MAYBE

The TRS could not be proven terminating. The proof attempt took 345 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 was processed with processor SubtermCriterion (0ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (53ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (38ms), DependencyGraph (1ms), ReductionPairSAT (26ms), DependencyGraph (1ms), SizeChangePrinciple (7ms)].
 | – Problem 5 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

from#(X) → from#(s(X))

Rewrite Rules

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

Original Signature

Termination of terms over the following signature is verified: posrecip, negrecip, plus, rnil, from, rcons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

square#(X) → times#(X, X)2ndspos#(s(N), cons(X, cons(Y, Z))) → 2ndsneg#(N, Z)
from#(X) → from#(s(X))plus#(s(X), Y) → plus#(X, Y)
pi#(X) → 2ndspos#(X, from(0))pi#(X) → from#(0)
times#(s(X), Y) → plus#(Y, times(X, Y))times#(s(X), Y) → times#(X, Y)
2ndsneg#(s(N), cons(X, cons(Y, Z))) → 2ndspos#(N, Z)

Rewrite Rules

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

Original Signature

Termination of terms over the following signature is verified: posrecip, negrecip, plus, rnil, from, rcons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons

Strategy


The following SCCs where found

2ndspos#(s(N), cons(X, cons(Y, Z))) → 2ndsneg#(N, Z)2ndsneg#(s(N), cons(X, cons(Y, Z))) → 2ndspos#(N, Z)

from#(X) → from#(s(X))

plus#(s(X), Y) → plus#(X, Y)

times#(s(X), Y) → times#(X, Y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(s(X), Y) → plus#(X, Y)

Rewrite Rules

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

Original Signature

Termination of terms over the following signature is verified: posrecip, negrecip, plus, rnil, from, rcons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(s(X), Y) → plus#(X, Y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

2ndspos#(s(N), cons(X, cons(Y, Z))) → 2ndsneg#(N, Z)2ndsneg#(s(N), cons(X, cons(Y, Z))) → 2ndspos#(N, Z)

Rewrite Rules

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

Original Signature

Termination of terms over the following signature is verified: posrecip, negrecip, plus, rnil, from, rcons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

2ndspos#(s(N), cons(X, cons(Y, Z))) → 2ndsneg#(N, Z)2ndsneg#(s(N), cons(X, cons(Y, Z))) → 2ndspos#(N, Z)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

times#(s(X), Y) → times#(X, Y)

Rewrite Rules

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

Original Signature

Termination of terms over the following signature is verified: posrecip, negrecip, plus, rnil, from, rcons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

times#(s(X), Y) → times#(X, Y)