YES

The TRS could be proven terminating. The proof took 209 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (119ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 5 was processed with processor DependencyGraph (2ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

2ndsneg#(s(N), cons2(X, cons(Y, Z)))activate#(Z)2ndspos#(s(N), cons(X, Z))activate#(Z)
2ndspos#(s(N), cons2(X, cons(Y, Z)))2ndsneg#(N, activate(Z))2ndspos#(s(N), cons2(X, cons(Y, Z)))activate#(Z)
activate#(n__from(X))from#(X)pi#(X)2ndspos#(X, from(0))
pi#(X)from#(0)times#(s(X), Y)plus#(Y, times(X, Y))
2ndspos#(s(N), cons(X, Z))2ndspos#(s(N), cons2(X, activate(Z)))times#(s(X), Y)times#(X, Y)
square#(X)times#(X, X)2ndsneg#(s(N), cons(X, Z))2ndsneg#(s(N), cons2(X, activate(Z)))
2ndsneg#(s(N), cons2(X, cons(Y, Z)))2ndspos#(N, activate(Z))2ndsneg#(s(N), cons(X, Z))activate#(Z)
plus#(s(X), Y)plus#(X, Y)

Rewrite 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)rnil2ndsneg(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)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)from(X)n__from(X)
activate(n__from(X))from(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: plus, posrecip, negrecip, cons2, n__from, rnil, from, rcons, activate, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons

Strategy


The following SCCs where found

2ndsneg#(s(N), cons(X, Z)) → 2ndsneg#(s(N), cons2(X, activate(Z)))2ndspos#(s(N), cons2(X, cons(Y, Z))) → 2ndsneg#(N, activate(Z))
2ndsneg#(s(N), cons2(X, cons(Y, Z))) → 2ndspos#(N, activate(Z))2ndspos#(s(N), cons(X, Z)) → 2ndspos#(s(N), cons2(X, activate(Z)))

plus#(s(X), Y) → plus#(X, Y)

times#(s(X), Y) → times#(X, Y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

times#(s(X), Y)times#(X, Y)

Rewrite 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)rnil2ndsneg(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)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)from(X)n__from(X)
activate(n__from(X))from(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: plus, posrecip, negrecip, cons2, n__from, rnil, from, rcons, activate, 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)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(s(X), Y)plus#(X, Y)

Rewrite 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)rnil2ndsneg(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)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)from(X)n__from(X)
activate(n__from(X))from(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: plus, posrecip, negrecip, cons2, n__from, rnil, from, rcons, activate, 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 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

2ndsneg#(s(N), cons(X, Z))2ndsneg#(s(N), cons2(X, activate(Z)))2ndspos#(s(N), cons2(X, cons(Y, Z)))2ndsneg#(N, activate(Z))
2ndsneg#(s(N), cons2(X, cons(Y, Z)))2ndspos#(N, activate(Z))2ndspos#(s(N), cons(X, Z))2ndspos#(s(N), cons2(X, activate(Z)))

Rewrite 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)rnil2ndsneg(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)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)from(X)n__from(X)
activate(n__from(X))from(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: plus, posrecip, negrecip, cons2, n__from, rnil, from, rcons, activate, 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), cons2(X, cons(Y, Z)))2ndsneg#(N, activate(Z))2ndsneg#(s(N), cons2(X, cons(Y, Z)))2ndspos#(N, activate(Z))

Problem 5: DependencyGraph



Dependency Pair Problem

Dependency Pairs

2ndsneg#(s(N), cons(X, Z))2ndsneg#(s(N), cons2(X, activate(Z)))2ndspos#(s(N), cons(X, Z))2ndspos#(s(N), cons2(X, activate(Z)))

Rewrite 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)rnil2ndsneg(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)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)from(X)n__from(X)
activate(n__from(X))from(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: plus, posrecip, negrecip, cons2, n__from, rnil, from, rcons, activate, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons

Strategy


There are no SCCs!