YES
The TRS could be proven terminating. The proof took 270 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (17ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (225ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | min#(X, Y) | | plus#(s(X), Y) | → | plus#(X, Y) |
| min#(min(X, Y), Z) | → | min#(X, plus(Y, Z)) | | quot#(s(X), s(Y)) | → | quot#(min(X, Y), s(Y)) |
| min#(s(X), s(Y)) | → | min#(X, Y) | | min#(min(X, Y), Z) | → | plus#(Y, Z) |
Rewrite Rules
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| min(min(X, Y), Z) | → | min(X, plus(Y, Z)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(min(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: min, plus, 0, s, quot, Z
Strategy
The following SCCs where found
| min#(min(X, Y), Z) → min#(X, plus(Y, Z)) | min#(s(X), s(Y)) → min#(X, Y) |
| quot#(s(X), s(Y)) → quot#(min(X, Y), s(Y)) |
| plus#(s(X), Y) → plus#(X, Y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(X), Y) | → | plus#(X, Y) |
Rewrite Rules
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| min(min(X, Y), Z) | → | min(X, plus(Y, Z)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(min(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: min, plus, 0, s, quot, Z
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#(min(X, Y), s(Y)) |
Rewrite Rules
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| min(min(X, Y), Z) | → | min(X, plus(Y, Z)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(min(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: min, plus, 0, s, quot, Z
Strategy
Polynomial Interpretation
- 0: 0
- Z: 2
- min(x,y): x
- plus(x,y): 0
- quot(x,y): 0
- quot#(x,y): x
- s(x): 2x + 1
Improved Usable rules
| min(min(X, Y), Z) | → | min(X, plus(Y, Z)) | | min(s(X), s(Y)) | → | min(X, Y) |
| min(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#(min(X, Y), s(Y)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| min#(min(X, Y), Z) | → | min#(X, plus(Y, Z)) | | min#(s(X), s(Y)) | → | min#(X, Y) |
Rewrite Rules
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| min(min(X, Y), Z) | → | min(X, plus(Y, Z)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(min(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: min, plus, 0, s, quot, Z
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| min#(min(X, Y), Z) | → | min#(X, plus(Y, Z)) | | min#(s(X), s(Y)) | → | min#(X, Y) |