YES
The TRS could be proven terminating. The proof took 56 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| +#(s(x), y) | → | +#(x, y) | | -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| -(0, y) | → | 0 | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) |
Original Signature
Termination of terms over the following signature is verified: 0, s, +, -
Strategy
The following SCCs where found
| -#(s(x), s(y)) → -#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| -(0, y) | → | 0 | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, 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
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| -(0, y) | → | 0 | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, 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:
| -#(s(x), s(y)) | → | -#(x, y) |