YES
The TRS could be proven terminating. The proof took 575 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (163ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| k#(x, h(x), a) | → | h#(x) | | h#(g(x)) | → | f#(x) |
| h#(g(x)) | → | h#(f(x)) | | f#(a) | → | h#(a) |
| k#(f(x), y, x) | → | f#(x) |
Rewrite Rules
| f(a) | → | g(h(a)) | | h(g(x)) | → | g(h(f(x))) |
| k(x, h(x), a) | → | h(x) | | k(f(x), y, x) | → | f(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, a, k, h
Strategy
The following SCCs where found
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(a) | → | g(h(a)) | | h(g(x)) | → | g(h(f(x))) |
| k(x, h(x), a) | → | h(x) | | k(f(x), y, x) | → | f(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, a, k, h
Strategy
Polynomial Interpretation
- a: 1
- f(x): x + 1
- g(x): 2x + 2
- h(x): 2x - 2
- h#(x): 2x - 2
- k(x,y,z): -2
Improved Usable rules
| h(g(x)) | → | g(h(f(x))) | | f(a) | → | g(h(a)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: