YES
The TRS could be proven terminating. The proof took 356 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (136ms).
| – Problem 2 was processed with processor DependencyGraph (2ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (37ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(c, c) | → | f#(a, a) | | f#(a, b) | → | f#(s(a), c) |
| f#(a, a) | → | f#(a, b) | | f#(s(X), c) | → | f#(X, c) |
Rewrite Rules
| f(a, a) | → | f(a, b) | | f(a, b) | → | f(s(a), c) |
| f(s(X), c) | → | f(X, c) | | f(c, c) | → | f(a, a) |
Original Signature
Termination of terms over the following signature is verified: f, b, s, c, a
Strategy
Polynomial Interpretation
- a: 0
- b: 3
- c: 1
- f(x,y): 0
- f#(x,y): x + 1
- s(x): x
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 2: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(a, b) | → | f#(s(a), c) | | f#(a, a) | → | f#(a, b) |
| f#(s(X), c) | → | f#(X, c) |
Rewrite Rules
| f(a, a) | → | f(a, b) | | f(a, b) | → | f(s(a), c) |
| f(s(X), c) | → | f(X, c) | | f(c, c) | → | f(a, a) |
Original Signature
Termination of terms over the following signature is verified: f, s, b, c, a
Strategy
The following SCCs where found
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(a, a) | → | f(a, b) | | f(a, b) | → | f(s(a), c) |
| f(s(X), c) | → | f(X, c) | | f(c, c) | → | f(a, a) |
Original Signature
Termination of terms over the following signature is verified: f, s, b, c, a
Strategy
Polynomial Interpretation
- a: 0
- b: 0
- c: 1
- f(x,y): 0
- f#(x,y): 2y + x
- 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: