MAYBE
The TRS could not be proven terminating. The proof attempt took 372 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 6 was processed with processor DependencyGraph (0ms).
| Problem 4 was processed with processor SubtermCriterion (0ms).
| Problem 5 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (66ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (36ms), DependencyGraph (1ms), ReductionPairSAT (26ms), DependencyGraph (1ms), SizeChangePrinciple (5ms)].
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
Rewrite Rules
from(X) | → | cons(X, from(s(X))) | | 2ndspos(0, Z) | → | rnil |
2ndspos(s(N), cons(X, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
2ndspos#(s(N), cons2(X, cons(Y, Z))) | → | 2ndsneg#(N, Z) | | square#(X) | → | times#(X, X) |
from#(X) | → | from#(s(X)) | | 2ndsneg#(s(N), cons(X, Z)) | → | 2ndsneg#(s(N), cons2(X, Z)) |
plus#(s(X), Y) | → | plus#(X, Y) | | 2ndspos#(s(N), cons(X, Z)) | → | 2ndspos#(s(N), cons2(X, Z)) |
pi#(X) | → | 2ndspos#(X, from(0)) | | pi#(X) | → | from#(0) |
times#(s(X), Y) | → | plus#(Y, times(X, Y)) | | 2ndsneg#(s(N), cons2(X, cons(Y, Z))) | → | 2ndspos#(N, Z) |
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, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons
Strategy
The following SCCs where found
2ndspos#(s(N), cons2(X, cons(Y, Z))) → 2ndsneg#(N, Z) | 2ndsneg#(s(N), cons(X, Z)) → 2ndsneg#(s(N), cons2(X, Z)) |
2ndspos#(s(N), cons(X, Z)) → 2ndspos#(s(N), cons2(X, Z)) | 2ndsneg#(s(N), cons2(X, cons(Y, Z))) → 2ndspos#(N, 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, from(s(X))) | | 2ndspos(0, Z) | → | rnil |
2ndspos(s(N), cons(X, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, 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
2ndspos#(s(N), cons2(X, cons(Y, Z))) | → | 2ndsneg#(N, Z) | | 2ndsneg#(s(N), cons(X, Z)) | → | 2ndsneg#(s(N), cons2(X, Z)) |
2ndspos#(s(N), cons(X, Z)) | → | 2ndspos#(s(N), cons2(X, Z)) | | 2ndsneg#(s(N), cons2(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, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons
Strategy
Projection
The following projection was used:
- π (2ndspos#): 1
- π (2ndsneg#): 1
Thus, the following dependency pairs are removed:
2ndspos#(s(N), cons2(X, cons(Y, Z))) | → | 2ndsneg#(N, Z) | | 2ndsneg#(s(N), cons2(X, cons(Y, Z))) | → | 2ndspos#(N, Z) |
Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
2ndsneg#(s(N), cons(X, Z)) | → | 2ndsneg#(s(N), cons2(X, Z)) | | 2ndspos#(s(N), cons(X, Z)) | → | 2ndspos#(s(N), cons2(X, Z)) |
Rewrite Rules
from(X) | → | cons(X, from(s(X))) | | 2ndspos(0, Z) | → | rnil |
2ndspos(s(N), cons(X, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons
Strategy
There are no SCCs!
Problem 4: 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, Z)) | → | 2ndspos(s(N), cons2(X, Z)) | | 2ndspos(s(N), cons2(X, cons(Y, Z))) | → | rcons(posrecip(Y), 2ndsneg(N, Z)) |
2ndsneg(0, Z) | → | rnil | | 2ndsneg(s(N), cons(X, Z)) | → | 2ndsneg(s(N), cons2(X, Z)) |
2ndsneg(s(N), cons2(X, cons(Y, Z))) | → | rcons(negrecip(Y), 2ndspos(N, 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) |
Original Signature
Termination of terms over the following signature is verified: posrecip, negrecip, plus, cons2, rnil, from, rcons, 2ndspos, 0, s, times, 2ndsneg, square, pi, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
plus#(s(X), Y) | → | plus#(X, Y) |