MAYBE
The TRS could not be proven terminating. The proof attempt took 512 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (52ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (71ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (35ms), DependencyGraph (1ms), ReductionPairSAT (21ms), DependencyGraph (1ms), SizeChangePrinciple (7ms)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (0ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
| – Problem 6 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, zWquot, from, quot, sel, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) | | minus#(s(X), s(Y)) | → | minus#(X, Y) |
| from#(X) | → | from#(s(X)) | | quot#(s(X), s(Y)) | → | minus#(X, Y) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | quot#(X, Y) | | sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | zWquot#(XS, YS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, zWquot, from, sel, quot, cons, nil
Strategy
The following SCCs where found
| quot#(s(X), s(Y)) → quot#(minus(X, Y), s(Y)) |
| minus#(s(X), s(Y)) → minus#(X, Y) |
| sel#(s(N), cons(X, XS)) → sel#(N, XS) |
| zWquot#(cons(X, XS), cons(Y, YS)) → zWquot#(XS, YS) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, zWquot, from, sel, quot, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, zWquot, from, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- cons(x,y): 0
- from(x): 0
- minus(x,y): 0
- nil: 0
- quot(x,y): 0
- quot#(x,y): x + 1
- s(x): 1
- sel(x,y): 0
- zWquot(x,y): 0
Improved Usable rules
| minus(s(X), s(Y)) | → | minus(X, Y) | | minus(X, 0) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| zWquot#(cons(X, XS), cons(Y, YS)) | → | zWquot#(XS, YS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, zWquot, from, sel, quot, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| zWquot#(cons(X, XS), cons(Y, YS)) | → | zWquot#(XS, YS) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) | | minus(X, 0) | → | 0 |
| minus(s(X), s(Y)) | → | minus(X, Y) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), zWquot(XS, YS)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, zWquot, from, sel, quot, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(X), s(Y)) | → | minus#(X, Y) |