YES
The TRS could be proven terminating. The proof took 468 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (310ms).
| – Problem 2 was processed with processor DependencyGraph (1ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| b#(b(a(x))) | → | b#(b(x)) | | a#(a(x)) | → | b#(b(x)) |
| b#(b(a(x))) | → | a#(b(b(x))) | | b#(b(a(x))) | → | b#(x) |
| a#(a(x)) | → | b#(x) |
Rewrite Rules
| a(a(x)) | → | b(b(x)) | | b(b(a(x))) | → | a(b(b(x))) |
Original Signature
Termination of terms over the following signature is verified: b, a
Strategy
Polynomial Interpretation
- a(x): x + 1
- a#(x): x
- b(x): x
- b#(x): x + 1
Improved Usable rules
| b(b(a(x))) | → | a(b(b(x))) | | a(a(x)) | → | b(b(x)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| b#(b(a(x))) | → | b#(b(x)) | | b#(b(a(x))) | → | a#(b(b(x))) |
| b#(b(a(x))) | → | b#(x) |
Problem 2: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| a#(a(x)) | → | b#(b(x)) | | a#(a(x)) | → | b#(x) |
Rewrite Rules
| a(a(x)) | → | b(b(x)) | | b(b(a(x))) | → | a(b(b(x))) |
Original Signature
Termination of terms over the following signature is verified: b, a
Strategy
There are no SCCs!