YES
The TRS could be proven terminating. The proof took 198 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (173ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (2ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__from(X)) | → | activate#(X) |
| 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndspos#(N, activate(Z)) | | pi#(X) | → | 2ndspos#(X, from(0)) |
| 2ndspos#(s(N), cons(X, n__cons(Y, Z))) | → | activate#(Z) | | pi#(X) | → | from#(0) |
| times#(s(X), Y) | → | plus#(Y, times(X, Y)) | | times#(s(X), Y) | → | times#(X, Y) |
| activate#(n__cons(X1, X2)) | → | cons#(activate(X1), X2) | | square#(X) | → | times#(X, X) |
| activate#(n__cons(X1, X2)) | → | activate#(X1) | | 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) | → | activate#(Z) |
| from#(X) | → | cons#(X, n__from(n__s(X))) | | 2ndspos#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndsneg#(N, activate(Z)) |
| plus#(s(X), Y) | → | s#(plus(X, Y)) | | 2ndspos#(s(N), cons(X, n__cons(Y, Z))) | → | activate#(Y) |
| activate#(n__from(X)) | → | from#(activate(X)) | | plus#(s(X), Y) | → | plus#(X, Y) |
| activate#(n__s(X)) | → | s#(activate(X)) | | 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) | → | activate#(Y) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | 2ndspos(0, Z) | → | rnil |
| 2ndspos(s(N), cons(X, n__cons(Y, Z))) | → | rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) | | 2ndsneg(0, Z) | → | rnil |
| 2ndsneg(s(N), cons(X, n__cons(Y, Z))) | → | rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) | | pi(X) | → | 2ndspos(X, from(0)) |
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| times(0, Y) | → | 0 | | times(s(X), Y) | → | plus(Y, times(X, Y)) |
| square(X) | → | times(X, X) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | cons(X1, X2) | → | n__cons(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__cons(X1, X2)) | → | cons(activate(X1), X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, negrecip, posrecip, n__from, rnil, from, rcons, n__s, activate, n__cons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons
Strategy
The following SCCs where found
| plus#(s(X), Y) → plus#(X, Y) |
| activate#(n__cons(X1, X2)) → activate#(X1) | activate#(n__s(X)) → activate#(X) |
| activate#(n__from(X)) → activate#(X) |
| 2ndspos#(s(N), cons(X, n__cons(Y, Z))) → 2ndsneg#(N, activate(Z)) | 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) → 2ndspos#(N, activate(Z)) |
| times#(s(X), Y) → times#(X, Y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| times#(s(X), Y) | → | times#(X, Y) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | 2ndspos(0, Z) | → | rnil |
| 2ndspos(s(N), cons(X, n__cons(Y, Z))) | → | rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) | | 2ndsneg(0, Z) | → | rnil |
| 2ndsneg(s(N), cons(X, n__cons(Y, Z))) | → | rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) | | pi(X) | → | 2ndspos(X, from(0)) |
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| times(0, Y) | → | 0 | | times(s(X), Y) | → | plus(Y, times(X, Y)) |
| square(X) | → | times(X, X) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | cons(X1, X2) | → | n__cons(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__cons(X1, X2)) | → | cons(activate(X1), X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, negrecip, posrecip, n__from, rnil, from, rcons, n__s, activate, n__cons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| times#(s(X), Y) | → | times#(X, Y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| 2ndspos#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndsneg#(N, activate(Z)) | | 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndspos#(N, activate(Z)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | 2ndspos(0, Z) | → | rnil |
| 2ndspos(s(N), cons(X, n__cons(Y, Z))) | → | rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) | | 2ndsneg(0, Z) | → | rnil |
| 2ndsneg(s(N), cons(X, n__cons(Y, Z))) | → | rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) | | pi(X) | → | 2ndspos(X, from(0)) |
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| times(0, Y) | → | 0 | | times(s(X), Y) | → | plus(Y, times(X, Y)) |
| square(X) | → | times(X, X) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | cons(X1, X2) | → | n__cons(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__cons(X1, X2)) | → | cons(activate(X1), X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, negrecip, posrecip, n__from, rnil, from, rcons, n__s, activate, n__cons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons
Strategy
Projection
The following projection was used:
- π (2ndspos#): 1
- π (2ndsneg#): 1
Thus, the following dependency pairs are removed:
| 2ndspos#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndsneg#(N, activate(Z)) | | 2ndsneg#(s(N), cons(X, n__cons(Y, Z))) | → | 2ndspos#(N, activate(Z)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__cons(X1, X2)) | → | activate#(X1) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | 2ndspos(0, Z) | → | rnil |
| 2ndspos(s(N), cons(X, n__cons(Y, Z))) | → | rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) | | 2ndsneg(0, Z) | → | rnil |
| 2ndsneg(s(N), cons(X, n__cons(Y, Z))) | → | rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) | | pi(X) | → | 2ndspos(X, from(0)) |
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| times(0, Y) | → | 0 | | times(s(X), Y) | → | plus(Y, times(X, Y)) |
| square(X) | → | times(X, X) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | cons(X1, X2) | → | n__cons(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__cons(X1, X2)) | → | cons(activate(X1), X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, negrecip, posrecip, n__from, rnil, from, rcons, n__s, activate, n__cons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__cons(X1, X2)) | → | activate#(X1) | | activate#(n__s(X)) | → | activate#(X) |
| activate#(n__from(X)) | → | activate#(X) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(X), Y) | → | plus#(X, Y) |
Rewrite Rules
| from(X) | → | cons(X, n__from(n__s(X))) | | 2ndspos(0, Z) | → | rnil |
| 2ndspos(s(N), cons(X, n__cons(Y, Z))) | → | rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) | | 2ndsneg(0, Z) | → | rnil |
| 2ndsneg(s(N), cons(X, n__cons(Y, Z))) | → | rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) | | pi(X) | → | 2ndspos(X, from(0)) |
| plus(0, Y) | → | Y | | plus(s(X), Y) | → | s(plus(X, Y)) |
| times(0, Y) | → | 0 | | times(s(X), Y) | → | plus(Y, times(X, Y)) |
| square(X) | → | times(X, X) | | from(X) | → | n__from(X) |
| s(X) | → | n__s(X) | | cons(X1, X2) | → | n__cons(X1, X2) |
| activate(n__from(X)) | → | from(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__cons(X1, X2)) | → | cons(activate(X1), X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: plus, negrecip, posrecip, n__from, rnil, from, rcons, n__s, activate, n__cons, 2ndspos, 0, s, 2ndsneg, times, square, pi, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(s(X), Y) | → | plus#(X, Y) |