YES
The TRS could be proven terminating. The proof took 465 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (169ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (81ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(s(x), y) | → | f#(f(x, y), y) | | f#(s(x), y) | → | f#(x, y) |
Rewrite Rules
| f(0, y) | → | 0 | | f(s(x), y) | → | f(f(x, y), y) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Polynomial Interpretation
- 0: 1
- f(x,y): 1
- f#(x,y): 2x
- s(x): x + 1
Improved Usable rules
| f(0, y) | → | 0 | | f(s(x), y) | → | f(f(x, y), y) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(s(x), y) | → | f#(f(x, y), y) |
Rewrite Rules
| f(0, y) | → | 0 | | f(s(x), y) | → | f(f(x, y), y) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Polynomial Interpretation
- 0: 1
- f(x,y): 1
- f#(x,y): x + 1
- s(x): 2
Improved Usable rules
| f(0, y) | → | 0 | | f(s(x), y) | → | f(f(x, y), y) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(s(x), y) | → | f#(f(x, y), y) |