YES
The TRS could be proven terminating. The proof took 248 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (24ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (158ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | double#(y) | | plus#(s(x), y) | → | plus#(x, y) |
| plus#(s(x), y) | → | minus#(x, y) | | minus#(s(x), s(y)) | → | minus#(x, y) |
| double#(s(x)) | → | double#(x) | | plus#(s(x), y) | → | plus#(minus(x, y), double(y)) |
| plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(s(x), y) | → | plus(x, s(y)) | | plus(s(x), y) | → | s(plus(minus(x, y), double(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, double
Strategy
The following SCCs where found
| minus#(s(x), s(y)) → minus#(x, y) |
| plus#(s(x), y) → plus#(x, y) | plus#(s(x), y) → plus#(minus(x, y), double(y)) |
| plus#(s(x), y) → plus#(x, s(y)) |
| double#(s(x)) → double#(x) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) | | plus#(s(x), y) | → | plus#(minus(x, y), double(y)) |
| plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(s(x), y) | → | plus(x, s(y)) | | plus(s(x), y) | → | s(plus(minus(x, y), double(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, double
Strategy
Polynomial Interpretation
- 0: 1
- double(x): 2x
- minus(x,y): 2x
- plus(x,y): 0
- plus#(x,y): x
- s(x): 2x + 1
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(s(x), y) | → | plus#(x, y) | | plus#(s(x), y) | → | plus#(minus(x, y), double(y)) |
| plus#(s(x), y) | → | plus#(x, s(y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| double#(s(x)) | → | double#(x) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(s(x), y) | → | plus(x, s(y)) | | plus(s(x), y) | → | s(plus(minus(x, y), double(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, double
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| double#(s(x)) | → | double#(x) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(s(x), y) | → | plus(x, s(y)) | | plus(s(x), y) | → | s(plus(minus(x, y), double(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, double
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |