Show that the decision problem for “the property $P$ such that $P(z)$ holds iff $lambda x.(left {z right...
$begingroup$
Show that the decision problem for "the property $P$ such that $P(z)$ holds iff $lambda x.(left {z right }(x))$ is total" is unsolvable.
Here is my work so far, and I am not sure whether it is correct:
Assume such a property $P$ is recursive. Then there exists a total recursive binary function $F$ universal for all recursive sequences $G$. Then for any recursive sequence $G$, there exists a number $e$ such that $G=lambda x[F(x,e)]$. But then for the sequence $lambda x[F(x,x)+1]$, if $x=e$, then $G(e)=F(e,e)=F(e,e)+1$, contradiction.
proof-verification logic recursion computability
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
$endgroup$
add a comment |
$begingroup$
Show that the decision problem for "the property $P$ such that $P(z)$ holds iff $lambda x.(left {z right }(x))$ is total" is unsolvable.
Here is my work so far, and I am not sure whether it is correct:
Assume such a property $P$ is recursive. Then there exists a total recursive binary function $F$ universal for all recursive sequences $G$. Then for any recursive sequence $G$, there exists a number $e$ such that $G=lambda x[F(x,e)]$. But then for the sequence $lambda x[F(x,x)+1]$, if $x=e$, then $G(e)=F(e,e)=F(e,e)+1$, contradiction.
proof-verification logic recursion computability
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
$endgroup$
$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04
add a comment |
$begingroup$
Show that the decision problem for "the property $P$ such that $P(z)$ holds iff $lambda x.(left {z right }(x))$ is total" is unsolvable.
Here is my work so far, and I am not sure whether it is correct:
Assume such a property $P$ is recursive. Then there exists a total recursive binary function $F$ universal for all recursive sequences $G$. Then for any recursive sequence $G$, there exists a number $e$ such that $G=lambda x[F(x,e)]$. But then for the sequence $lambda x[F(x,x)+1]$, if $x=e$, then $G(e)=F(e,e)=F(e,e)+1$, contradiction.
proof-verification logic recursion computability
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
$endgroup$
Show that the decision problem for "the property $P$ such that $P(z)$ holds iff $lambda x.(left {z right }(x))$ is total" is unsolvable.
Here is my work so far, and I am not sure whether it is correct:
Assume such a property $P$ is recursive. Then there exists a total recursive binary function $F$ universal for all recursive sequences $G$. Then for any recursive sequence $G$, there exists a number $e$ such that $G=lambda x[F(x,e)]$. But then for the sequence $lambda x[F(x,x)+1]$, if $x=e$, then $G(e)=F(e,e)=F(e,e)+1$, contradiction.
proof-verification logic recursion computability
proof-verification logic recursion computability
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
asked Dec 5 '18 at 2:03
numericalorangenumericalorange
1,733311
1,733311
$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04
add a comment |
$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04
$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04
$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04
add a comment |
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$begingroup$
You haven't explained how you are using the existence of $P$ to conclude the existence of $F$. I can't see any way of fixing that. An alternative approach is to use such a $P$ to solve the halting problem.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:04