MAYBE
The TRS could not be proven terminating. The proof attempt took 485 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 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (97ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (159ms), DependencyGraph (2ms), ReductionPairSAT (77ms), DependencyGraph (0ms), SizeChangePrinciple (9ms)].
| – Problem 4 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| fib1#(X, Y) | → | fib1#(Y, add(X, Y)) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| 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: 0, s, add, fib1, sel, fib, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) | | fib#(N) | → | sel#(N, fib1(s(0), s(0))) |
| fib#(N) | → | fib1#(s(0), s(0)) | | fib1#(X, Y) | → | fib1#(Y, add(X, Y)) |
| fib1#(X, Y) | → | add#(X, Y) | | sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| 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: 0, s, fib1, add, sel, fib, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| fib1#(X, Y) → fib1#(Y, add(X, Y)) |
| sel#(s(N), cons(X, XS)) → sel#(N, XS) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| 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: 0, s, fib1, add, sel, fib, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| 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: 0, s, fib1, add, sel, fib, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(N), cons(X, XS)) | → | sel#(N, XS) |