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Prove that A = x : φx = λx.42 is not recursive; i.e., x ∈ A is unsolvable/undecidable. [closed]

Show that the language TOT= M is a Turing Machine that halts with all inputs is not recursively enumerable nor its complement.Prove that $overlineL$ is not recognizable by showing that $B_TM le_m L$Clarification of the argument for the set of total recursive functions not being recursively enumerable?Do there exist infinite sets of non-halting programs such that every program in the set computes every other program in the set?How to argue that a set is recursive or recursively enumerable?Using reductions of turing machines properlyWhy do we get a contradiction for $x=n$?Non recursive set that intersects every recursively enumerable setProve that set of all perfect numbers is primitive recursiveFunctions corresponding to Turing machines that might not halt but consume bounded tape

-1

$begingroup$

please help to understand. Here is a similar example:
hint: (via S-m-n) that
K ≤ A.
Definition. (Complete Index Sets) Let C ⊆ P and A = x : φx ∈ C.
A is thus the set of ALL programs (known by their addresses) x that compute
any unary f ∈ C: Indeed, let λx.f(x) ∈ C. Thus f = φi for some i. Then i ∈ A.
But this is true of all φm that equal f.
We call A a complete (all) index (programs) set.
example:
The set A = x : ran(φx) = ∅ is not semi-recursive.
Recall that “range” for λx.f(x), denoted by ran(f), is defined by
x : (∃y)f(y) = x
We will try to show that
K ≤ A
ψ(x, y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
Here is how to compute ψ:
Given x, y, ignore y. Fetch machine M at address x from the standard listing,
and call it on input x. If it ever halts, then print “0” and halt everything. If
it never halts, then you will never return from the call, which is the correct
specified in (2) behavior for ψ(x, y).
By CT, ψ is in P, so, by the S-m-n Theorem, there is a recursive h such that
ψ(x, y) = φh(x)(y), for all x, y
You may NOT use S-m-n UNTIL after you have proved that your
“λxy.ψ(x, y)” is in P (partial recursive).
We can rewrite this as,
φh(x)(y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
or, rewriting (3) without arguments (as equality of functions, not equality of
function calls)
φh(x) =

{λy.0 if φx(x) ↓
{∅ if φx(x) ↑

In 3, ∅ stands for λy. ↑, the empty function.
Thus,
h(x) ∈ A iff ran(φh(x)) = ∅
bottom case in
h(x) ∈ A iff ran(φh(x)) = ∅ iff φx(x) ↑
The above says x ∈ K ≡ h(x) ∈ A, hence K ≤ A and
thus A (don't belong to) P.

computability recursive-algorithms turing-machines

share|cite|improve this question

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

$endgroup$

closed as unclear what you're asking by Shailesh, mrtaurho, Vinyl_cape_jawa, José Carlos Santos, Riccardo.Alestra Mar 15 at 15:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A reference for MathJax, a means of formatting your math text and such on this site -- math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Eevee Trainer
  Mar 15 at 3:04
  

  
  
  
  

add a comment | 

-1

$begingroup$

please help to understand. Here is a similar example:
hint: (via S-m-n) that
K ≤ A.
Definition. (Complete Index Sets) Let C ⊆ P and A = x : φx ∈ C.
A is thus the set of ALL programs (known by their addresses) x that compute
any unary f ∈ C: Indeed, let λx.f(x) ∈ C. Thus f = φi for some i. Then i ∈ A.
But this is true of all φm that equal f.
We call A a complete (all) index (programs) set.
example:
The set A = x : ran(φx) = ∅ is not semi-recursive.
Recall that “range” for λx.f(x), denoted by ran(f), is defined by
x : (∃y)f(y) = x
We will try to show that
K ≤ A
ψ(x, y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
Here is how to compute ψ:
Given x, y, ignore y. Fetch machine M at address x from the standard listing,
and call it on input x. If it ever halts, then print “0” and halt everything. If
it never halts, then you will never return from the call, which is the correct
specified in (2) behavior for ψ(x, y).
By CT, ψ is in P, so, by the S-m-n Theorem, there is a recursive h such that
ψ(x, y) = φh(x)(y), for all x, y
You may NOT use S-m-n UNTIL after you have proved that your
“λxy.ψ(x, y)” is in P (partial recursive).
We can rewrite this as,
φh(x)(y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
or, rewriting (3) without arguments (as equality of functions, not equality of
function calls)
φh(x) =

{λy.0 if φx(x) ↓
{∅ if φx(x) ↑

In 3, ∅ stands for λy. ↑, the empty function.
Thus,
h(x) ∈ A iff ran(φh(x)) = ∅
bottom case in
h(x) ∈ A iff ran(φh(x)) = ∅ iff φx(x) ↑
The above says x ∈ K ≡ h(x) ∈ A, hence K ≤ A and
thus A (don't belong to) P.

computability recursive-algorithms turing-machines

share|cite|improve this question

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

$endgroup$

closed as unclear what you're asking by Shailesh, mrtaurho, Vinyl_cape_jawa, José Carlos Santos, Riccardo.Alestra Mar 15 at 15:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A reference for MathJax, a means of formatting your math text and such on this site -- math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Eevee Trainer
  Mar 15 at 3:04
  

  
  
  
  

add a comment | 

$begingroup$

please help to understand. Here is a similar example:
hint: (via S-m-n) that
K ≤ A.
Definition. (Complete Index Sets) Let C ⊆ P and A = x : φx ∈ C.
A is thus the set of ALL programs (known by their addresses) x that compute
any unary f ∈ C: Indeed, let λx.f(x) ∈ C. Thus f = φi for some i. Then i ∈ A.
But this is true of all φm that equal f.
We call A a complete (all) index (programs) set.
example:
The set A = x : ran(φx) = ∅ is not semi-recursive.
Recall that “range” for λx.f(x), denoted by ran(f), is defined by
x : (∃y)f(y) = x
We will try to show that
K ≤ A
ψ(x, y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
Here is how to compute ψ:
Given x, y, ignore y. Fetch machine M at address x from the standard listing,
and call it on input x. If it ever halts, then print “0” and halt everything. If
it never halts, then you will never return from the call, which is the correct
specified in (2) behavior for ψ(x, y).
By CT, ψ is in P, so, by the S-m-n Theorem, there is a recursive h such that
ψ(x, y) = φh(x)(y), for all x, y
You may NOT use S-m-n UNTIL after you have proved that your
“λxy.ψ(x, y)” is in P (partial recursive).
We can rewrite this as,
φh(x)(y) = (
0 if φx(x) ↓
↑ if φx(x) ↑
or, rewriting (3) without arguments (as equality of functions, not equality of
function calls)
φh(x) =

{λy.0 if φx(x) ↓
{∅ if φx(x) ↑

In 3, ∅ stands for λy. ↑, the empty function.
Thus,
h(x) ∈ A iff ran(φh(x)) = ∅
bottom case in
h(x) ∈ A iff ran(φh(x)) = ∅ iff φx(x) ↑
The above says x ∈ K ≡ h(x) ∈ A, hence K ≤ A and
thus A (don't belong to) P.

computability recursive-algorithms turing-machines

share|cite|improve this question

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

$endgroup$

share|cite|improve this question

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

share|cite|improve this question

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1

asked Mar 15 at 3:03

Alexey UgadovAlexey Ugadov

1