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Mar 24, 2018 at 14:27 vote accept Untitled
Jun 24, 2013 at 7:39 comment added András Salamon If $\phi \Rightarrow \psi$ then $\psi$ is weaker than $\phi$.
Jun 24, 2013 at 3:19 comment added Untitled Interesting. So this problem is equivalent to the problem of $L=NP=P=NPC$. Could you explain why you say $L\neq NP$ is a weak assumption?
Jun 22, 2013 at 22:24 history edited András Salamon CC BY-SA 3.0
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Jun 22, 2013 at 21:32 history edited András Salamon CC BY-SA 3.0
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Jun 21, 2013 at 19:22 history answered András Salamon CC BY-SA 3.0