YES
The TRS could be proven terminating. The proof took 683 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (223ms).
| – Problem 2 was processed with processor DependencyGraph (1ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (89ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (82ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(c(x, s(y))) | → | g#(c(s(x), y)) | | f#(c(s(x), s(y))) | → | g#(c(x, y)) |
| g#(c(s(x), s(y))) | → | f#(c(x, y)) | | f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Rewrite Rules
| f(c(s(x), y)) | → | f(c(x, s(y))) | | f(c(s(x), s(y))) | → | g(c(x, y)) |
| g(c(x, s(y))) | → | g(c(s(x), y)) | | g(c(s(x), s(y))) | → | f(c(x, y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, s, c
Strategy
Polynomial Interpretation
- c(x,y): y + x
- f(x): 0
- f#(x): x + 1
- 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(s(x), s(y))) | → | f#(c(x, y)) | | f#(c(s(x), s(y))) | → | g#(c(x, y)) |
Problem 2: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(c(x, s(y))) | → | g#(c(s(x), y)) | | f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Rewrite Rules
| f(c(s(x), y)) | → | f(c(x, s(y))) | | f(c(s(x), s(y))) | → | g(c(x, y)) |
| g(c(x, s(y))) | → | g(c(s(x), y)) | | g(c(s(x), s(y))) | → | f(c(x, y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 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 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(c(s(x), y)) | → | f#(c(x, s(y))) |
Rewrite Rules
| f(c(s(x), y)) | → | f(c(x, s(y))) | | f(c(s(x), s(y))) | → | g(c(x, y)) |
| g(c(x, s(y))) | → | g(c(s(x), y)) | | g(c(s(x), s(y))) | → | f(c(x, y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, s, c
Strategy
Polynomial Interpretation
- c(x,y): 2x
- 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 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(c(x, s(y))) | → | g#(c(s(x), y)) |
Rewrite Rules
| f(c(s(x), y)) | → | f(c(x, s(y))) | | f(c(s(x), s(y))) | → | g(c(x, y)) |
| g(c(x, s(y))) | → | g(c(s(x), y)) | | g(c(s(x), s(y))) | → | f(c(x, y)) |
Original Signature
Termination of terms over the following signature is verified: f, g, s, c
Strategy
Polynomial Interpretation
- c(x,y): y
- f(x): 0
- g(x): 0
- g#(x): x + 1
- s(x): 2x + 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)) |