YES
The TRS could be proven terminating. The proof took 610 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (10ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (223ms).
| | – Problem 3 was processed with processor DependencyGraph (1ms).
| | | – Problem 4 was processed with processor PolynomialLinearRange4iUR (152ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__tail#(cons(X, XS)) | → | mark#(XS) | | mark#(tail(X)) | → | mark#(X) |
| mark#(zeros) | → | a__zeros# | | mark#(cons(X1, X2)) | → | mark#(X1) |
| mark#(tail(X)) | → | a__tail#(mark(X)) |
Rewrite Rules
| a__zeros | → | cons(0, zeros) | | a__tail(cons(X, XS)) | → | mark(XS) |
| mark(zeros) | → | a__zeros | | mark(tail(X)) | → | a__tail(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| a__zeros | → | zeros | | a__tail(X) | → | tail(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, 0, zeros, mark, tail, cons, a__tail
Strategy
The following SCCs where found
| a__tail#(cons(X, XS)) → mark#(XS) | mark#(tail(X)) → mark#(X) |
| mark#(cons(X1, X2)) → mark#(X1) | mark#(tail(X)) → a__tail#(mark(X)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| a__tail#(cons(X, XS)) | → | mark#(XS) | | mark#(tail(X)) | → | mark#(X) |
| mark#(cons(X1, X2)) | → | mark#(X1) | | mark#(tail(X)) | → | a__tail#(mark(X)) |
Rewrite Rules
| a__zeros | → | cons(0, zeros) | | a__tail(cons(X, XS)) | → | mark(XS) |
| mark(zeros) | → | a__zeros | | mark(tail(X)) | → | a__tail(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| a__zeros | → | zeros | | a__tail(X) | → | tail(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, 0, zeros, mark, tail, cons, a__tail
Strategy
Polynomial Interpretation
- 0: 0
- a__tail(x): 2x + 2
- a__tail#(x): 2x
- a__zeros: 0
- cons(x,y): y + x
- mark(x): x
- mark#(x): x
- tail(x): 2x + 2
- zeros: 0
Improved Usable rules
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| mark(tail(X)) | → | a__tail(mark(X)) | | a__tail(X) | → | tail(X) |
| a__zeros | → | cons(0, zeros) | | a__zeros | → | zeros |
| mark(zeros) | → | a__zeros | | a__tail(cons(X, XS)) | → | mark(XS) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(tail(X)) | → | mark#(X) | | mark#(tail(X)) | → | a__tail#(mark(X)) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a__tail#(cons(X, XS)) | → | mark#(XS) | | mark#(cons(X1, X2)) | → | mark#(X1) |
Rewrite Rules
| a__zeros | → | cons(0, zeros) | | a__tail(cons(X, XS)) | → | mark(XS) |
| mark(zeros) | → | a__zeros | | mark(tail(X)) | → | a__tail(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| a__zeros | → | zeros | | a__tail(X) | → | tail(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, 0, zeros, mark, tail, a__tail, cons
Strategy
The following SCCs where found
| mark#(cons(X1, X2)) → mark#(X1) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| mark#(cons(X1, X2)) | → | mark#(X1) |
Rewrite Rules
| a__zeros | → | cons(0, zeros) | | a__tail(cons(X, XS)) | → | mark(XS) |
| mark(zeros) | → | a__zeros | | mark(tail(X)) | → | a__tail(mark(X)) |
| mark(cons(X1, X2)) | → | cons(mark(X1), X2) | | mark(0) | → | 0 |
| a__zeros | → | zeros | | a__tail(X) | → | tail(X) |
Original Signature
Termination of terms over the following signature is verified: a__zeros, 0, zeros, mark, tail, a__tail, cons
Strategy
Polynomial Interpretation
- 0: 0
- a__tail(x): 0
- a__zeros: 0
- cons(x,y): x + 2
- mark(x): 0
- mark#(x): x + 1
- tail(x): 0
- zeros: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| mark#(cons(X1, X2)) | → | mark#(X1) |