YES
The TRS could be proven terminating. The proof took 200 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (26ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (103ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(p(X)) | → | a__p#(mark(X)) | | mark#(p(X)) | → | mark#(X) |
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(f(X)) | → | a__f#(mark(X)) |
| mark#(s(X)) | → | mark#(X) | | mark#(f(X)) | → | mark#(X) |
| a__f#(s(0)) | → | a__p#(s(0)) | | a__f#(s(0)) | → | a__f#(a__p(s(0))) |
Rewrite Rules
| a__f(0) | → | cons(0, f(s(0))) | | a__f(s(0)) | → | a__f(a__p(s(0))) |
| a__p(s(0)) | → | 0 | | mark(f(X)) | → | a__f(mark(X)) |
| mark(p(X)) | → | a__p(mark(X)) | | mark(0) | → | 0 |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(s(X)) | → | s(mark(X)) |
| a__f(X) | → | f(X) | | a__p(X) | → | p(X) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons
Strategy
The following SCCs where found
| mark#(p(X)) → mark#(X) | mark#(cons(X1, X2)) → mark#(X1) |
| mark#(s(X)) → mark#(X) | mark#(f(X)) → mark#(X) |
| a__f#(s(0)) → a__f#(a__p(s(0))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| mark#(p(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) | | mark#(f(X)) | → | mark#(X) |
Rewrite Rules
| a__f(0) | → | cons(0, f(s(0))) | | a__f(s(0)) | → | a__f(a__p(s(0))) |
| a__p(s(0)) | → | 0 | | mark(f(X)) | → | a__f(mark(X)) |
| mark(p(X)) | → | a__p(mark(X)) | | mark(0) | → | 0 |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(s(X)) | → | s(mark(X)) |
| a__f(X) | → | f(X) | | a__p(X) | → | p(X) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| mark#(p(X)) | → | mark#(X) | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(s(X)) | → | mark#(X) | | mark#(f(X)) | → | mark#(X) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| a__f#(s(0)) | → | a__f#(a__p(s(0))) |
Rewrite Rules
| a__f(0) | → | cons(0, f(s(0))) | | a__f(s(0)) | → | a__f(a__p(s(0))) |
| a__p(s(0)) | → | 0 | | mark(f(X)) | → | a__f(mark(X)) |
| mark(p(X)) | → | a__p(mark(X)) | | mark(0) | → | 0 |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(s(X)) | → | s(mark(X)) |
| a__f(X) | → | f(X) | | a__p(X) | → | p(X) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, a__p, p, mark, a__f, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__f(x): 0
- a__f#(x): x + 1
- a__p(x): 0
- cons(x,y): 0
- f(x): 0
- mark(x): 0
- p(x): 0
- s(x): 1
Improved Usable rules
| a__p(X) | → | p(X) | | a__p(s(0)) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| a__f#(s(0)) | → | a__f#(a__p(s(0))) |