MAYBE
The TRS could not be proven terminating. The proof attempt took 279 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 (67ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (31ms), DependencyGraph (1ms), ReductionPairSAT (24ms), DependencyGraph (1ms), SizeChangePrinciple (4ms)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (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
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sqr#(s(X)) | → | add#(sqr(X), dbl(X)) | | add#(s(X), Y) | → | add#(X, Y) |
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) | | dbl#(s(X)) | → | dbl#(X) |
| terms#(N) | → | sqr#(N) | | terms#(N) | → | terms#(s(N)) |
| sqr#(s(X)) | → | dbl#(X) | | sqr#(s(X)) | → | sqr#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| first#(s(X), cons(Y, Z)) → first#(X, Z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), terms(s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |