Gödel's Incompleteness Theorems¶

Construct a Turing machine \(M\) prints "M" itself on its tape.

A -- Write B on the tape
B -- Write A on the tape and swap it with B function q : \(\Sigma^* \rightarrow \Sigma^*\) such that \(q(w) = "M_w"\)
\(M_w\) = on any input x: 1. Print w on the tape

Given any string w" 1. Construct \(M_w\) as above 2. Return \(M_w\)

B = on input w 1. Compute \(q(w)\) 2. write \(q(w)\) on the tape and swap it with w

A = on input w 1. Write "B" on the tape

Recursion Theorem¶

For any TM T, there exists a TM R such that for any input w, the computation of R on w is equivalent to the computation of T on \(<R, w>\).

T = on input M,w: ...

R = on input w :

Run T on

Proof¶

Another Proof of \(A_{TM}\) not recursive¶

Gödel's Incompleteness Theorems¶

\(V(x,t) = \begin{cases} 1 & \text{if t is a valid for x} \\ 0 & \text{otherwise} \end{cases}\)

Let \(\Tau\) be a language (the set of statements) A proof system for \(\Tau\) is a TM V such that

1. Effectiveness : For x,y \(\in \Sigma^*\), V either accepts or rejects \(V(x,y)\)
1. Soundness : For x \(\notin \Tau\), \(\forall y, V(x,y) = 0\)

V is a complete proof system for \(\Tau\) if for x \(\in \Tau\), \(\exists y, V(x,y) = 1\)

Theorem 1¶

Some language \(\Tau\) does not have a complete proof system.

Lemma 1¶

A has a complete proof system if and only if A is recursively enumerable

\(\Rightarrow\) : \(\exists\) a complete proof system V for A

M = on input x:

For y \(\in \Sigma^*\) in increasing order of length:

If \(V(x,y) = 1\), accept

\(\Leftarrow\) : \(\exists\) a TM M that semi-decides A

V = on input x,y:

Run M on x for y steps

If M accepts x within y steps, accept

Theorem 2¶

\(\overline{A_{TM}}\) is not recursively enumerable

Or \(\overline{A_{TM}}\) does not have a complete proof system

Proof¶

If \(\overline{A_{TM}}\) is recursively enumerable, then \(\overline{A_{TM}}\) has a complete proof system

Suppose \(\overline{A_{TM}}\) has a complete proof system V

R = on input x:

Obtain "R"

For y \(\in \Sigma^*\) in increasing order of length:

If \(V("R"x,y) = 1\), halt and accept x

If V is a complete proof system for \(\overline{A_{TM}}\), then R is a complete proof system for \(A_{TM}\)

\(R\) accepts x if and only if \(\exists y, V("R"x,y) = 1\)
\(V("R"x,y) = 1\) if and only if \("R"x\) \(\in \overline{A_{TM}}\)

Enumerator¶

We say a TM M enumerates a language L if for some state q, \(L = \{w: (s\triangleright U) \vdash^*_M (q,DUw)\}\) where q is output state and w is output w

Theorem 3¶

A is Turing enumerable if and only if A is recursively enumerable

Proof¶

Assume that A is a finite set.

\(\Rightarrow\) : By definition, A is recursively enumerable \(\Leftarrow\) :

\(\exists\) a TM M that semi-decides A

for s \(\in \Sigma^*\) in increasing order of length: 1. Run M on s 2. If M accepts s, output s

For i = 1,2,3,...
    For j = 1,2,3,...,i
        Run M on s_j for i steps
        If M accepts s_j, output s_j

Theorem 4¶

A TM M is minimal if \(|N|\) < \(|M|\) implies \(L(N) \neq L(M)\)

最后更新: 2024年12月16日 00:12:48
创建日期: 2024年12月16日 00:12:48