration of M on x, with x1,x2,…,xn and then some blanks (value=k) of length g.START(v,c)=/\i=0..nDIGIT(v,pi,xi) /\ pn<g /\y(POWERp(y) /\ pn<y<gDIGIT(v,y,k)) • v has a halt. Set H corresponds to symbols containing some halt state.HALT(v,h)=y(POWERp(y)/\y<h/\\/bHDIGIT(v,y,b)

  12. Finally • VALCOMPM,X(v)=gh(POWERp(g)/\g<h/\ LENGTH(v,h)/\START(v,g)/\ MOVE(v,g,h)/\HALT(v,h))

  13. Goedel’s proof (Sketch) • Goedel’s proof is more difficult! • It encodes numbers [] that represent formulas . • X([]) represents that the property X holds for formula . • Then it encodes proofs (again, a sequence can be represented as a number)! • It forms a formula  such that  <-->  PROVABLE([])

  14. Read more… • Automata and Computability Dexter Kozen, Springer Verlag, 282-298

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