YES
The TRS could be proven terminating. The proof took 172 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor PolynomialLinearRange4 (131ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| and#(tt, X) | → | T(X) | | __#(__(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 | | and(tt, X) | → | X |
| isNePal(__(I, __(P, I))) | → | tt |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, nil, and
Strategy
Context-sensitive strategy:
μ(T) = μ(tt) = μ(nil) = ∅
μ(isNePal#) = μ(isNePal) = μ(and#) = μ(and) = {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 | | and(tt, X) | → | X |
| isNePal(__(I, __(P, I))) | → | tt |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, nil, and
Strategy
Context-sensitive strategy:
μ(T) = μ(tt) = μ(nil) = ∅
μ(isNePal#) = μ(isNePal) = μ(and#) = μ(and) = {1}
μ(__#) = μ(__) = {1, 2}
Polynomial Interpretation
- __(x,y): y + 2x + 1
- __#(x,y): 2x + 1
- and(x,y): 0
- 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) |