YES
The TRS could be proven terminating. The proof took 725 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (55ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (132ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor PolynomialLinearRange4iUR (88ms).
| | – Problem 6 was processed with processor DependencyGraph (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) |
| minus#(s(X), s(Y)) | → | minus#(X, Y) | | activate#(n__zWquot(X1, X2)) | → | zWquot#(X1, X2) |
| activate#(n__from(X)) | → | from#(X) | | sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
| quot#(s(X), s(Y)) | → | minus#(X, Y) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | quot#(X, Y) |
| sel#(s(N), cons(X, XS)) | → | activate#(XS) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, 0, minus, s, n__from, zWquot, from, sel, quot, cons, nil
Strategy
The following SCCs where found
| quot#(s(X), s(Y)) → quot#(minus(X, Y), s(Y)) |
| minus#(s(X), s(Y)) → minus#(X, Y) |
| zWquot#(cons(X, XS), cons(Y, YS)) → activate#(YS) | activate#(n__zWquot(X1, X2)) → zWquot#(X1, X2) |
| zWquot#(cons(X, XS), cons(Y, YS)) → activate#(XS) |
| sel#(s(N), cons(X, XS)) → sel#(N, activate(XS)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, 0, minus, s, n__from, zWquot, from, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- cons(x,y): 0
- from(x): 0
- minus(x,y): 0
- n__from(x): 0
- n__zWquot(x,y): 0
- nil: 0
- quot(x,y): 0
- quot#(x,y): x
- s(x): 1
- sel(x,y): 0
- zWquot(x,y): 0
Improved Usable rules
| minus(s(X), s(Y)) | → | minus(X, Y) | | minus(X, 0) | → | 0 |
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 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, 0, minus, s, n__from, zWquot, from, sel, quot, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, 0, minus, s, n__from, zWquot, from, sel, quot, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | activate#(n__zWquot(X1, X2)) | → | zWquot#(X1, X2) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, 0, minus, s, n__from, zWquot, from, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): x
- cons(x,y): 2y
- from(x): 0
- minus(x,y): 0
- n__from(x): 0
- n__zWquot(x,y): 2y + 2x + 2
- nil: 0
- quot(x,y): 0
- s(x): 0
- sel(x,y): 0
- zWquot(x,y): 0
- zWquot#(x,y): y + 2x
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__zWquot(X1, X2)) | → | zWquot#(X1, X2) |
Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | sel(0, cons(X, XS)) | → | X |
| sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) | | minus(X, 0) | → | 0 |
| 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))) | | zWquot(XS, nil) | → | nil |
| zWquot(nil, XS) | → | nil | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| from(X) | → | n__from(X) | | zWquot(X1, X2) | → | n__zWquot(X1, X2) |
| activate(n__from(X)) | → | from(X) | | activate(n__zWquot(X1, X2)) | → | zWquot(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, n__zWquot, minus, 0, n__from, s, zWquot, from, quot, sel, nil, cons
Strategy
There are no SCCs!