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    Squaring a polynomial still yields a polynomial, so NPSPA... — Carmelics
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    Supports→PSPACE equals NPSPACE

    Squaring a polynomial still yields a polynomial, so NPSPACE does not exceed PSPACE

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    By Savitch's Theorem, NSPACE(s(n)) is a subset of SPACE((s(n))^2) for any space ...PSPACE equals NPSPACE

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    SEP: computational-complexity
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    As \(\phi \in \sc{SAT}\) just in case a satisfying valuation exists, this is a correct method for deciding \(\sc{SAT}\) relative to conventions (i)–(iii) from above. This means that \(\sc{SAT}\) can be solved in polynomial time relative to \(\mathfrak{N}\). This example also illustrates why adding non-determinism to the original deterministic model \(\mathfrak{T}\) does not enlarge the class of decidable problems. [12] It is evident that if \(N\) has time complexity \(f(n)\), then \(T_N\) must

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