YES
The TRS could be proven terminating. The proof took 170 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| Problem 2 was processed with processor PolynomialLinearRange4iUR (135ms).
| Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
f#(a, h(x)) | → | g#(x) | | h#(g(x)) | → | h#(a) |
f#(a, h(x)) | → | f#(g(x), h(x)) | | g#(h(x)) | → | g#(x) |
f#(a, h(x)) | → | h#(x) |
Rewrite Rules
f(a, h(x)) | → | f(g(x), h(x)) | | h(g(x)) | → | h(a) |
g(h(x)) | → | g(x) | | h(h(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
The following SCCs where found
f#(a, h(x)) → f#(g(x), h(x)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
f#(a, h(x)) | → | f#(g(x), h(x)) |
Rewrite Rules
f(a, h(x)) | → | f(g(x), h(x)) | | h(g(x)) | → | h(a) |
g(h(x)) | → | g(x) | | h(h(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
Polynomial Interpretation
- a: 2
- f(x,y): 0
- f#(x,y): x + 1
- g(x): 1
- h(x): 1
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
f#(a, h(x)) | → | f#(g(x), h(x)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
f(a, h(x)) | → | f(g(x), h(x)) | | h(g(x)) | → | h(a) |
g(h(x)) | → | g(x) | | h(h(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: