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Is there an oracle $A$ with $P^A = NP^A$, but $EXP^A \not= NEXP^A$ ?

I found a proof with padding arguments (wikipedia), that $$ P = NP \Rightarrow EXP = NEXP $$ If an oracle $A$ exists with $P=NP$ and $EXP\not=NEXP$ relative to $A$ then the padding technique would be a proof methode, that circumvents the Relativization Barrier.

Maybe padding arguments can help to solve problems like P vs. NP.

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Yes, there is an oracle with $P^A = NP^A$ and $EXP^A \not= NEXP^A$.

It is an exercise in Odifreddi: Classical Recursion Theory Vol. II on page 253. An example is in Kurtz : Sparce sets in NP - P: relativization S.I.A.M J. Comp. 14(1985) 113-119

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