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Contradiction proof for inequality of P and NP?



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Proof for P-complete is not closed under intersectionProof of sum of powerset?Contradiction between best-case running time of insertion sort and $nlog n$ lower bound?bounded length CoNP proofLogarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofIs it always possible to have one part of the reduction?Is this language NP Hard?Testing algorithm for a modified sieve of EratosthenesFinding a complexity by solving inequality










1












$begingroup$


I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.




We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.




Is there something wrong with my proof? I was struggling for hours before asking this, though!










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.




    We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.




    Is there something wrong with my proof? I was struggling for hours before asking this, though!










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.




      We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.




      Is there something wrong with my proof? I was struggling for hours before asking this, though!










      share|cite|improve this question









      $endgroup$




      I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.




      We start off by assuming that $P=NP$. Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.




      Is there something wrong with my proof? I was struggling for hours before asking this, though!







      complexity-theory time-complexity






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      inverted_indexinverted_index

      1384




      1384




















          1 Answer
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          $begingroup$


          Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.




          Sure.




          As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.




          No. Polynomial time reductions aren't free. We can say takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.



          And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.






          share|cite|improve this answer











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            $begingroup$


            Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.




            Sure.




            As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.




            No. Polynomial time reductions aren't free. We can say takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.



            And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.






            share|cite|improve this answer











            $endgroup$

















              4












              $begingroup$


              Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.




              Sure.




              As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.




              No. Polynomial time reductions aren't free. We can say takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.



              And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.






              share|cite|improve this answer











              $endgroup$















                4












                4








                4





                $begingroup$


                Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.




                Sure.




                As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.




                No. Polynomial time reductions aren't free. We can say takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.



                And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.






                share|cite|improve this answer











                $endgroup$




                Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.




                Sure.




                As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.




                No. Polynomial time reductions aren't free. We can say takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.



                And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 27 mins ago

























                answered 58 mins ago









                orlporlp

                6,1251826




                6,1251826



























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