User:IssaRice/Computability and logic/Diagonalization lemma: Difference between revisions

From Machinelearning
Line 52: Line 52:
|-
|-
| Composition applied to own index (i.e. diagonalization of the composition) || <math>f\circ d(i)</math> || <math>B(\ulcorner B\urcorner)</math>
| Composition applied to own index (i.e. diagonalization of the composition) || <math>f\circ d(i)</math> || <math>B(\ulcorner B\urcorner)</math>
|-
| || <math>f(d(i)) = \varphi_i(i)</math> || <math>A(\mathrm{diag}(\ulcorner B\urcorner)) \leftrightarrow B(\ulcorner B\urcorner)</math>
|-
|-
| Explicitly showing previous composition || <math>f(d(i))</math> || <math>A(\mathrm{diag}(\ulcorner B\urcorner))</math>
| Explicitly showing previous composition || <math>f(d(i))</math> || <math>A(\mathrm{diag}(\ulcorner B\urcorner))</math>

Revision as of 04:30, 8 February 2019

The diagonalization lemma, also called the Godel-Carnap fixed point theorem, is a fixed point theorem in logic.

Rogers's fixed point theorem

Let f be a total computable function. Then there exists an index e such that φeφf(e).

(simplified)

Define d(x)=φx(x) (this is actually slightly wrong, but it brings out the analogy better).

Consider the function fd. This is partial recursive, so fdφi for some index i.

Now φf(d(i))φφi(i) since fdφi. This is equivalent to φd(i) by definition of d. Thus, we may take e=d(i) to complete the proof.

Diagonalization lemma

(semantic version)

Let A be a formula with one free variable. Then there exists a sentence G such that G iff A(G).

Define diag(x) to be C(C) where x=C. In other words, given a number x, the function diag finds the formula with that Godel number, then diagonalizes it (i.e. substitutes the Godel number of the formula into the formula itself), then returns the Godel number of the resulting sentence.

Let B be A(diag(x)), and let G be B(B).

Then G is A(diag(B)), by substituting x=B in the definition of B.

We also have diag(B)=B(B) by definition of diag. By definition of G, this is G, so we have diag(B)=G.

To complete the proof, apply A to both sides of the final equality to obtain A(diag(B)) iff A(G); this simplifies to G iff A(G).

Comparison table

Step Rogers's fixed point theorem Diagonalization lemma
Theorem statement φeφf(e) GA(G)
Given mapping f A
Definition of diagonal function d(x)=φx(x) diag(C)=C(C)
Composition of given mapping with diagonal function (givendiagonal) f(d(x)) A(diag(x))
Naming the composition fd B
Index of composition i B
d(i) diag(B)
Expanding using definition of diagonal d(i)=φi(i) diag(B)=B(B)
Composition applied to own index (i.e. diagonalization of the composition) fd(i) B(B)
f(d(i))=φi(i) A(diag(B))B(B)
Explicitly showing previous composition f(d(i)) A(diag(B))
Definition of G G is A(diag(B))
Renaming index e=d(i) G=diag(B)
Leibniz law to previous row φe=φd(i) A(G)A(diag(B))
f(e) A(G)
Substituting definition of G A(G)G