YES
The TRS could be proven terminating. The proof took 661 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (16ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (609ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(g(x), g(y)) | → | p#(f(g(x), s(y))) | | f#(g(x), g(y)) | → | f#(g(x), s(y)) |
| f#(g(x), g(y)) | → | f#(p(f(g(x), s(y))), g(s(p(x)))) | | f#(g(x), g(y)) | → | g#(s(p(x))) |
| f#(g(x), g(y)) | → | g#(x) | | g#(s(p(x))) | → | p#(x) |
| p#(0) | → | g#(0) | | f#(g(x), g(y)) | → | p#(x) |
Rewrite Rules
| f(g(x), g(y)) | → | f(p(f(g(x), s(y))), g(s(p(x)))) | | p(0) | → | g(0) |
| g(s(p(x))) | → | p(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, p
Strategy
The following SCCs where found
| f#(g(x), g(y)) → f#(p(f(g(x), s(y))), g(s(p(x)))) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(g(x), g(y)) | → | f#(p(f(g(x), s(y))), g(s(p(x)))) |
Rewrite Rules
| f(g(x), g(y)) | → | f(p(f(g(x), s(y))), g(s(p(x)))) | | p(0) | → | g(0) |
| g(s(p(x))) | → | p(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, p
Strategy
Polynomial Interpretation
- 0: 1
- f(x,y): 0
- f#(x,y): 2x
- g(x): 2x + 1
- p(x): 3x
- s(x): x + 1
Improved Usable rules
| p(0) | → | g(0) | | g(s(p(x))) | → | p(x) |
| f(g(x), g(y)) | → | f(p(f(g(x), s(y))), g(s(p(x)))) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(g(x), g(y)) | → | f#(p(f(g(x), s(y))), g(s(p(x)))) |