YES
The TRS could be proven terminating. The proof took 19 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| and#(tt, X) | → | activate#(X) | | x#(N, s(M)) | → | plus#(x(N, M), N) |
| x#(N, s(M)) | → | x#(N, M) | | plus#(N, s(M)) | → | plus#(N, M) |
Rewrite Rules
| and(tt, X) | → | activate(X) | | plus(N, 0) | → | N |
| plus(N, s(M)) | → | s(plus(N, M)) | | x(N, 0) | → | 0 |
| x(N, s(M)) | → | plus(x(N, M), N) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, and, x
Strategy
The following SCCs where found
| plus#(N, s(M)) → plus#(N, M) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(N, s(M)) | → | plus#(N, M) |
Rewrite Rules
| and(tt, X) | → | activate(X) | | plus(N, 0) | → | N |
| plus(N, s(M)) | → | s(plus(N, M)) | | x(N, 0) | → | 0 |
| x(N, s(M)) | → | plus(x(N, M), N) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, and, x
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(N, s(M)) | → | plus#(N, M) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| and(tt, X) | → | activate(X) | | plus(N, 0) | → | N |
| plus(N, s(M)) | → | s(plus(N, M)) | | x(N, 0) | → | 0 |
| x(N, s(M)) | → | plus(x(N, M), N) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, plus, 0, s, tt, and, x
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: