MAYBE
The TRS could not be proven terminating. The proof attempt took 299 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 (61ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (67ms), DependencyGraph (1ms), ReductionPairSAT (18ms), DependencyGraph (3ms), SizeChangePrinciple (4ms)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 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))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(XS) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, take(N, XS)) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) |
Original Signature
Termination of terms over the following signature is verified: 2nd, 0, s, take, from, head, sel, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| 2nd#(cons(X, XS)) | → | head#(XS) | | take#(s(N), cons(X, XS)) | → | take#(N, XS) |
| from#(X) | → | from#(s(X)) | | sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(XS) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, take(N, XS)) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) |
Original Signature
Termination of terms over the following signature is verified: 2nd, 0, s, take, from, head, sel, cons, nil
Strategy
The following SCCs where found
| take#(s(N), cons(X, XS)) → take#(N, XS) |
| sel#(s(N), cons(X, XS)) → sel#(N, XS) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| take#(s(N), cons(X, XS)) | → | take#(N, XS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(XS) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, take(N, XS)) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) |
Original Signature
Termination of terms over the following signature is verified: 2nd, 0, s, take, from, head, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| take#(s(N), cons(X, XS)) | → | take#(N, XS) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Rewrite Rules
| from(X) | → | cons(X, from(s(X))) | | head(cons(X, XS)) | → | X |
| 2nd(cons(X, XS)) | → | head(XS) | | take(0, XS) | → | nil |
| take(s(N), cons(X, XS)) | → | cons(X, take(N, XS)) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, XS) |
Original Signature
Termination of terms over the following signature is verified: 2nd, 0, s, take, from, head, sel, 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) |