YES
The TRS could be proven terminating. The proof took 158 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor PolynomialLinearRange4 (119ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |
| U11#(tt) | → | U12#(tt) | | isNePal#(__(I, __(P, I))) | → | U11#(tt) |
Rewrite Rules
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | | __(X, nil) | → | X |
| __(nil, X) | → | X | | U11(tt) | → | U12(tt) |
| U12(tt) | → | tt | | isNePal(__(I, __(P, I))) | → | U11(tt) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, U11, U12, nil
Strategy
Context-sensitive strategy:
μ(T) = μ(tt) = μ(nil) = ∅
μ(U11#) = μ(U12#) = μ(isNePal#) = μ(isNePal) = μ(U11) = μ(U12) = {1}
μ(__#) = μ(__) = {1, 2}
The following SCCs where found
| __#(__(X, Y), Z) → __#(X, __(Y, Z)) | __#(__(X, Y), Z) → __#(Y, Z) |
Problem 2: PolynomialLinearRange4
Dependency Pair Problem
Dependency Pairs
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |
Rewrite Rules
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | | __(X, nil) | → | X |
| __(nil, X) | → | X | | U11(tt) | → | U12(tt) |
| U12(tt) | → | tt | | isNePal(__(I, __(P, I))) | → | U11(tt) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, U11, U12, nil
Strategy
Context-sensitive strategy:
μ(T) = μ(tt) = μ(nil) = ∅
μ(U11#) = μ(U12#) = μ(isNePal#) = μ(isNePal) = μ(U11) = μ(U12) = {1}
μ(__#) = μ(__) = {1, 2}
Polynomial Interpretation
- U11(x): 0
- U12(x): 0
- __(x,y): y + 2x + 1
- __#(x,y): 2x
- isNePal(x): 0
- nil: 0
- tt: 0
Standard Usable rules
| __(__(X, Y), Z) | → | __(X, __(Y, Z)) | | __(X, nil) | → | X |
| __(nil, X) | → | X |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| __#(__(X, Y), Z) | → | __#(X, __(Y, Z)) | | __#(__(X, Y), Z) | → | __#(Y, Z) |