YES
The TRS could be proven terminating. The proof took 178 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (49ms).
| – Problem 2 was processed with processor SubtermCriterion (4ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (21ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sqr#(s(X)) | → | add#(sqr(X), dbl(X)) | | add#(s(X), Y) | → | add#(X, Y) |
| activate#(n__terms(X)) | → | terms#(X) | | first#(s(X), cons(Y, Z)) | → | activate#(Z) |
| dbl#(s(X)) | → | dbl#(X) | | half#(s(s(X))) | → | half#(X) |
| terms#(N) | → | sqr#(N) | | sqr#(s(X)) | → | dbl#(X) |
| activate#(n__first(X1, X2)) | → | first#(X1, X2) | | sqr#(s(X)) | → | sqr#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, activate, 0, s, n__first, first, nil, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| half#(s(s(X))) → half#(X) |
| first#(s(X), cons(Y, Z)) → activate#(Z) | activate#(n__first(X1, X2)) → first#(X1, X2) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(X))) | → | half#(X) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, 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 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, 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
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, 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 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__first(X1, X2)) | → | first#(X1, X2) |
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
- π (activate#): 1
- π (first#): 2
Thus, the following dependency pairs are removed:
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__first(X1, X2)) | → | first#(X1, X2) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| terms(N) | → | cons(recip(sqr(N)), n__terms(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) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__terms(X)) | → | terms(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: terms, sqr, half, dbl, recip, n__terms, add, activate, 0, s, n__first, first, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: