YES
The TRS could be proven terminating. The proof took 2515 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (143ms).
| – Problem 2 was processed with processor SubtermCriterion (6ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (187ms).
| – Problem 4 was processed with processor SubtermCriterion (2ms).
| – Problem 5 was processed with processor PolynomialLinearRange4iUR (987ms).
| | – Problem 6 was processed with processor PolynomialLinearRange4iUR (489ms).
| | | – Problem 7 was processed with processor PolynomialLinearRange4iUR (500ms).
| | | | – Problem 8 was processed with processor DependencyGraph (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__zWquot(X1, X2)) | → | activate#(X2) | | activate#(n__zWquot(X1, X2)) | → | zWquot#(activate(X1), activate(X2)) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | minus#(s(X), s(Y)) | → | minus#(X, Y) |
| activate#(n__s(X)) | → | activate#(X) | | quot#(s(X), s(Y)) | → | s#(Y) |
| activate#(n__from(X)) | → | activate#(X) | | sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
| quot#(s(X), s(Y)) | → | s#(quot(minus(X, Y), s(Y))) | | quot#(s(X), s(Y)) | → | minus#(X, Y) |
| quot#(s(X), s(Y)) | → | quot#(minus(X, Y), s(Y)) | | activate#(n__from(X)) | → | from#(activate(X)) |
| activate#(n__s(X)) | → | s#(activate(X)) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | quot#(X, Y) |
| sel#(s(N), cons(X, XS)) | → | activate#(XS) | | activate#(n__zWquot(X1, X2)) | → | activate#(X1) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, 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) |
| sel#(s(N), cons(X, XS)) → sel#(N, activate(XS)) |
| activate#(n__zWquot(X1, X2)) → zWquot#(activate(X1), activate(X2)) | activate#(n__zWquot(X1, X2)) → activate#(X2) |
| zWquot#(cons(X, XS), cons(Y, YS)) → activate#(YS) | activate#(n__s(X)) → activate#(X) |
| activate#(n__from(X)) → activate#(X) | activate#(n__zWquot(X1, X2)) → activate#(X1) |
| zWquot#(cons(X, XS), cons(Y, YS)) → activate#(XS) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(X), s(Y)) | → | minus#(X, Y) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, 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 3: 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(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, 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__s(x): 3
- n__zWquot(x,y): 0
- nil: 0
- quot(x,y): 0
- quot#(x,y): 2x + 1
- 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 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(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, 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
| activate#(n__zWquot(X1, X2)) | → | zWquot#(activate(X1), activate(X2)) | | activate#(n__zWquot(X1, X2)) | → | activate#(X2) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) | | activate#(n__zWquot(X1, X2)) | → | activate#(X1) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 1
- activate(x): x
- activate#(x): x
- cons(x,y): y
- from(x): x + 1
- minus(x,y): 3
- n__from(x): x + 1
- n__s(x): x
- n__zWquot(x,y): y + 2x
- nil: 0
- quot(x,y): 3
- s(x): x
- sel(x,y): 0
- zWquot(x,y): y + 2x
- zWquot#(x,y): y + 2x
Improved Usable rules
| zWquot(XS, nil) | → | nil | | from(X) | → | cons(X, n__from(n__s(X))) |
| s(X) | → | n__s(X) | | from(X) | → | n__from(X) |
| activate(X) | → | X | | activate(n__from(X)) | → | from(activate(X)) |
| zWquot(nil, XS) | → | nil | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__from(X)) | → | activate#(X) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__zWquot(X1, X2)) | → | activate#(X2) | | activate#(n__zWquot(X1, X2)) | → | zWquot#(activate(X1), activate(X2)) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__zWquot(X1, X2)) | → | activate#(X1) | | zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): x
- activate#(x): 2x
- cons(x,y): 2y
- from(x): 0
- minus(x,y): 1
- n__from(x): 0
- n__s(x): x + 1
- n__zWquot(x,y): y + 2x
- nil: 0
- quot(x,y): 1
- s(x): x + 1
- sel(x,y): 0
- zWquot(x,y): y + 2x
- zWquot#(x,y): 2y + x
Improved Usable rules
| zWquot(XS, nil) | → | nil | | from(X) | → | cons(X, n__from(n__s(X))) |
| s(X) | → | n__s(X) | | from(X) | → | n__from(X) |
| activate(X) | → | X | | activate(n__from(X)) | → | from(activate(X)) |
| zWquot(nil, XS) | → | nil | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__s(X)) | → | activate#(X) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__zWquot(X1, X2)) | → | zWquot#(activate(X1), activate(X2)) | | activate#(n__zWquot(X1, X2)) | → | activate#(X2) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(YS) | | activate#(n__zWquot(X1, X2)) | → | activate#(X1) |
| zWquot#(cons(X, XS), cons(Y, YS)) | → | activate#(XS) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, sel, quot, cons, nil
Strategy
Polynomial Interpretation
- 0: 1
- activate(x): x
- activate#(x): 2x
- cons(x,y): y
- from(x): 2
- minus(x,y): 1
- n__from(x): 2
- n__s(x): 0
- n__zWquot(x,y): y + x + 1
- nil: 0
- quot(x,y): 3
- s(x): 0
- sel(x,y): 0
- zWquot(x,y): y + x + 1
- zWquot#(x,y): 2y + 2x
Improved Usable rules
| zWquot(XS, nil) | → | nil | | from(X) | → | cons(X, n__from(n__s(X))) |
| s(X) | → | n__s(X) | | from(X) | → | n__from(X) |
| activate(X) | → | X | | activate(n__from(X)) | → | from(activate(X)) |
| zWquot(nil, XS) | → | nil | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) | | zWquot(cons(X, XS), cons(Y, YS)) | → | cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__zWquot(X1, X2)) | → | activate#(X2) | | activate#(n__zWquot(X1, X2)) | → | zWquot#(activate(X1), activate(X2)) |
| activate#(n__zWquot(X1, X2)) | → | activate#(X1) |
Problem 8: 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(n__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) | | s(X) | → | n__s(X) |
| zWquot(X1, X2) | → | n__zWquot(X1, X2) | | activate(n__from(X)) | → | from(activate(X)) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__zWquot(X1, X2)) | → | zWquot(activate(X1), activate(X2)) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: minus, n__from, from, n__s, activate, n__zWquot, 0, s, zWquot, sel, quot, cons, nil
Strategy
There are no SCCs!