YES
The TRS could be proven terminating. The proof took 366 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (8ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (124ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (72ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(x, s(y)) | → | g#(f(x, y), 0) | | g#(f(x, y), 0) | → | g#(x, 0) |
| g#(f(x, y), 0) | → | g#(y, 0) | | g#(s(x), y) | → | g#(f(x, y), 0) |
Rewrite Rules
| g(0, f(x, x)) | → | x | | g(x, s(y)) | → | g(f(x, y), 0) |
| g(s(x), y) | → | g(f(x, y), 0) | | g(f(x, y), 0) | → | f(g(x, 0), g(y, 0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
The following SCCs where found
| g#(f(x, y), 0) → g#(x, 0) | g#(f(x, y), 0) → g#(y, 0) |
| g#(s(x), y) → g#(f(x, y), 0) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(f(x, y), 0) | → | g#(x, 0) | | g#(f(x, y), 0) | → | g#(y, 0) |
| g#(s(x), y) | → | g#(f(x, y), 0) |
Rewrite Rules
| g(0, f(x, x)) | → | x | | g(x, s(y)) | → | g(f(x, y), 0) |
| g(s(x), y) | → | g(f(x, y), 0) | | g(f(x, y), 0) | → | f(g(x, 0), g(y, 0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Polynomial Interpretation
- 0: 0
- f(x,y): y + 2x
- g(x,y): 0
- g#(x,y): 2y + 2x + 1
- s(x): 3x + 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#(s(x), y) | → | g#(f(x, y), 0) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(f(x, y), 0) | → | g#(y, 0) | | g#(f(x, y), 0) | → | g#(x, 0) |
Rewrite Rules
| g(0, f(x, x)) | → | x | | g(x, s(y)) | → | g(f(x, y), 0) |
| g(s(x), y) | → | g(f(x, y), 0) | | g(f(x, y), 0) | → | f(g(x, 0), g(y, 0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Polynomial Interpretation
- 0: 2
- f(x,y): y + 2x + 1
- g(x,y): 0
- g#(x,y): 2x + 1
- s(x): 0
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#(f(x, y), 0) | → | g#(x, 0) | | g#(f(x, y), 0) | → | g#(y, 0) |