The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

0 CpxTRS
↳1 NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CpxTRS
↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 8 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtUsableRulesProof (⇔, 0 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 51 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 BOUNDS(1, 1)

(0) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Rewrite Strategy: INNERMOST

(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)

The following defined symbols can occur below the 0th argument of from: s, from, activate
The following defined symbols can occur below the 0th argument of s: s, from, activate
The following defined symbols can occur below the 1th argument of 2ndspos: from

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
plus(s(X), Y) → s(plus(X, Y))
times(s(X), Y) → plus(Y, times(X, Y))

(2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).


The TRS R consists of the following rules:

square(X) → times(X, X)
activate(n__from(X)) → from(activate(X))
from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
from(X) → n__from(X)
s(X) → n__s(X)
activate(X) → X
times(0, Y) → 0
2ndsneg(0, Z) → rnil
activate(n__s(X)) → s(activate(X))

Rewrite Strategy: INNERMOST

(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

square(z0) → times(z0, z0)
activate(n__from(z0)) → from(activate(z0))
activate(z0) → z0
activate(n__s(z0)) → s(activate(z0))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
s(z0) → n__s(z0)
times(0, z0) → 0
2ndsneg(0, z0) → rnil
Tuples:

SQUARE(z0) → c(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c1(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c2
ACTIVATE(n__s(z0)) → c3(S(activate(z0)), ACTIVATE(z0))
FROM(z0) → c4
FROM(z0) → c5
2NDSPOS(0, z0) → c6
PI(z0) → c7(2NDSPOS(z0, from(0)), FROM(0))
PLUS(0, z0) → c8
S(z0) → c9
TIMES(0, z0) → c10
2NDSNEG(0, z0) → c11
S tuples:

SQUARE(z0) → c(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c1(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c2
ACTIVATE(n__s(z0)) → c3(S(activate(z0)), ACTIVATE(z0))
FROM(z0) → c4
FROM(z0) → c5
2NDSPOS(0, z0) → c6
PI(z0) → c7(2NDSPOS(z0, from(0)), FROM(0))
PLUS(0, z0) → c8
S(z0) → c9
TIMES(0, z0) → c10
2NDSNEG(0, z0) → c11
K tuples:none
Defined Rule Symbols:

square, activate, from, 2ndspos, pi, plus, s, times, 2ndsneg

Defined Pair Symbols:

SQUARE, ACTIVATE, FROM, 2NDSPOS, PI, PLUS, S, TIMES, 2NDSNEG

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 10 trailing nodes:

SQUARE(z0) → c(TIMES(z0, z0))
2NDSPOS(0, z0) → c6
PLUS(0, z0) → c8
FROM(z0) → c4
FROM(z0) → c5
S(z0) → c9
PI(z0) → c7(2NDSPOS(z0, from(0)), FROM(0))
2NDSNEG(0, z0) → c11
TIMES(0, z0) → c10
ACTIVATE(z0) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

square(z0) → times(z0, z0)
activate(n__from(z0)) → from(activate(z0))
activate(z0) → z0
activate(n__s(z0)) → s(activate(z0))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
s(z0) → n__s(z0)
times(0, z0) → 0
2ndsneg(0, z0) → rnil
Tuples:

ACTIVATE(n__from(z0)) → c1(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(S(activate(z0)), ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c1(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

square, activate, from, 2ndspos, pi, plus, s, times, 2ndsneg

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c1, c3

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

square(z0) → times(z0, z0)
activate(n__from(z0)) → from(activate(z0))
activate(z0) → z0
activate(n__s(z0)) → s(activate(z0))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
s(z0) → n__s(z0)
times(0, z0) → 0
2ndsneg(0, z0) → rnil
Tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

square, activate, from, 2ndspos, pi, plus, s, times, 2ndsneg

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c1, c3

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

square(z0) → times(z0, z0)
activate(n__from(z0)) → from(activate(z0))
activate(z0) → z0
activate(n__s(z0)) → s(activate(z0))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
s(z0) → n__s(z0)
times(0, z0) → 0
2ndsneg(0, z0) → rnil

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c1, c3

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = [2]x1   
POL(c1(x1)) = x1   
POL(c3(x1)) = x1   
POL(n__from(x1)) = [1] + x1   
POL(n__s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__from(z0)) → c1(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c3(ACTIVATE(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c1, c3

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(14) BOUNDS(1, 1)