YES
The TRS could be proven terminating. The proof took 36 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sum#(s(x)) | → | sum#(x) | | sum1#(s(x)) | → | sum1#(x) |
Rewrite Rules
| sum(0) | → | 0 | | sum(s(x)) | → | +(sum(x), s(x)) |
| sum1(0) | → | 0 | | sum1(s(x)) | → | s(+(sum1(x), +(x, x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sum1, sum, +
Strategy
The following SCCs where found
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| sum(0) | → | 0 | | sum(s(x)) | → | +(sum(x), s(x)) |
| sum1(0) | → | 0 | | sum1(s(x)) | → | s(+(sum1(x), +(x, x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sum1, sum, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| sum(0) | → | 0 | | sum(s(x)) | → | +(sum(x), s(x)) |
| sum1(0) | → | 0 | | sum1(s(x)) | → | s(+(sum1(x), +(x, x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sum1, sum, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: