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

(0) Obligation:

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


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 Cpx (relative) TRS 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:

FROM(z0) → c
FROM(z0) → c1
2NDSPOS(0, z0) → c2
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(0, z0) → c5
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(0, z0) → c9
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(0, z0) → c11
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))
ACTIVATE(z0) → c15

S tuples:

FROM(z0) → c
FROM(z0) → c1
2NDSPOS(0, z0) → c2
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(0, z0) → c5
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(0, z0) → c9
PLUS(s(z0), z1) → c10(PLUS(z0, z1))
TIMES(0, z0) → c11
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))
ACTIVATE(z0) → c15

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

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

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14, c15

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

SQUARE(z0) → c13(TIMES(z0, z0))

Removed 8 trailing nodes:

FROM(z0) → c1
2NDSNEG(0, z0) → c5
TIMES(0, z0) → c11
FROM(z0) → c
2NDSPOS(0, z0) → c2
PLUS(0, z0) → c9
ACTIVATE(n__from(z0)) → c14(FROM(z0))
ACTIVATE(z0) → c15

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

Removed 5 trailing tuple parts

(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:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

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

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

PI(z0) → c8(2NDSPOS(z0, from(0)))

(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:

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

S tuples:

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

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)))

Defined Rule Symbols:

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

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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))
square(z0) → times(z0, z0)

(10) Obligation:

Complexity Dependency Tuples Problem
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))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

Tuples:

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

S tuples:

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

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)))

Defined Rule Symbols:

times, plus, activate, from

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

(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.

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

We considered the (Usable) Rules:none
And the Tuples:

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

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(PI(x₁)) = x₁   
POL(PLUS(x₁, x₂)) = 0   
POL(TIMES(x₁, x₂)) = x₁   
POL(activate(x₁)) = [1] + x₁   
POL(c10(x₁)) = x₁   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(cons(x₁, x₂)) = 0   
POL(cons2(x₁, x₂)) = 0   
POL(from(x₁)) = [1]   
POL(n__from(x₁)) = 0   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   
POL(times(x₁, x₂)) = 0   

(12) Obligation:

Complexity Dependency Tuples Problem
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))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

Tuples:

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

S tuples:

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

K tuples:

PI(z0) → c8(2NDSPOS(z0, from(0)))
TIMES(s(z0), z1) → c12(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

Defined Rule Symbols:

times, plus, activate, from

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(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)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))

We considered the (Usable) Rules:none
And the Tuples:

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

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

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

(14) Obligation:

Complexity Dependency Tuples Problem
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))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

Tuples:

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

S tuples:

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

K tuples:

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

Defined Rule Symbols:

times, plus, activate, from

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

(15) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) 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))))
2NDSNEG(s(z0), cons(z1, z2)) → c6(2NDSNEG(s(z0), cons2(z1, activate(z2))))
2NDSPOS(s(z0), cons2(z1, cons(z2, z3))) → c4(2NDSNEG(z0, activate(z3)))
2NDSNEG(s(z0), cons2(z1, cons(z2, z3))) → c7(2NDSPOS(z0, activate(z3)))

(16) Obligation:

Complexity Dependency Tuples Problem
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))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

Tuples:

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

S tuples:

PLUS(s(z0), z1) → c10(PLUS(z0, z1))

K tuples:

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

Defined Rule Symbols:

times, plus, activate, from

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(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:none
And the Tuples:

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

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

POL(0) = [1]   
POL(2NDSNEG(x₁, x₂)) = [2] + x₁   
POL(2NDSPOS(x₁, x₂)) = [1] + x₁   
POL(PI(x₁)) = [2] + [2]x₁ + x₁²   
POL(PLUS(x₁, x₂)) = [1] + [2]x₁   
POL(TIMES(x₁, x₂)) = [2]x₁ + x₁·x₂ + x₁²   
POL(activate(x₁)) = x₁ + [2]x₁²   
POL(c10(x₁)) = x₁   
POL(c12(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(c8(x₁)) = x₁   
POL(cons(x₁, x₂)) = [1]   
POL(cons2(x₁, x₂)) = [1]   
POL(from(x₁)) = [2] + [2]x₁   
POL(n__from(x₁)) = x₁   
POL(plus(x₁, x₂)) = [2]x₁ + [2]x₂ + [2]x₂² + [2]x₁·x₂ + [2]x₁²   
POL(s(x₁)) = [2] + x₁   
POL(times(x₁, x₂)) = [2] + [2]x₂² + [2]x₁·x₂ + x₁²   

(18) Obligation:

Complexity Dependency Tuples Problem
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))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

times, plus, activate, from

Defined Pair Symbols:

PLUS, TIMES, 2NDSPOS, 2NDSNEG, PI

Compound Symbols:

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

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

The set S is empty

(20) BOUNDS(1, 1)