CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

CS420

Introduction to the Theory of Computation

Lecture 26: Mapping reducibility

Tiago Cogumbreiro

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

How to write a 21^st century proof

Leslie Lamport

A method of writing proofs is described that makes it harder to prove things that are not true. The method, based on hierarchical structuring, is simple and practical. The author’s twenty years of experience writing such proofs is discussed.

Source: lamport.azurewebsites.net/pubs/proof.pdf

Why should we read this?

Structured proofs are just a method of displaying your proofs in a brief, yet rigorous way
I will be doing structured proofs in this lesson, you can use this method of presenting proofs in the test!

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Today we will learn…

Computable functions
Mapping reducible
Mapping reducibility and decidability/undecidability
Mapping reducibility and Turing recognition/unrecognition

Section 5.3.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof. If $A$ is regular, then let $C$ be the DFA that recognizes $A$ . Let intersect be the implementation of $\cap$ and E_DFA the decider of $E_{DFA}$ . The following is the decider of $X_A$ .

def X_A(D):
  return not E_DFA(intersect(C, D))

We reduced the problem of checking if $X_A$ is decidable in terms of checking if $E_{DFA}$ .

Can we generalize this process?

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.
For any $D$ we have that $L_1(D)$ is regular. (Proof?)

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.
For any $D$ we have that $L_1(D)$ is regular. (Proof?)
Let $D_{DA}$ be the DFA that recognizes $L_1(D)$ . (Proof?)

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.
For any $D$ we have that $L_1(D)$ is regular. (Proof?)
Let $D_{DA}$ be the DFA that recognizes $L_1(D)$ . (Proof?)
$\langle D \rangle \in X_A$ iff $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ (Proof?) $^\dagger$

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.
For any $D$ we have that $L_1(D)$ is regular. (Proof?)
Let $D_{DA}$ be the DFA that recognizes $L_1(D)$ . (Proof?)
$\langle D \rangle \in X_A$ iff $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ (Proof?) $^\dagger$
The test $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ is decidable, and equivalent to testing $\langle D \rangle \in X_A$ , so the latter is decidable?

$^\dagger$ : Recall that if $A$ decidable, then $\overline A$ decidable (Lesson 21).

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Mapping reducibility

Intuition

If we can establish an equivalence up to some function, then we can reduce a problem into another known problem solved in another language.

Example

(Mapping-reducibility): $\langle D \rangle \in X_A$ iff $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$
(Decidability): The test $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ is decidable, and equivalent to testing $\langle D \rangle \in X_A$ , so the latter is decidable?

We will now implement a framework on reducibility

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Mapping reducibility

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Computable function

Definition 5.17

We say that

f: \Sigma^\star \longrightarrow \Sigma^\star

is a computable function if there exists a Turing Machine $M$ that when given $w$ halts and results in $f(w)$ on its tape.

Intuition

This is a total function (terminates for all inputs) encoded in terms of a Turing Machine.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Mapping reducible

Definition 5.20

Language $A$ is mapping reducible to language $B$ , notation $A \le_\mathrm{m} B$ if there is a computable function $f$ , where for every $w$ ,

w \in A \iff f(w) \in B

What can we do with mapping reducible?

Convert membership testing in $A$ into membership testing in $B$

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

E_{\mathrm{DFA}} = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) = \emptyset \}

We show that $X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$ .

Given a DFA $D$ let $L_1(D)$ be the DFA that recognizes $L(D) \cap A$ , where $A$ is regular.
We show that $X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example

X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}

E_{\mathrm{DFA}} = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) = \emptyset \}

We show that $X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$ .

Given a DFA $D$ let $L_1(D)$ be the DFA that recognizes $L(D) \cap A$ , where $A$ is regular.
We show that $X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$
- Show: If $\langle D \rangle \in X_A$ , then $\langle L_1(D) \rangle \in \overline E_{\mathrm{DFA}}$ .
  1. $L(D) \cap A \neq \emptyset$ (assumption)
  2. $\langle L_1(D) \rangle \in \overline E_{\mathrm{DFA}}$ (by 2.1)
- Show: If $\langle L_1(D) \rangle \in \overline E_{\mathrm{DFA}}$ , then $\langle D \rangle \in X_A$ .
  1. $\langle D\rangle \in X_A$ by $L_1(D) \neq \emptyset$ (assumption)

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Ungraded homework exercises

Show that $\le_\mathrm{m}$ is a reflexive relation.
Show that $\le_\mathrm{m}$ is a transitive relation.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Motivation

If $A$ is regular, then $X_A$ decidable.

Proof (2^nd try).

Let $L_1(D) = L(D) \cap A$ where $D$ is a DFA.
For any $D$ we have that $L_1(D)$ is regular. (Proof?)
Let $D$ be the DFA that recognizes $L_1$ . (Proof?)
$\color{purple} X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$ (Before: $\langle D \rangle \in X_A$ iff $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ )
The test $\langle L_1(D)\rangle \in \overline E_{\mathrm{DFA}}$ is decidable, and equivalent to testing $\langle D \rangle \in X_A$ , so the latter is decidable?

We will now generalize the (5) step

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

If $A \le_\mathrm{m} B$ and $B$ is decidable, then $A$ is decidable.

Proof in cogumbreiro/turing.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Completing the running example

If $A$ regular, then $X_A = \{ \langle D \rangle \mid D \text{ is a DFA} \land L(D) \cap A \neq \emptyset \}$ decidable.

Proof (3^rd try).

For any $D$ we have that $L_{DA} = L(D) \cap A$ is regular, since:
- For any DFA $D$ we have that $L(D) \cap A$ is regular, since regular langs are closed for $\cap$ , $L(D)$ is regular (def of reg langs), and $A$ is regular (assumption).
$X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$ , by Slide 9
$\overline E_{\mathrm{DFA}}$ is decidable, by Lemma R.4 and $E_{\mathrm{DFA}}$ decidable (Theorem 4.4)
$X_A$ is decidable, by Theorem 5.22, $\overline E_{\mathrm{DFA}}$ is decidable, and $X_A \le_\mathrm{m} \overline E_{\mathrm{DFA}}$ (2)

Lemma R.4 (Lesson 21). If $A$ decidable, then $\overline A$ decidable.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Undecidability on membership reducible

Corollary 5.23

If $A \le_\mathrm{m} B$ and $A$ is undecidable, then $B$ is undecidable.

Proof.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Undecidability on membership reducible

Corollary 5.23

If $A \le_\mathrm{m} B$ and $A$ is undecidable, then $B$ is undecidable.

Proof.

$B$ is decidable, by contradiction.
$A$ is decidable, by Theorem 5.22 and $A \le_\mathrm{m} B$ (assumption) and $B$ decidable (1)
We reach a contradiction: $A$ is decidable (2) and undecidable (assumption).

(1)

H0: A <=m B
H1: ~ Decidable A
----------------
~ Decidable B

(2)

H0: A <=m B
H1: ~ Decidable A
H2: Decidable B
----------------
False

(3)

H0: A <=m B
H1: ~ Decidable A
H2: Decidable B
H3: Decidable A
----------------
False

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24

$A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ holds $^\dagger$ .
Show that $HALT_\mathsf{TM}$ is undecidable.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24

$A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ holds $^\dagger$ .
Show that $HALT_\mathsf{TM}$ is undecidable.

Apply Corollary 5.23 since $A_\mathsf{TM}$ is undecidable (Theorem 4.11) and $A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ (hypothesis).

$^\dagger$ Proof in cogumbreiro/turing.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Exercise

$A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$

Show that $A_\mathsf{TM}$ is recognizable via mapping reducibility.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Exercise

$A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$

Show that $A_\mathsf{TM}$ is recognizable via mapping reducibility.

Proof.

Give a program that recognizes $HALT_\mathsf{TM}$ (homework!)
$A_\mathsf{TM}$ , by Theorem 5.28, $A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ , and (1).

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Corollary 5.29

If $A$ is unrecognizable and $A \le_{\mathrm{m}} B$ , then $B$ is unrecognizable.

Theorem R.1

If $A \le_{\mathrm{m}} B$ , then $\overline A \le_{\mathrm{m}} \overline B$ .

Exercise

Show that $\overline{HALT}_\mathsf{TM}$ is unrecognizable.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Corollary 5.29

If $A$ is unrecognizable and $A \le_{\mathrm{m}} B$ , then $B$ is unrecognizable.

Theorem R.1

If $A \le_{\mathrm{m}} B$ , then $\overline A \le_{\mathrm{m}} \overline B$ .

Exercise

Show that $\overline{HALT}_\mathsf{TM}$ is unrecognizable.

Proof.

$\overline{A}_\mathsf{TM} \le_{\mathrm{m}} \overline{HALT}_\mathsf{TM}$ , by Theorem R.1 and $A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ (exercise 5.24)
$\overline{HALT}_\mathsf{TM}$ is unrecognizable, by Corollary 5.29, $\overline{A}_\mathsf{TM} \le_{\mathrm{m}} \overline{HALT}_\mathsf{TM}$ (1), and $\overline{A}_\mathsf{TM}$ is unrecognizable (Corollary 4.23)

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Extra proofs

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

If $A \le_\mathrm{m} B$ and $B$ is decidable, then $A$ is decidable.

Proof.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

If $A \le_\mathrm{m} B$ and $B$ is decidable, then $A$ is decidable.

Proof.

$B$ is decidable, so let $M_B$ be its decider.
Let $M_A$ be a turing machine defined as: $M_A(w) = M_B(f(w))$
Run $M_B$ with input $f(w)$ . If $M_B$ accepts, $M_A$ accepts. If $M_B$ rejects, $M_A$ rejects.
Correctness: Prove that $L(M_A) = A$ (next slide)
Termination: Prove that $M_A$ halts for every input:

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

If $A \le_\mathrm{m} B$ and $B$ is decidable, then $A$ is decidable.

Proof.

$B$ is decidable, so let $M_B$ be its decider.
Let $M_A$ be a turing machine defined as: $M_A(w) = M_B(f(w))$
Run $M_B$ with input $f(w)$ . If $M_B$ accepts, $M_A$ accepts. If $M_B$ rejects, $M_A$ rejects.
Correctness: Prove that $L(M_A) = A$ (next slide)
Termination: Prove that $M_A$ halts for every input: $M_A$ just runs $M_B$ , which halts for every input.

Our goal is show that there exists a Turing machine that decides $A$ , so we must prove that it does recognize $A$ (correctness) and that it decides $A$ (termination).

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

Proof (Continuation). Show that $L(M_A) = A$ .

We do a case analysis on the result of executing $M_A$ with input $w$ and show that $w$ is (not) in $A$ :

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

Proof (Continuation). Show that $L(M_A) = A$ .

We do a case analysis on the result of executing $M_A$ with input $w$ and show that $w$ is (not) in $A$ :

If $M_A$ accepts some $w$ we must show that $w \in A$ . From $M_B$ , we get that $f(w) \in L(B)$ , thus, from Def 5.20, we have $w \in A$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Decidability on membership reducible

Theorem 5.22

Proof (Continuation). Show that $L(M_A) = A$ .

We do a case analysis on the result of executing $M_A$ with input $w$ and show that $w$ is (not) in $A$ :

If $M_A$ accepts some $w$ we must show that $w \in A$ . From $M_B$ , we get that $f(w) \in L(B)$ , thus, from Def 5.20, we have $w \in A$ .
If $M_A$ rejects some $w$ we must show that $w \notin A$ . If reject, then $f(w) \notin L(B)$ , thus, from Def 5.20, we have $w \notin A$ .

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TODO: WHY NOT $w \in A$ implies $w$ accepted by $M_A$

CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Proof.

We show that $A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ with $f$ , where $f(\langle M, w \rangle) = \langle M', w \rangle$ and $M'$ runs $M(w)$ if $M$ rejects, then loop, otherwise accept.
Since $A_\mathsf{TM}$ is undecidable, then $HALT_\mathsf{TM}$ is undecidable (Corollary 5.23).

Unfold Def 5.20:

\langle M,w\rangle \in A_\mathsf{TM} \iff f(\langle M,w \rangle) \in HALT_\mathsf{TM}

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Proof.

We show that $A_\mathsf{TM} \le_{\mathrm{m}} HALT_\mathsf{TM}$ with $f$ , where $f(\langle M, w \rangle) = \langle M', w \rangle$ and $M'$ runs $M(w)$ if $M$ rejects, then loop, otherwise accept.
Since $A_\mathsf{TM}$ is undecidable, then $HALT_\mathsf{TM}$ is undecidable (Corollary 5.23).

Unfold Def 5.20:

\langle M,w\rangle \in A_\mathsf{TM} \iff f(\langle M,w \rangle) \in HALT_\mathsf{TM}

Step 1: $\langle M,w\rangle \in A_\mathsf{TM} \implies f(\langle M,w \rangle) \in HALT_\mathsf{TM}$

Step 2: $f(\langle M,w \rangle) \in HALT_\mathsf{TM} \implies \langle M,w\rangle \in A_\mathsf{TM}$

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}

and that $M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 1. $\langle M,w\rangle \in A_\mathsf{TM} \implies f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}

and that $M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 1. $\langle M,w\rangle \in A_\mathsf{TM} \implies f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ .

Since $\langle M,w\rangle \in A_\mathsf{TM}$ , then $M$ accepts $w$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}

and that $M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 1. $\langle M,w\rangle \in A_\mathsf{TM} \implies f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ .

Since $\langle M,w\rangle \in A_\mathsf{TM}$ , then $M$ accepts $w$ .
Thus, $M'$ halts, and therefore $\langle M', w \rangle \in HALT_\mathsf{TM}$

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

$HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}$
$M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 2. We have $f(\langle M,w \rangle) = \langle M', w \rangle \in HALT_\mathsf{TM}$ and must show $\langle M,w\rangle \in A_\mathsf{TM}$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

$HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}$
$M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 2. We have $f(\langle M,w \rangle) = \langle M', w \rangle \in HALT_\mathsf{TM}$ and must show $\langle M,w\rangle \in A_\mathsf{TM}$ .

Since $f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ and (1), then $M'$ halts.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

$HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}$
$M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 2. We have $f(\langle M,w \rangle) = \langle M', w \rangle \in HALT_\mathsf{TM}$ and must show $\langle M,w\rangle \in A_\mathsf{TM}$ .

Since $f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ and (1), then $M'$ halts.
Thus, $M'$ accepts,

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Example 5.24

$HALT_\mathsf{TM}$ is undecidable

Recall that:

$HALT_{\mathsf{TM}} = \{ \langle M, w \rangle \mid M \text{ is a TM and } M \text{ halts on input } w \}$
$M'$ runs $M(w)$ if $M$ reject, then loop, otherwise accept.

Proof (continuation).

Step 2. We have $f(\langle M,w \rangle) = \langle M', w \rangle \in HALT_\mathsf{TM}$ and must show $\langle M,w\rangle \in A_\mathsf{TM}$ .

Since $f(\langle M,w \rangle) \in HALT_\mathsf{TM}$ and (1), then $M'$ halts.
Thus, $M'$ accepts, and since $M'$ only accepts when $M$ accepts $w$ , we conclude our proof.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Detailed proof.

We must show that there exists some $M_A$ that recognizes $A$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Detailed proof.

We must show that there exists some $M_A$ that recognizes $A$ .

Let $M_A(w) = M_B(f(w))$ .
That is, machine $M_A$ given $w$ computes $f(w)$ and accepts.
Show that $L(M_A)=A$ .
- Step 1: If $M_A$ accepts $w$ , then $w \in A$ .
- Step 2: If $w \in A$ , then $M_A$ accepts $w$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Proof (Step 1).

	Hypothesis
H1	$M_A$ accepts $w$
H2	$w \in A \iff f(w) \in B$
H3	$M_B$ recognizes $B$
H4	$M_A(w) = M_B(f(w))$

Goal: show that $w \in A$

Since (H1) $M_A$ accept $w$ and H4, we have that $M_B$ accepts $f(w)$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Proof (Step 1).

	Hypothesis
H1	$M_A$ accepts $w$
H2	$w \in A \iff f(w) \in B$
H3	$M_B$ recognizes $B$
H4	$M_A(w) = M_B(f(w))$

Goal: show that $w \in A$

Since (H1) $M_A$ accept $w$ and H4, we have that $M_B$ accepts $f(w)$ .
From $M_B$ accepts $f(w)$ and H3, we get $f(w) \in B$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Proof (Step 1).

	Hypothesis
H1	$M_A$ accepts $w$
H2	$w \in A \iff f(w) \in B$
H3	$M_B$ recognizes $B$
H4	$M_A(w) = M_B(f(w))$

Goal: show that $w \in A$

Since (H1) $M_A$ accept $w$ and H4, we have that $M_B$ accepts $f(w)$ .
From $M_B$ accepts $f(w)$ and H3, we get $f(w) \in B$ .
Since $f(w) \in B$ and H2, then $w \in A$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Proof. (Step 2)

	Hypothesis
H0	$w \in A$
H1	$w \in A \iff f(w) \in B$
H2	$M_B$ recognizes $B$
H3	$M_A(w) = M_B(f(w))$

Goal: show that $M_A$ accepts $w$ .

27 / 35

CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Proof. (Step 2)

	Hypothesis
H0	$w \in A$
H1	$w \in A \iff f(w) \in B$
H2	$M_B$ recognizes $B$
H3	$M_A(w) = M_B(f(w))$

Goal: show that $M_A$ accepts $w$ .

From (H1) $w \in A$ and H2, we have that $f(w) \in B$ .
From $f(w) \in B$ and H3, we have that $M_B$ accepts $f(w)$
From $M_B$ accepts $f(w)$ and H4, we have that $M_A$ accepts $w$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Detailed proof.

We must show that there exists some $M_A$ that recognizes $A$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Theorem 5.28

If $A \le_{\mathrm{m}} B$ and $B$ is recognizable, then $A$ is recognizable.

Detailed proof.

We must show that there exists some $M_A$ that recognizes $A$ .

Let $M_A(w) = M_B(f(w))$ .
That is, machine $M_A$ given $w$ computes $f(w)$ and accepts.
Show that $L(M_A)=A$ .
- Step 1: If $M_A$ accepts $w$ , then $w \in A$ .
- Step 2: If $w \in A$ , then $M_A$ accepts $w$ .

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Homework 8

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24 (Proof)

Use Corollary 5.23.

$A_{\mathsf{TM}} \le_{\mathrm{m}} HALT_{\mathsf{TM}}$ (next slide)
$A_{\mathsf{TM}}$ is undecidable by Theorem 4.11 (Lesson 21)

Theorem example_5_24:
  ~ Decidable HALT_tm.
Proof.
  apply reducible_undecidable with (A:=A_tm).
  - apply A_tm_red_HALT_tm.
  - apply a_tm_undecidable.
Qed.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24 (Proof)

The mapping function is given in the book. We separate the construction of $M'$ into its own function just so we can prove theorems more simply.

(* Construct the following machine *)
Definition A_tm_looper M := Build (
  fun x => (* On input x: *)
  mlet r <- Call M x in (* 1. Run M on x         *)
  if r then ACCEPT  (* 2. If M accepts, accept   *)
  else LOOP).  (* 3. If M rejects, enter a loop. *)
Definition A_tm_to_HALT_tm p:=
    (* On input <[ M, w ]> *)
    let (M, w) := decode_machine_input p in
    (* 1. Construct the following machine M' *)
    let M' := A_tm_looper M w in
    (* 2. Output <[ M', w ]> *)
    <[ M', w ]>.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24 (Proof)

Show that if $F(w) \in HALT_{TM}$ , then $w \in A_{TM}$ .
Show that if $w \in A_{TM}$ , then $F(w) \in HALT_{TM}$ .

Theorem A_tm_red_HALT_tm: A_tm <=m HALT_tm.
Proof.
  apply reducible_iff with A_tm_to_HALT_tm.
  split; intros.
  - apply A_tm_red_HALT_tm_1; auto.
  - apply A_tm_red_HALT_tm_2; auto.
Qed.

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24 (Proof)

Structure of the proof

Simplify assumption $w \in A_{TM}$ (with inversion theorem)
- We get that $w = \langle M, i\rangle$ for some $M$ and $i$
- We get that $M$ accepts $i$
Show $F(\langle M, i\rangle) \in HALT_{TM}$ (with constructor _def theorem)
- $F(\langle M, i \rangle) = \langle looper(M), i\rangle$
- Show that $looper(M)$ does not loop when $M$ accepts

Lemma A_tm_red_HALT_tm_1:
  forall w,
  A_tm w ->
  HALT_tm (A_tm_to_HALT_tm w).

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CS420 ☽ Mapping reducibility ☽ Lecture 26 ☽ Tiago Cogumbreiro

Exercise 5.24 (Proof)

Simplify assumption $F(w) \in HALT_{TM}$ (by inverting with _inv)
- Obtain $F(w) = \langle M', i\rangle$ for some $M'$ and $i$ (invert)
  - Obtain that $w = \langle M, i \rangle$ for some $M$
  - Obtain that $M' = looper(M)$
- Obtain that $M'$ does not loop with $i$
  - Thus, $M$ accepts $i$
Show $w \in A_{TM}$ (construct goal using _def theorem)
- Since $M$ accepts $i$ and $w = \langle M, i\rangle$

Lemma A_tm_red_HALT_tm_2:
  forall w,
  HALT_tm (A_tm_to_HALT_tm w) ->
  A_tm w.

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CS420

Introduction to the Theory of Computation

Lecture 26: Mapping reducibility

Tiago Cogumbreiro

How to write a 21st century proof

Leslie Lamport

Why should we read this?

Today we will learn…

Motivation

Can we generalize this process?

Motivation

Motivation

Motivation

Motivation

Motivation

Mapping reducibility

Intuition

Example

We will now implement a framework on reducibility

Mapping reducibility

Computable function

Definition 5.17

Intuition

Mapping reducible

Definition 5.20

What can we do with mapping reducible?

Example

Example

Ungraded homework exercises

Motivation

We will now generalize the (5) step

Decidability on membership reducible

Theorem 5.22

Completing the running example

Undecidability on membership reducible

Corollary 5.23

Undecidability on membership reducible

Corollary 5.23

Exercise 5.24

Exercise 5.24

Theorem 5.28

Exercise

Theorem 5.28

Exercise

Corollary 5.29

Theorem R.1

Exercise

Corollary 5.29

Theorem R.1

Exercise

Extra proofs

Decidability on membership reducible

Theorem 5.22

Decidability on membership reducible

Theorem 5.22

Decidability on membership reducible

Theorem 5.22

Decidability on membership reducible

Theorem 5.22

Decidability on membership reducible

Theorem 5.22

Decidability on membership reducible

Theorem 5.22

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

HALTTMHALT_\mathsf{TM}HALTTM​ is undecidable

Example 5.24

How to write a 21^st century proof

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

$HALT_\mathsf{TM}$ is undecidable

How to write a 21^st century proof