YES
The TRS could be proven terminating. The proof took 245 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (128ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (30ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(x) | | g#(c(x, s(y))) | → | g#(c(s(x), y)) |
| f#(f(x)) | → | f#(d(f(x))) | | f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Rewrite Rules
| g(c(x, s(y))) | → | g(c(s(x), y)) | | f(c(s(x), y)) | → | f(c(x, s(y))) |
| f(f(x)) | → | f(d(f(x))) | | f(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, d, s, c
Strategy
The following SCCs where found
| g#(c(x, s(y))) → g#(c(s(x), y)) |
| f#(c(s(x), y)) → f#(c(x, s(y))) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Rewrite Rules
| g(c(x, s(y))) | → | g(c(s(x), y)) | | f(c(s(x), y)) | → | f(c(x, s(y))) |
| f(f(x)) | → | f(d(f(x))) | | f(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, d, s, c
Strategy
Polynomial Interpretation
- c(x,y): x
- d(x): 0
- f(x): 0
- f#(x): x
- g(x): 0
- s(x): x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(c(x, s(y))) | → | g#(c(s(x), y)) |
Rewrite Rules
| g(c(x, s(y))) | → | g(c(s(x), y)) | | f(c(s(x), y)) | → | f(c(x, s(y))) |
| f(f(x)) | → | f(d(f(x))) | | f(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, d, s, c
Strategy
Polynomial Interpretation
- c(x,y): 2y + x
- d(x): 0
- f(x): 0
- g(x): 0
- g#(x): x + 1
- s(x): x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| g#(c(x, s(y))) | → | g#(c(s(x), y)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| g(c(x, s(y))) | → | g(c(s(x), y)) | | f(c(s(x), y)) | → | f(c(x, s(y))) |
| f(f(x)) | → | f(d(f(x))) | | f(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, g, d, s, c
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: