YES
The TRS could be proven terminating. The proof took 485 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (19ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| | – Problem 5 was processed with processor DependencyGraph (0ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (300ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) | | ifMinus#(false, s(X), Y) | → | minus#(X, Y) |
| minus#(s(X), Y) | → | ifMinus#(le(s(X), Y), s(X), Y) | | minus#(s(X), Y) | → | le#(s(X), Y) |
| quot#(s(X), s(Y)) | → | minus#(X, Y) | | le#(s(X), s(Y)) | → | le#(X, Y) |
Rewrite Rules
| le(0, Y) | → | true | | le(s(X), 0) | → | false |
| le(s(X), s(Y)) | → | le(X, Y) | | minus(0, Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(true, s(X), Y) | → | 0 |
| ifMinus(false, s(X), Y) | → | s(minus(X, Y)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, true, false, ifMinus, quot
Strategy
The following SCCs where found
| quot#(s(X), s(Y)) → quot#(minus(X, Y), s(Y)) |
| ifMinus#(false, s(X), Y) → minus#(X, Y) | minus#(s(X), Y) → ifMinus#(le(s(X), Y), s(X), Y) |
| le#(s(X), s(Y)) → le#(X, Y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ifMinus#(false, s(X), Y) | → | minus#(X, Y) | | minus#(s(X), Y) | → | ifMinus#(le(s(X), Y), s(X), Y) |
Rewrite Rules
| le(0, Y) | → | true | | le(s(X), 0) | → | false |
| le(s(X), s(Y)) | → | le(X, Y) | | minus(0, Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(true, s(X), Y) | → | 0 |
| ifMinus(false, s(X), Y) | → | s(minus(X, Y)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, true, false, ifMinus, quot
Strategy
Projection
The following projection was used:
- π (ifMinus#): 2
- π (minus#): 1
Thus, the following dependency pairs are removed:
| ifMinus#(false, s(X), Y) | → | minus#(X, Y) |
Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), Y) | → | ifMinus#(le(s(X), Y), s(X), Y) |
Rewrite Rules
| le(0, Y) | → | true | | le(s(X), 0) | → | false |
| le(s(X), s(Y)) | → | le(X, Y) | | minus(0, Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(true, s(X), Y) | → | 0 |
| ifMinus(false, s(X), Y) | → | s(minus(X, Y)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, false, true, quot, ifMinus
Strategy
There are no SCCs!
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) |
Rewrite Rules
| le(0, Y) | → | true | | le(s(X), 0) | → | false |
| le(s(X), s(Y)) | → | le(X, Y) | | minus(0, Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(true, s(X), Y) | → | 0 |
| ifMinus(false, s(X), Y) | → | s(minus(X, Y)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, true, false, ifMinus, quot
Strategy
Polynomial Interpretation
- 0: 1
- false: 0
- ifMinus(x,y,z): y
- le(x,y): x
- minus(x,y): x
- quot(x,y): 0
- quot#(x,y): 2x
- s(x): 2x + 1
- true: 0
Improved Usable rules
| minus(0, Y) | → | 0 | | ifMinus(true, s(X), Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(false, s(X), Y) | → | s(minus(X, Y)) |
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
| le#(s(X), s(Y)) | → | le#(X, Y) |
Rewrite Rules
| le(0, Y) | → | true | | le(s(X), 0) | → | false |
| le(s(X), s(Y)) | → | le(X, Y) | | minus(0, Y) | → | 0 |
| minus(s(X), Y) | → | ifMinus(le(s(X), Y), s(X), Y) | | ifMinus(true, s(X), Y) | → | 0 |
| ifMinus(false, s(X), Y) | → | s(minus(X, Y)) | | quot(0, s(Y)) | → | 0 |
| quot(s(X), s(Y)) | → | s(quot(minus(X, Y), s(Y))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, true, false, ifMinus, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(X), s(Y)) | → | le#(X, Y) |