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

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(Comparison table)
(Comparison table)
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! Step !! Rogers's fixed point theorem !! Diagonalization lemma
 
! Step !! Rogers's fixed point theorem !! Diagonalization lemma
 
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| Theorem statement || <math>\varphi_e \simeq \varphi_{f(e)}</math> || <math>G \leftrightarrow A(\ulcorner G\urcorner)</math>
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| Theorem statement || <math>\forall f \exists e \ \varphi_e \simeq \varphi_{f(e)}</math> || <math>\forall A \exists G\ G \leftrightarrow A(\ulcorner G\urcorner)</math>
 
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| Given mapping || <math>f</math> || <math>A</math>
 
| Given mapping || <math>f</math> || <math>A</math>

Revision as of 05:14, 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 \varphi_e \simeq \varphi_{f(e)}.

(simplified)

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

Consider the function f\circ d. This is partial recursive, so f\circ d \simeq \varphi_i for some index i.

Now \varphi_{f(d(i))} \simeq \varphi_{\varphi_i(i)} since f\circ d \simeq \varphi_i. This is equivalent to \varphi_{d(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(\ulcorner G\urcorner).

Define \mathrm{diag}(x) to be \ulcorner C(\ulcorner C\urcorner)\urcorner where x = \ulcorner C\urcorner. In other words, given a number x, the function \mathrm{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(\mathrm{diag}(x)), and let G be B(\ulcorner B\urcorner).

Then G is A(\mathrm{diag}(\ulcorner B\urcorner)), by substituting x = \ulcorner B\urcorner in the definition of B.

We also have \mathrm{diag}(\ulcorner B\urcorner) = \ulcorner B(\ulcorner B\urcorner)\urcorner by definition of \mathrm{diag}. By definition of G, this is \ulcorner G\urcorner, so we have \mathrm{diag}(\ulcorner B\urcorner) = \ulcorner G\urcorner.

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

Comparison table

Step Rogers's fixed point theorem Diagonalization lemma
Theorem statement \forall f \exists e \ \varphi_e \simeq \varphi_{f(e)} \forall A \exists G\ G \leftrightarrow A(\ulcorner G\urcorner)
Given mapping f A
Definition of diagonal function d(x) = \varphi_x(x) \mathrm{diag}(\ulcorner C\urcorner) = \ulcorner C(\ulcorner C\urcorner)\urcorner
Composition of given mapping with diagonal function (\mathrm{given} \circ \mathrm{diagonal}) f(d(x)) A(\mathrm{diag}(x))
Naming the composition f\circ d B
Index of composition i \ulcorner B\urcorner
d(i) \mathrm{diag}(\ulcorner B\urcorner)
Expanding using definition of diagonal d(i) = \varphi_i(i) \mathrm{diag}(\ulcorner B\urcorner) = \ulcorner B(\ulcorner B\urcorner) \urcorner
Composition applied to own index (i.e. diagonalization of the composition) f\circ d(i) B(\ulcorner B\urcorner)
G defined \varphi_i(i) (no equivalent definition) G is B(\ulcorner B\urcorner)
f(d(i)) = \varphi_i(i) A(\mathrm{diag}(\ulcorner B\urcorner)) \leftrightarrow B(\ulcorner B\urcorner)
Explicitly showing previous composition f(d(i)) A(\mathrm{diag}(\ulcorner B\urcorner))
Renaming index e = d(i) \ulcorner G\urcorner = \mathrm{diag}(\ulcorner B\urcorner)
Leibniz law to previous row Apply \varphi_{f(\cdot)} to obtain \varphi_{f(e)} = \varphi_{f(d(i))} Apply A(\cdot) to obtain A(\ulcorner G\urcorner) \leftrightarrow A(\mathrm{diag}(\ulcorner B\urcorner))
Use definition of G \varphi_{f(e)} = \varphi_{\varphi_i(i)} = \varphi_e A(\ulcorner G\urcorner) \leftrightarrow B(\ulcorner B\urcorner) \leftrightarrow G
(Definition of G)? \varphi_i(i) is f(d(i)) G is A(\mathrm{diag}(\ulcorner B\urcorner))