YES
The TRS could be proven terminating. The proof took 20 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(+(x, s(0))) | → | f#(x) | | f#(+(x, y)) | → | f#(x) |
| f#(s(0)) | → | f#(0) | | f#(+(x, y)) | → | f#(y) |
Rewrite Rules
| f(0) | → | s(0) | | f(s(0)) | → | s(s(0)) |
| f(s(0)) | → | *(s(s(0)), f(0)) | | f(+(x, s(0))) | → | +(s(s(0)), f(x)) |
| f(+(x, y)) | → | *(f(x), f(y)) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, *, +
Strategy
The following SCCs where found
| f#(+(x, s(0))) → f#(x) | f#(+(x, y)) → f#(x) |
| f#(+(x, y)) → f#(y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(+(x, s(0))) | → | f#(x) | | f#(+(x, y)) | → | f#(x) |
| f#(+(x, y)) | → | f#(y) |
Rewrite Rules
| f(0) | → | s(0) | | f(s(0)) | → | s(s(0)) |
| f(s(0)) | → | *(s(s(0)), f(0)) | | f(+(x, s(0))) | → | +(s(s(0)), f(x)) |
| f(+(x, y)) | → | *(f(x), f(y)) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, *, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(+(x, y)) | → | f#(x) | | f#(+(x, s(0))) | → | f#(x) |
| f#(+(x, y)) | → | f#(y) |