Runtime Complexity (innermost) proof of /tmp/tmpHjeheD/ExAppendixB_AEL03

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS

↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)

↳2 CdtProblem

↳3 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)

↳4 CdtProblem

↳5 CdtKnowledgeProof (⇔, 0 ms)

↳6 CdtProblem

↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 393 ms)

↳8 CdtProblem

↳9 CdtKnowledgeProof (⇔, 0 ms)

↳10 CdtProblem

↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 417 ms)

↳12 CdtProblem

↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 1146 ms)

↳14 CdtProblem

↳15 SIsEmptyProof (⇔, 0 ms)

↳16 BOUNDS(O(1), O(1))

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

from(X) → cons(X, n__from(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)
activate(n__from(X)) → from(X)
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c13(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(z0))

S tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c13(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(z0))

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

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

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12, c13, c14

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

SQUARE(z0) → c13(TIMES(z0, z0))

Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c14(FROM(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))

Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))

We considered the (Usable) Rules:

times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

And the Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]
POL(2NDSNEG(x₁, x₂)) = x₁
POL(2NDSPOS(x₁, x₂)) = x₁
POL(ACTIVATE(x₁)) = 0
POL(FROM(x₁)) = [3]
POL(PI(x₁)) = [4] + [4]x₁
POL(PLUS(x₁, x₂)) = 0
POL(TIMES(x₁, x₂)) = 0
POL(activate(x₁)) = [2]
POL(c10(x₁)) = x₁
POL(c12(x₁, x₂)) = x₁ + x₂
POL(c3(x₁, x₂)) = x₁ + x₂
POL(c4(x₁, x₂)) = x₁ + x₂
POL(c6(x₁, x₂)) = x₁ + x₂
POL(c7(x₁, x₂)) = x₁ + x₂
POL(c8(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = x₁
POL(cons2(x₁, x₂)) = [3]
POL(from(x₁)) = [4] + [2]x₁
POL(n__from(x₁)) = [3] + x₁
POL(plus(x₁, x₂)) = [2] + [2]x₂
POL(s(x₁)) = [3] + x₁
POL(times(x₁, x₂)) = [5]x₂

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))

Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:

PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))

Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

We considered the (Usable) Rules:

times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

And the Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]
POL(2NDSNEG(x₁, x₂)) = [1] + [2]x₁
POL(2NDSPOS(x₁, x₂)) = [2] + [2]x₁
POL(ACTIVATE(x₁)) = 0
POL(FROM(x₁)) = 0
POL(PI(x₁)) = [5] + [5]x₁
POL(PLUS(x₁, x₂)) = [4]
POL(TIMES(x₁, x₂)) = [4]x₁
POL(activate(x₁)) = [5]
POL(c10(x₁)) = x₁
POL(c12(x₁, x₂)) = x₁ + x₂
POL(c3(x₁, x₂)) = x₁ + x₂
POL(c4(x₁, x₂)) = x₁ + x₂
POL(c6(x₁, x₂)) = x₁ + x₂
POL(c7(x₁, x₂)) = x₁ + x₂
POL(c8(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = [3]
POL(cons2(x₁, x₂)) = [4]
POL(from(x₁)) = [4] + x₁
POL(n__from(x₁)) = [3] + x₁
POL(plus(x₁, x₂)) = [2] + [2]x₁ + [3]x₂
POL(s(x₁)) = [4] + x₁
POL(times(x₁, x₂)) = [2]

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:

PLUS(s(z0), z1) → c10(PLUS(z0, z1))

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

PLUS(s(z0), z1) → c10(PLUS(z0, z1))

We considered the (Usable) Rules:

times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

And the Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(2NDSNEG(x₁, x₂)) = 0
POL(2NDSPOS(x₁, x₂)) = 0
POL(ACTIVATE(x₁)) = 0
POL(FROM(x₁)) = [1]
POL(PI(x₁)) = [2] + [3]x₁
POL(PLUS(x₁, x₂)) = [2] + x₁
POL(TIMES(x₁, x₂)) = x₁·x₂ + x₁²
POL(activate(x₁)) = 0
POL(c10(x₁)) = x₁
POL(c12(x₁, x₂)) = x₁ + x₂
POL(c3(x₁, x₂)) = x₁ + x₂
POL(c4(x₁, x₂)) = x₁ + x₂
POL(c6(x₁, x₂)) = x₁ + x₂
POL(c7(x₁, x₂)) = x₁ + x₂
POL(c8(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = 0
POL(cons2(x₁, x₂)) = 0
POL(from(x₁)) = 0
POL(n__from(x₁)) = 0
POL(plus(x₁, x₂)) = 0
POL(s(x₁)) = [2] + x₁
POL(times(x₁, x₂)) = 0

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, z2)) → 2ndspos(s(z0), cons2(z1, activate(z2)))
2ndspos(s(z0), cons2(z1, cons(z2, z3))) → rcons(posrecip(z2), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, z2)) → 2ndsneg(s(z0), cons2(z1, activate(z2)))
2ndsneg(s(z0), cons2(z1, cons(z2, z3))) → rcons(negrecip(z2), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

S tuples:none
K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)), FROM(0))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
2NDSPOS(s(z0), cons(z1, z2)) → c3(2NDSPOS(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))), ACTIVATE(z2))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
PLUS(s(z0), z1) → c10(PLUS(z0, z1))

Defined Rule Symbols:

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

Defined Pair Symbols:

2NDSPOS, 2NDSNEG, PI, PLUS, TIMES

Compound Symbols:

c3, c4, c6, c7, c8, c10, c12

(15) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(0) Obligation:

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

(2) Obligation:

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

(4) Obligation:

(5) CdtKnowledgeProof (EQUIVALENT transformation)

(6) Obligation:

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

(8) Obligation:

(9) CdtKnowledgeProof (EQUIVALENT transformation)

(10) Obligation:

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

(12) Obligation:

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

(14) Obligation:

(15) SIsEmptyProof (EQUIVALENT transformation)

(16) BOUNDS(O(1), O(1))