YES
The TRS could be proven terminating. The proof took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (221ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (2ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor PolynomialLinearRange4iUR (1324ms).
| | – Problem 7 was processed with processor PolynomialLinearRange4iUR (19ms).
| | | – Problem 8 was processed with processor PolynomialLinearRange4iUR (16ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sqr#(s(X)) | → | add#(sqr(X), dbl(X)) | | sqr#(s(X)) | → | s#(add(sqr(X), dbl(X))) |
| activate#(n__s(X)) | → | activate#(X) | | dbl#(s(X)) | → | dbl#(X) |
| half#(s(s(X))) | → | s#(half(X)) | | dbl#(s(X)) | → | s#(dbl(X)) |
| half#(s(s(X))) | → | half#(X) | | add#(s(X), Y) | → | s#(add(X, Y)) |
| terms#(N) | → | sqr#(N) | | sqr#(s(X)) | → | sqr#(X) |
| activate#(n__first(X1, X2)) | → | activate#(X2) | | activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) |
| add#(s(X), Y) | → | add#(X, Y) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__terms(X)) | → | activate#(X) |
| dbl#(s(X)) | → | s#(s(dbl(X))) | | sqr#(s(X)) | → | dbl#(X) |
| activate#(n__terms(X)) | → | terms#(activate(X)) | | activate#(n__s(X)) | → | s#(activate(X)) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| activate#(n__first(X1, X2)) → first#(activate(X1), activate(X2)) | activate#(n__first(X1, X2)) → activate#(X1) |
| activate#(n__terms(X)) → activate#(X) | first#(s(X), cons(Y, Z)) → activate#(Z) |
| activate#(n__s(X)) → activate#(X) | activate#(n__first(X1, X2)) → activate#(X2) |
| half#(s(s(X))) → half#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(X))) | → | half#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(X))) | → | half#(X) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| activate#(n__terms(X)) | → | activate#(X) | | first#(s(X), cons(Y, Z)) | → | activate#(Z) |
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__first(X1, X2)) | → | activate#(X2) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Polynomial Interpretation
- 0: 1
- activate(x): x
- activate#(x): 2x
- add(x,y): 0
- cons(x,y): y
- dbl(x): 0
- first(x,y): y + x + 1
- first#(x,y): 2y + 2x + 1
- half(x): 0
- n__first(x,y): y + x + 1
- n__s(x): x
- n__terms(x): x
- nil: 1
- recip(x): 3
- s(x): x
- sqr(x): 3
- terms(x): x
Improved Usable rules
| first(X1, X2) | → | n__first(X1, X2) | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | activate(X) | → | X |
| activate(n__terms(X)) | → | terms(activate(X)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) |
| activate(n__s(X)) | → | s(activate(X)) | | activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| activate#(n__first(X1, X2)) | → | first#(activate(X1), activate(X2)) | | activate#(n__first(X1, X2)) | → | activate#(X1) |
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__first(X1, X2)) | → | activate#(X2) |
Problem 7: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__terms(X)) | → | activate#(X) | | activate#(n__s(X)) | → | activate#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, cons, nil
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): 3x
- add(x,y): 0
- cons(x,y): 0
- dbl(x): 0
- first(x,y): 0
- half(x): 0
- n__first(x,y): 0
- n__s(x): 3x
- n__terms(x): x + 1
- nil: 0
- recip(x): 0
- s(x): 0
- sqr(x): 0
- terms(x): 0
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__terms(X)) | → | activate#(X) |
Problem 8: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(n__s(N))) | | sqr(0) | → | 0 |
| sqr(s(X)) | → | s(add(sqr(X), dbl(X))) | | dbl(0) | → | 0 |
| dbl(s(X)) | → | s(s(dbl(X))) | | add(0, X) | → | X |
| add(s(X), Y) | → | s(add(X, Y)) | | first(0, X) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | half(0) | → | 0 |
| half(s(0)) | → | 0 | | half(s(s(X))) | → | s(half(X)) |
| half(dbl(X)) | → | X | | terms(X) | → | n__terms(X) |
| s(X) | → | n__s(X) | | first(X1, X2) | → | n__first(X1, X2) |
| activate(n__terms(X)) | → | terms(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__first(X1, X2)) | → | first(activate(X1), activate(X2)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, n__s, activate, 0, s, n__first, first, nil, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- activate#(x): 2x + 1
- add(x,y): 0
- cons(x,y): 0
- dbl(x): 0
- first(x,y): 0
- half(x): 0
- n__first(x,y): 0
- n__s(x): x + 1
- n__terms(x): 0
- nil: 0
- recip(x): 0
- s(x): 0
- sqr(x): 0
- terms(x): 0
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__s(X)) | → | activate#(X) |