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 (6ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| +#(0, s(y)) | → | s#(y) | | +#(x, s(y)) | → | +#(x, y) |
| s#(+(0, y)) | → | s#(y) | | +#(x, s(y)) | → | s#(+(x, y)) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, s(y)) | → | s(y) | | s(+(0, y)) | → | s(y) |
Original Signature
Termination of terms over the following signature is verified: 0, s, +
Strategy
The following SCCs where found
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, s(y)) | → | s(y) | | s(+(0, y)) | → | s(y) |
Original Signature
Termination of terms over the following signature is verified: 0, s, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, s(y)) | → | s(y) | | s(+(0, y)) | → | s(y) |
Original Signature
Termination of terms over the following signature is verified: 0, s, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: