YES
The TRS could be proven terminating. The proof took 67 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (14ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | sel#(s(X), cons(Y, Z)) | → | activate#(Z) |
| activate#(n__from(X)) | → | from#(X) | | activate#(n__first(X1, X2)) | → | first#(X1, X2) |
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, n__from, n__first, from, first, sel, cons, nil
Strategy
The following SCCs where found
| first#(s(X), cons(Y, Z)) → activate#(Z) | activate#(n__first(X1, X2)) → first#(X1, X2) |
| sel#(s(X), cons(Y, Z)) → sel#(X, activate(Z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__first(X1, X2)) | → | first#(X1, X2) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, n__from, n__first, from, first, sel, cons, nil
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 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Rewrite Rules
| from(X) | → | cons(X, n__from(s(X))) | | first(0, Z) | → | nil |
| first(s(X), cons(Y, Z)) | → | cons(Y, n__first(X, activate(Z))) | | sel(0, cons(X, Z)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | from(X) | → | n__from(X) |
| first(X1, X2) | → | n__first(X1, X2) | | activate(n__from(X)) | → | from(X) |
| activate(n__first(X1, X2)) | → | first(X1, X2) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, n__from, n__first, from, first, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |