YES

The TRS could be proven terminating. The proof took 24 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (7ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).

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)XU11(tt)U12(tt)
U12(tt)ttisNePal(__(I, __(P, I)))U11(tt)

Original Signature

Termination of terms over the following signature is verified: tt, isNePal, __, U11, U12, nil

Strategy


The following SCCs where found

__#(__(X, Y), Z) → __#(X, __(Y, Z))__#(__(X, Y), Z) → __#(Y, Z)

Problem 2: SubtermCriterion



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)XU11(tt)U12(tt)
U12(tt)ttisNePal(__(I, __(P, I)))U11(tt)

Original Signature

Termination of terms over the following signature is verified: tt, isNePal, __, U11, U12, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

__#(__(X, Y), Z)__#(X, __(Y, Z))__#(__(X, Y), Z)__#(Y, Z)