YES
The TRS could be proven terminating. The proof took 43 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (21ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(s(X)) | → | g#(X) | | activate#(n__g(X)) | → | activate#(X) |
| sel#(s(X), cons(Y, Z)) | → | activate#(Z) | | activate#(n__f(X)) | → | f#(activate(X)) |
| activate#(n__g(X)) | → | g#(activate(X)) | | activate#(n__f(X)) | → | activate#(X) |
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Rewrite Rules
| f(X) | → | cons(X, n__f(n__g(X))) | | g(0) | → | s(0) |
| g(s(X)) | → | s(s(g(X))) | | sel(0, cons(X, Y)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | f(X) | → | n__f(X) |
| g(X) | → | n__g(X) | | activate(n__f(X)) | → | f(activate(X)) |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons, n__g
Strategy
The following SCCs where found
| activate#(n__g(X)) → activate#(X) | activate#(n__f(X)) → activate#(X) |
| sel#(s(X), cons(Y, Z)) → sel#(X, activate(Z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__g(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |
Rewrite Rules
| f(X) | → | cons(X, n__f(n__g(X))) | | g(0) | → | s(0) |
| g(s(X)) | → | s(s(g(X))) | | sel(0, cons(X, Y)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | f(X) | → | n__f(X) |
| g(X) | → | n__g(X) | | activate(n__f(X)) | → | f(activate(X)) |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons, n__g
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__g(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Rewrite Rules
| f(X) | → | cons(X, n__f(n__g(X))) | | g(0) | → | s(0) |
| g(s(X)) | → | s(s(g(X))) | | sel(0, cons(X, Y)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | f(X) | → | n__f(X) |
| g(X) | → | n__g(X) | | activate(n__f(X)) | → | f(activate(X)) |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons, n__g
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(X), cons(Y, Z)) | → | sel#(X, activate(Z)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(X) | → | cons(X, n__f(n__g(X))) | | g(0) | → | s(0) |
| g(s(X)) | → | s(s(g(X))) | | sel(0, cons(X, Y)) | → | X |
| sel(s(X), cons(Y, Z)) | → | sel(X, activate(Z)) | | f(X) | → | n__f(X) |
| g(X) | → | n__g(X) | | activate(n__f(X)) | → | f(activate(X)) |
| activate(n__g(X)) | → | g(activate(X)) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons, n__g
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: