YES
The TRS could be proven terminating. The proof took 169 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (15ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (136ms).
| | – Problem 3 was processed with processor DependencyGraph (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| U12#(tt, M, N) | → | activate#(N) | | U12#(tt, M, N) | → | activate#(M) |
| U12#(tt, M, N) | → | plus#(activate(N), activate(M)) | | U11#(tt, M, N) | → | U12#(tt, activate(M), activate(N)) |
| U11#(tt, M, N) | → | activate#(M) | | U11#(tt, M, N) | → | activate#(N) |
| plus#(N, s(M)) | → | U11#(tt, M, N) |
Rewrite Rules
| U11(tt, M, N) | → | U12(tt, activate(M), activate(N)) | | U12(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| plus(N, 0) | → | N | | plus(N, s(M)) | → | U11(tt, M, N) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, U11, U12
Strategy
The following SCCs where found
| U12#(tt, M, N) → plus#(activate(N), activate(M)) | U11#(tt, M, N) → U12#(tt, activate(M), activate(N)) |
| plus#(N, s(M)) → U11#(tt, M, N) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| U12#(tt, M, N) | → | plus#(activate(N), activate(M)) | | U11#(tt, M, N) | → | U12#(tt, activate(M), activate(N)) |
| plus#(N, s(M)) | → | U11#(tt, M, N) |
Rewrite Rules
| U11(tt, M, N) | → | U12(tt, activate(M), activate(N)) | | U12(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| plus(N, 0) | → | N | | plus(N, s(M)) | → | U11(tt, M, N) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, U11, U12
Strategy
Polynomial Interpretation
- 0: 0
- U11(x,y,z): 0
- U11#(x,y,z): 2y + 2
- U12(x,y,z): 0
- U12#(x,y,z): 2y
- activate(x): x
- plus(x,y): 0
- plus#(x,y): 2y
- s(x): x + 1
- tt: 0
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| U11#(tt, M, N) | → | U12#(tt, activate(M), activate(N)) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| U12#(tt, M, N) | → | plus#(activate(N), activate(M)) | | plus#(N, s(M)) | → | U11#(tt, M, N) |
Rewrite Rules
| U11(tt, M, N) | → | U12(tt, activate(M), activate(N)) | | U12(tt, M, N) | → | s(plus(activate(N), activate(M))) |
| plus(N, 0) | → | N | | plus(N, s(M)) | → | U11(tt, M, N) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, U11, U12
Strategy
There are no SCCs!