YES
The TRS could be proven terminating. The proof took 300 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (18ms).
| – Problem 2 was processed with processor SubtermCriterion (3ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (216ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | plus#(y, z) |
| plus#(s(x), y) | → | plus#(x, y) | | quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
The following SCCs where found
| minus#(s(x), s(y)) → minus#(x, y) | minus#(minus(x, y), z) → minus#(x, plus(y, z)) |
| plus#(s(x), y) → plus#(x, y) |
| quot#(s(x), s(y)) → quot#(minus(x, y), s(y)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(s(x), y) | → | plus#(x, y) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Polynomial Interpretation
- 0: 0
- minus(x,y): x
- plus(x,y): 1
- quot(x,y): 0
- quot#(x,y): 2x
- s(x): x + 1
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | minus(minus(x, y), z) | → | minus(x, plus(y, z)) |
| minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| minus(minus(x, y), z) | → | minus(x, plus(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) | | minus#(minus(x, y), z) | → | minus#(x, plus(y, z)) |