The Proof (continued)

Notations

• Suppose M is a TM accepting L.

• Write L(M) = L.

– In particular, if M (x) = for all x, then L(M) = ∅.

• If M(x) is never “yes” nor  (as required by the definition of acceptance), we also let L(M ) = ∅.

Nontrivial Properties of Sets in RE

• A property of the recursively enumerable languages can be defined by the set C of all the recursively

enumerable languages that satisfy it.

– The property of finite recursively enumerable languages is

{ L : L = L(M) for a TM M, L is finite }.

• A property is trivial if C = RE or C = ∅.

– Answer to a trivial property is always “yes”or always

“no.”

Nontrivial Properties of Sets in RE (concluded)

• Here is a trivial property (always yes): Does the TM accept a recursively enumerable language?^a

• A property is nontrivial if C = RE and C = ∅.

– In other words, answer to a nontrivial property is

“yes” for some TMs and “no” for others.

• Here is a nontrivial property: Does the TM accept an empty language?^b

• Up to now, all nontrivial properties (of recursively

enumerable languages) are undecidable (pp. 156–157).

• In fact, Rice’s theorem confirms that.

aOr, L(M ) ∈ RE?

bOr, L(M ) = ∅?

Rice’s Theorem

Theorem 13 (Rice, 1956) Suppose C = ∅ is a proper subset of the set of all recursively enumerable languages.^a Then the question “L(M ) ∈ C?” is undecidable.

• Note that the input is a TM program M.

• Assume that ∅ ∈ C (otherwise, repeat the proof for the class of all recursively enumerable languages not in C).

• Let L ∈ C be accepted by TM M^L (recall that C = ∅).

• Let MH accept the undecidable language H.

– M_H exists (p. 139).

aA nontrivial property, i.e.

The Proof (continued)

• Construct machine Mx(y):

if M_H(x) = “yes” then M_L(y) else 

• On the next page, we will prove that

x ∈ H if and only if L(M^x) ∈ C. (1) – As a result, the halting problem is reduced to

deciding L(M_x) ∈ C.

– Hence L(M_x) ∈ C must be undecidable, and we are done.

The Proof (concluded)

• Suppose x ∈ H, i.e., M^H(x) = “yes.”

– M_x(y) determines this, and it either accepts y or never halts, depending on whether y ∈ L.

– Hence L(M_x) = L ∈ C.

• Suppose M^H(x) =.

– M_x never halts.

– L(M_x) = ∅ ∈ C.

Comments

• C must be arbitrary.

• The following Mx(y), though similar, will not work:

if M_L(y) = “yes” then M_H(x) else .

• Rice’s theorem is about properties of the languages accepted by Turing machines.

• It then says any nontrivial property is undecidable.

• Rice’s theorem is not about Turing machines

themselves, such as ”Does a TM contain 5 states?”

Consequences of Rice’s Theorem

Corollary 14 The following properties of recursively enumerative sets are undecidable.

• Emptiness.

• Finiteness.

• Recursiveness.

• Σ^∗.

• Regularity.^a

• Context-freedom.^b

aIs it a regular language?

bIs it a context-free language?

Undecidability in Logic and Mathematics

• First-order logic is undecidable (answer to Hilbert’s (1928) Entscheidungsproblem).^a

• Natural numbers with addition and multiplication is undecidable.^b

• Rational numbers with addition and multiplication is undecidable.^c

aChurch (1936).

bRosser (1937).

cRobinson (1948).

Undecidability in Logic and Mathematics (concluded)

• Natural numbers with addition and equality is decidable and complete.^a

• Elementary theory of groups is undecidable.^b

aPresburger’s Master’s thesis (1928), his only work in logic. The direction was suggested by Tarski. Moj¯zesz Presburger (1904–1943) died in a concentration camp during World War II.

bTarski (1949).

Julia Hall Bowman Robinson (1919–1985)

Alfred Tarski (1901–1983)

Boolean Logic

Christianity is either false or true.

— Girolamo Savonarola (1497) Both of us had said the very same thing.

Did we both speak the truth

—or one of us did

—or neither?

— Joseph Conrad (1857–1924), Lord Jim (1900)

Boolean Logic

Boolean variables: x1, x2, . . ..

Literals: xi, ¬xi.

Boolean connectives: ∨, ∧, ¬.

Boolean expressions: Boolean variables, ¬φ (negation), φ1 ∨ φ2 (disjunction), φ1 ∧ φ2 (conjunction).

• _n

i=1 φi stands for φ1 ∨ φ2 ∨ · · · ∨ φn.

• _n

i=1 φi stands for φ1 ∧ φ2 ∧ · · · ∧ φn.

Implications: φ1 ⇒ φ2 is a shorthand for ¬φ1 ∨ φ2.

Biconditionals: φ1 ⇔ φ2 is a shorthand for (φ1 ⇒ φ2) ∧ (φ2 ⇒ φ1).

aGeorge Boole (1815–1864) in 1847.

Truth Assignments

• A truth assignment T is a mapping from boolean variables to truth values true and false.

• A truth assignment is appropriate to boolean expression φ if it defines the truth value for every variable in φ.

– { x1 = true, x2 = false } is appropriate to x1 ∨ x2. – { x2 = true, x3 = false } is not appropriate to

x₁ ∨ x2.

Satisfaction

• T |= φ means boolean expression φ is true under T ; in other words, T satisfies φ.

• φ1 and φ₂ are equivalent, written φ₁ ≡ φ2,

if for any truth assignment T appropriate to both of them, T |= φ1 if and only if T |= φ2.

Truth Table

• Suppose φ has n boolean variables.

• A truth table contains 2ⁿ rows.

• Each row corresponds to one truth assignment of the n variables and records the truth value of φ under it.

• A truth table can be used to prove if two boolean expressions are equivalent.

– Just check if they give identical truth values under all appropriate truth assignments.

aPost (1921); Wittgenstein (1922). Here, 1 is used to denote true; 0 is used to denote false. This is called the standard representation (Beigel, 1993).

A Truth Table

p q p ∧ q

0 0 0

0 1 0

1 0 0

1 1 1

A Second Truth Table

p q p ∨ q

0 0 0

0 1 1

1 0 1

1 1 1

A Third Truth Table

p ¬p

0 1

1 0

Proof of Equivalency by the Truth Table:

p ⇒ q ≡ ¬q ⇒ ¬p

p q p ⇒ q ¬q ⇒ ¬p

0 0 1 1

0 1 1 1

1 0 0 0

1 1 1 1

De Morgan’s Laws

• De Morgan’s laws state that

¬(φ1 ∧ φ2) ≡ ¬φ1 ∨ ¬φ2,

¬(φ¹ ∨ φ²) ≡ ¬φ¹ ∧ ¬φ².

• Here is a proof of the first law:

φ₁ φ₂ ¬(φ1 ∧ φ2) ¬φ1 ∨ ¬φ2

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 0

aAugustus DeMorgan (1806–1871) or William of Ockham (1288–

Conjunctive Normal Forms

• A boolean expression φ is in conjunctive normal form (CNF) if

φ =

n i=1

C_i,

where each clause C_i is the disjunction of zero or more literals.^a

– For example,

(x₁ ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (x2 ∨ x3).

• Convention: An empty CNF is satisfiable, but a CNF containing an empty clause is not.

aImproved by Mr. Aufbu Huang (R95922070) on October 5, 2006.

Disjunctive Normal Forms

• A boolean expression φ is in disjunctive normal form (DNF) if

φ =

n i=1

D_i,

where each implicant^a or simply term D_i is the conjunction of zero or more literals.

– For example,

(x₁ ∧ x2) ∨ (x1 ∧ ¬x2) ∨ (x2 ∧ x3).

aD_i implies φ, thus the term.

Clauses and Implicants

• The 

of clauses yields a clause.

– For example,

(x₁ ∨ x2) ∨ (x1 ∨ ¬x2) ∨ (x2 ∨ x3)

= x₁ ∨ x2 ∨ x1 ∨ ¬x2 ∨ x2 ∨ x3.

• The 

of implicants yields an implicant.

– For example,

(x₁ ∧ x2) ∧ (x1 ∧ ¬x2) ∧ (x2 ∧ x3)

= x₁ ∧ x2 ∧ x1 ∧ ¬x2 ∧ x2 ∧ x3.

Any Expression φ Can Be Converted into CNFs and DNFs φ = x_j:

• This is trivially true.

φ = ¬φ1 and a CNF is sought:

• Turn φ1 into a DNF.

• Apply de Morgan’s laws to make a CNF for φ.

φ = ¬φ1 and a DNF is sought:

• Turn φ1 into a CNF.

• Apply de Morgan’s laws to make a DNF for φ.

Any Expression φ Can Be Converted into CNFs and DNFs (continued)

φ = φ₁ ∨ φ² and a DNF is sought:

• Make φ1 and φ₂ DNFs.

φ = φ₁ ∨ φ2 and a CNF is sought:

• Turn φ1 and φ₂ into CNFs,^a φ₁ =

n₁



i=1

A_i, φ₂ =

n₂



j=1

B_j.

• Set

φ =

n₁



i=1 n₂



j=1

(A_i ∨ B^j).

aCorrected by Mr. Chun-Jie Yang (R99922150) on November 9, 2010.

Any Expression φ Can Be Converted into CNFs and DNFs (concluded)

φ = φ₁ ∧ φ2 and a CNF is sought:

• Make φ1 and φ₂ CNFs.

φ = φ₁ ∧ φ2 and a DNF is sought:

• Turn φ1 and φ₂ into DNFs, φ₁ =

n₁



i=1

A_i, φ₂ =

n₂



j=1

B_j.

• Set

φ =

n₁



i=1 n₂



j=1

(A_i ∧ Bj).

An Example: Turn ¬((a ∧ y) ∨ (z ∨ w)) into a DNF

¬((a ∧ y) ∨ (z ∨ w))

¬(CNF∨CNF)

= ¬(((a) ∧ (y)) ∨ ((z ∨ w)))

¬(CNF)

= ¬((a ∨ z ∨ w) ∧ (y ∨ z ∨ w))

de Morgan

= ¬(a ∨ z ∨ w) ∨ ¬(y ∨ z ∨ w)

de Morgan

= (¬a ∧ ¬z ∧ ¬w) ∨ (¬y ∧ ¬z ∧ ¬w).

Functional Completeness

• A set of logical connectives is called functionally

complete if every boolean expression is equivalent to one involving only these connectives.

• The set { ¬, ∨, ∧ } is functionally complete.

– Every boolean expression can be turned into a CNF, which involves only ¬, ∨, and ∧.

• The sets { ¬, ∨ } and { ¬, ∧ } are functionally complete.^a – By the above result and de Morgan’s laws.

• { nand } and { nor } are functionally complete.^b

aPost (1921).

bPeirce (c. 1880); Sheffer (1913).

Satisfiability

• A boolean expression φ is satisfiable if there is a truth assignment T appropriate to it such that T |= φ.

• φ is valid or a tautology,^a written |= φ, if T |= φ for all T appropriate to φ.

aWittgenstein (1922). Wittgenstein is one of the most important philosophers of all time. Russell (1919), “The importance of ‘tautology’

for a definition of mathematics was pointed out to me by my former pupil Ludwig Wittgenstein, who was working on the problem. I do not know whether he has solved it, or even whether he is alive or dead.”

“God has arrived,” the great economist Keynes (1883–1946) said of him on January 18, 1928, “I met him on the 5:15 train.”

Satisfiability (concluded)

• φ is unsatisfiable or a contradiction if φ is false under all appropriate truth assignments.

– Or, equivalently, if ¬φ is valid (prove it).

• φ is a contingency if φ is neither a tautology nor a contradiction.

Ludwig Wittgenstein (1889–1951)

Wittgenstein (1922),

“Whereof one cannot speak, thereof one must be silent.”

satisfiability (sat)

• The length of a boolean expression is the length of the string encoding it.

• satisfiability (sat): Given a CNF φ, is it satisfiable?

• Solvable in exponential time on a TM by the truth table method.

• Solvable in polynomial time on an NTM, hence in NP (p. 121).

• A most important problem in settling the “P = NP”^? problem (p. 323).

unsatisfiability (unsat or sat complement) and validity

• unsat (sat complement): Given a boolean expression φ, is it unsatisfiable?

• validity: Given a boolean expression φ, is it valid?

– φ is valid if and only if ¬φ is unsatisfiable.

– φ and ¬φ are basically of the same length.

– So unsat and validity have the same complexity.

• Both are solvable in exponential time on a TM by the truth table method.

Relations among sat, unsat, and validity

Contingent

Valid Unsatisfiable

• The negation of an unsatisfiable expression is a valid expression.

• None of the three problems—satisfiability,

unsatisfiability, validity—are known to be in P.

Boolean Functions

• An n-ary boolean function is a function

f : { true, false }ⁿ → { true, false }.

• It can be represented by a truth table.

• There are 2²ⁿ such boolean functions.

– We can assign true or false to f for each of the 2ⁿ truth assignments.

Boolean Functions (continued)

Assignment Truth value 1 true or false 2 true or false

... ...

2ⁿ true or false

• A boolean expression expresses a boolean function.

– Think of its truth values under all possible truth assignments.

Boolean Functions (continued)

• A boolean function expresses a boolean expression.

– 

T |= φ, literal yi is true in “row” T (y₁ ∧ · · · ∧ yn).^a

∗ The implicant y1 ∧ · · · ∧ yn is called the minterm over { x1, . . . , x_n } for T .

– The size^b is ≤ n2ⁿ ≤ 2²ⁿ.

– This DNF is optimal for the parity function, for example.^c

aSimilar to programmable logic array. This is called the table lookup representation (Beigel, 1993).

bWe count only the literals here.

cDu & Ko (2000).

Boolean Functions (continued)

x₁ x₂ f(x₁, x₂)

0 0 1

0 1 1

1 0 0

1 1 1

The corresponding boolean expression:

(¬x¹ ∧ ¬x²) ∨ (¬x¹ ∧ x²) ∨ (x¹ ∧ x²).

Boolean Functions (concluded)

Corollary 15 Every n-ary boolean function can be expressed by a boolean expression of size O(n2ⁿ).

• In general, the exponential length in n cannot be avoided (p. 212).

• The size of the truth table is also O(n2ⁿ).^a

aThere are 2ⁿ n-bit strings.

Boolean Circuits

• A boolean circuit is a graph C whose nodes are the gates.

• There are no cycles in C.

• All nodes have indegree (number of incoming edges) equal to 0, 1, or 2.

• Each gate has a sort from

{ true, false, ∨, ∧, ¬, x¹, x₂, . . . }.

– There are n + 5 sorts.

Boolean Circuits (concluded)

• Gates with a sort from { true, false, x1, x₂, . . . } are the inputs of C and have an indegree of zero.

• The output gate(s) has no outgoing edges.

• A boolean circuit computes a boolean function.

• A boolean function can be realized by infinitely many equivalent boolean circuits.

Boolean Circuits and Expressions

• They are equivalent representations.

• One can construct one from the other:

¬ [_L ¬

[_L

[_L ∨ [_M ∨

[_L [_M

[_L ∧ [_M ∧

[_L [_M

An Example

((x₁ x₂ ) (x₃ x₄)) (x₃ x₄))

x₁ x₂ x₃ x₄

• Circuits are more economical because of the possibility of “sharing.”

circuit sat and circuit value

circuit sat: Given a circuit, is there a truth assignment such that the circuit outputs true?

• circuit sat ∈ NP: Guess a truth assignment and then evaluate the circuit.^a

circuit value: The same as circuit sat except that the circuit has no variable gates.

• circuit value ∈ P: Evaluate the circuit from the input gates gradually towards the output gate.

aEssentially the same algorithm as the one on p. 121.

Some

Boolean Functions Need Exponential Circuits

Theorem 16 For any n ≥ 2, there is an n-ary boolean function f such that no boolean circuits with 2ⁿ/(2n) or fewer gates can compute it.

• There are 2²ⁿ different n-ary boolean functions (p. 202).

• So it suffices to prove that the number of boolean circuits with 2ⁿ/(2n) or fewer gates is less than 2²ⁿ.

aCan be strengthened to “Almost all.”

bRiordan & Shannon (1942); Shannon (1949).

The Proof (concluded)

• There are at most ((n + 5) × m²)^m boolean circuits with m or fewer gates (see next page).

• But ((n + 5) × m²)^m < 2²ⁿ when m = 2ⁿ/(2n):

m log₂((n + 5) × m²)

= 2ⁿ



1 − log_{2 n+5}⁴ⁿ² 2n



< 2ⁿ for n ≥ 2.

m choices

n+5 choices

m choices

Claude Elwood Shannon (1916–2001)

Howard Gardner (1987), “[Shan- non’s master’s thesis is] possibly the most important, and also the most famous, master’s thesis of the cen- tury.”

Comments

• The lower bound 2ⁿ/(2n) is rather tight because an upper bound is n2ⁿ (p. 204).

• The proof counted the number of circuits.

– Some circuits may not be valid at all.

– Different circuits may also compute the same function.

• Both are fine because we only need an upper bound on the number of circuits.

• We do not need to consider the outgoing edges because they have been counted as incoming edges.^a

aIf you prove the theorem by considering outgoing edges, the bound will not be good. (Try it!)

Relations between Complexity Classes

It is, I own, not uncommon to be wrong in theory and right in practice.

— Edmund Burke (1729–1797), A Philosophical Enquiry into the Origin of Our Ideas of the Sublime and Beautiful (1757) The problem with QE is it works in practice, but it doesn’t work in theory.

— Ben Bernanke (2014)

Proper (Complexity) Functions

• We say that f : N → N is a proper (complexity) function if the following hold:

– f is nondecreasing.

– There is a k-string TM M_f such that M_f(x) = ^{f(| x |)} for any x.^a

– M_f halts after O(| x | + f(| x |)) steps.

– M_f uses O(f (| x |)) space besides its input x.

• Mf’s behavior depends only on | x | not x’s contents.

• Mf’s running time is bounded by f (n).

aThe textbook calls “” the quasi-blank symbol. The use of M_f(x) will become clear in Proposition 17 (p. 222).

Examples of Proper Functions

• Most “reasonable” functions are proper: c, log n, polynomials of n, 2ⁿ, √n , n!, etc.

• If f and g are proper, then so are f + g, fg, and 2^g.^a

• Nonproper functions when serving as the time bounds for complexity classes spoil “theory building.”

– For example, TIME(f (n)) = TIME(2^f(n)) for some recursive function f (the gap theorem).^b

• Only proper functions f will be used in TIME(f(n)), SPACE(f (n)), NTIME(f (n)), and NSPACE(f (n)).

aFor f (g(n)), we need to add f (n) ≥ n.

bTrakhtenbrot (1964); Borodin (1972). Theorem 7.3 on p. 145 of the textbook proves it.

Precise Turing Machines

• A TM M is precise if there are functions f and g such that for every n ∈ N, for every x of length n, and for every computation path of M ,

– M halts after precisely f(n) steps,^a and

– All of its strings are of length precisely^b g(n) at halting.^c

∗ Recall that if M is a TM with input and output, we exclude the first and last strings.

• M can be deterministic or nondeterministic.

aFully time constructible (Hopcroft & Ullman, 1979).

bThis strong requirement does not seem needed later.

cFully space constructible (Hopcroft & Ullman, 1979).

Precise TMs Are General

Proposition 17 Suppose a TM^a M decides L within time (space) f (n), where f is proper. Then there is a precise TM M^ which decides L in time O(n + f (n)) (space O(f (n)), respectively).

• M^ on input x first simulates the TM M_f associated with the proper function f on x.

• M^f’s output, of length f (| x |), will serve as a

“yardstick” or an “alarm clock.”

aIt can be deterministic or nondeterministic.

The Proof (continued)

• Then M^ simulates M (x).

• M^(x) halts when and only when the alarm clock runs out—even if M halts earlier.

• If f is a time bound:

– The simulation of each step of M on x is matched by advancing the cursor on the “clock” string.

– Because M^ stops at the moment the “clock” string is exhausted—even if M (x) stops earlier, it is precise.

– The time bound is therefore O(| x | + f(| x |)).

The Proof (concluded)

• If f is a space bound (sketch):

– M^ simulates M on the quasi-blanks of M_f’s output string.^a

– The total space, not counting the input string, is O(f(n)).

– But we still need a way to make sure there is no infinite loop even if M does not halt.^b

aThis is to make sure the space bound is precise.

bSee the proof of Theorem 24 (p. 241).

Important Complexity Classes

• We write expressions like n^k to denote the union of all complexity classes, one for each value of k.

• For example,

NTIME(n^k) =^Δ 

j>0

NTIME(n^j).

Important Complexity Classes (concluded)

P =^Δ TIME(n^k), NP =^Δ NTIME(n^k), PSPACE =^Δ SPACE(n^k), NPSPACE =^Δ NSPACE(n^k),

E =^Δ TIME(2^kn), EXP =^Δ TIME(2^nk), NEXP =^Δ NTIME(2^nk),

L =^Δ SPACE(log n), NL =^Δ NSPACE(log n).

Complements of Nondeterministic Classes

• Recall that the complement of L, or ¯L, is the language Σ^∗ − L.

– sat complement is the set of unsatisfiable boolean expressions.

• R, RE, and coRE are distinct (p. 161).

– Again, coRE contains the complements of languages in RE, not languages that are not in RE.

• How about coC when C is a complexity class?

The Co-Classes

• For any complexity class C, coC denotes the class { L : ¯L ∈ C }.

• Clearly, if C is a deterministic time or space complexity class, then C = coC.

– They are said to be closed under complement.

– A deterministic TM deciding L can be converted to one that decides ¯L within the same time or space bound by reversing the “yes” and “no” states.^a

• Whether nondeterministic classes for time are closed under complement is not known (see p. 113).

aSee p. 158.

Comments

• As

coC = { L : ¯L ∈ C }, L ∈ C if and only if ¯L ∈ coC.

• But it is not true that L ∈ C if and only if L ∈ coC.

– coC is not defined as ¯C.

• For example, suppose C = {{ 2, 4, 6, 8, 10, . . . }, . . . }.

• Then coC = {{ 1, 3, 5, 7, 9, . . . }, . . . }.

• But ¯C = 2{ 1,2,3,... } − {{ 2, 4, 6, 8, 10, . . . }, . . . }.

The Quantified Halting Problem

• Let f(n) ≥ n be proper.

• Define

H_f =^Δ { M; x : M accepts input x after at most f (| x |) steps }, where M is deterministic.

• Assume the input is binary as usual.

H

∈ TIME(f(n)

)

• For each input M; x, we simulate M on x with an alarm clock of length f (| x |).

– Use the single-string simulator (p. 85), the universal TM (p. 137), and the linear speedup theorem (p. 95).

– Our simulator accepts M ; x if and only if M accepts x before the alarm clock runs out.

• From p. 92, the total running time is O(Mk_M² f(n)²), where _M is the length to encode each symbol or state of M and k_M is M ’s number of strings.

• As Mk_M² = O(n), the running time is O(f (n)³), where the constant is independent of M .

H

∈ TIME(f(n/2))

• Suppose TM MH_f decides H_f in time f (n/2).

• Consider machine:

D_f(M ) {

if M_Hf(M ; M ) = “yes”

then “no”;

else “yes”;

}

The Proof (continued)

• MH_f(M ; M ) runs in time f (²ⁿ⁺¹₂ ) = f(n), where n = | M |.^a

• By construction, D^f(M ) runs in the same amount of time as M_Hf(M ; M ), i.e., f (n), where n = | M |.

aMr. Hsiao-Fei Liu (F92922019) and Mr. Hong-Lung Wang (F92922085) pointed out on October 6, 2004, that this estimation (and the text’s Lemma 7.2) forgets to include the time to write down M ; M .

The Proof (concluded)

• First, suppose D^f(D_f) = “yes”.

• This implies

D_f; D_f ∈ H^f.

• Thus D^f does not accept D_f within time f (| D^f |).

• But D^f(D_f) stops in time f (| D^f |) with an answer.

• Hence Df(D_f) = “no”, a contradiction

• Similarly, Df(D_f) = “no” ⇒ Df(D_f) = “yes.”

The Time Hierarchy Theorem

Theorem 18 If f (n) ≥ n is proper, then

TIME(f (n))  TIME(f(2n + 1)³).

• The quantified halting problem makes it so.

Corollary 19 P  E.

• P ⊆ TIME(2ⁿ) because poly(n) ≤ 2ⁿ for n large enough.

• But by Theorem 18,

TIME (2ⁿ)  TIME

(2²ⁿ⁺¹)³

⊆ E.

• So P  E.

The Space Hierarchy Theorem

Theorem 20 (Hennie & Stearns, 1966) If f (n) is proper, then

SPACE(f (n))  SPACE(f(n) log f(n)).

Corollary 21 L  PSPACE.

Nondeterministic Time Hierarchy Theorems

Theorem 22 (Cook, 1973) NTIME(n^r)  NTIME(n^s) whenever 1 ≤ r < s.

Theorem 23 (Seiferas, Fischer, & Meyer, 1978) If T₁(n) and T₂(n) are proper, then

NTIME(T₁(n))  NTIME(T2(n)) whenever T₁(n + 1) = o(T₂(n)).