YES
The TRS could be proven terminating. The proof took 24 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(a, f(f(a, x), a)) | → | f#(a, x) | | f#(a, f(f(a, x), a)) | → | f#(a, f(a, x)) |
| f#(a, f(f(a, x), a)) | → | f#(f(a, f(a, x)), a) |
Rewrite Rules
| f(a, f(f(a, x), a)) | → | f(f(a, f(a, x)), a) |
Original Signature
Termination of terms over the following signature is verified: f, a
Strategy
The following SCCs where found
| f#(a, f(f(a, x), a)) → f#(a, x) | f#(a, f(f(a, x), a)) → f#(a, f(a, x)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(a, f(f(a, x), a)) | → | f#(a, x) | | f#(a, f(f(a, x), a)) | → | f#(a, f(a, x)) |
Rewrite Rules
| f(a, f(f(a, x), a)) | → | f(f(a, f(a, x)), a) |
Original Signature
Termination of terms over the following signature is verified: f, a
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(a, f(f(a, x), a)) | → | f#(a, x) | | f#(a, f(f(a, x), a)) | → | f#(a, f(a, x)) |