YES
The TRS could be proven terminating. The proof took 27 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (13ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| :#(z, +(x, f(y))) | → | :#(g(z, y), +(x, a)) | | :#(:(x, y), z) | → | :#(x, :(y, z)) |
| :#(:(x, y), z) | → | :#(y, z) | | :#(+(x, y), z) | → | :#(y, z) |
| :#(+(x, y), z) | → | :#(x, z) |
Rewrite Rules
| :(:(x, y), z) | → | :(x, :(y, z)) | | :(+(x, y), z) | → | +(:(x, z), :(y, z)) |
| :(z, +(x, f(y))) | → | :(g(z, y), +(x, a)) |
Original Signature
Termination of terms over the following signature is verified: f, g, a, :, +
Strategy
The following SCCs where found
| :#(:(x, y), z) → :#(x, :(y, z)) | :#(:(x, y), z) → :#(y, z) |
| :#(+(x, y), z) → :#(y, z) | :#(+(x, y), z) → :#(x, z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| :#(:(x, y), z) | → | :#(x, :(y, z)) | | :#(:(x, y), z) | → | :#(y, z) |
| :#(+(x, y), z) | → | :#(y, z) | | :#(+(x, y), z) | → | :#(x, z) |
Rewrite Rules
| :(:(x, y), z) | → | :(x, :(y, z)) | | :(+(x, y), z) | → | +(:(x, z), :(y, z)) |
| :(z, +(x, f(y))) | → | :(g(z, y), +(x, a)) |
Original Signature
Termination of terms over the following signature is verified: f, g, a, :, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| :#(:(x, y), z) | → | :#(x, :(y, z)) | | :#(:(x, y), z) | → | :#(y, z) |
| :#(+(x, y), z) | → | :#(y, z) | | :#(+(x, y), z) | → | :#(x, z) |