MAYBE
The TRS could not be proven terminating. The proof attempt took 310 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 (55ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (74ms), DependencyGraph (0ms), ReductionPairSAT (20ms), DependencyGraph (1ms), SizeChangePrinciple (5ms)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, fst(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(len(Z)) |
Original Signature
Termination of terms over the following signature is verified: fst, 0, s, len, from, add, cons, nil
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) | | fst#(s(X), cons(Y, Z)) | → | fst#(X, Z) |
| len#(cons(X, Z)) | → | len#(Z) | | from#(X) | → | from#(s(X)) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, fst(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(len(Z)) |
Original Signature
Termination of terms over the following signature is verified: fst, 0, s, from, len, add, nil, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| fst#(s(X), cons(Y, Z)) → fst#(X, Z) |
| len#(cons(X, Z)) → len#(Z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| fst#(s(X), cons(Y, Z)) | → | fst#(X, Z) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, fst(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(len(Z)) |
Original Signature
Termination of terms over the following signature is verified: fst, 0, s, from, len, add, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| fst#(s(X), cons(Y, Z)) | → | fst#(X, Z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| len#(cons(X, Z)) | → | len#(Z) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, fst(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(len(Z)) |
Original Signature
Termination of terms over the following signature is verified: fst, 0, s, from, len, add, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| len#(cons(X, Z)) | → | len#(Z) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| fst(0, Z) | → | nil | | fst(s(X), cons(Y, Z)) | → | cons(Y, fst(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | len(nil) | → | 0 |
| len(cons(X, Z)) | → | s(len(Z)) |
Original Signature
Termination of terms over the following signature is verified: fst, 0, s, from, len, add, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |