Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(X, nil) → mark(X)
a____(nil, X) → mark(X)
a__U11(tt) → a__U12(tt)
a__U12(tt) → tt
a__isNePal(__(I, __(P, I))) → a__U11(tt)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(X) → U11(X)
a__U12(X) → U12(X)
a__isNePal(X) → isNePal(X)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))
A____(__(X, Y), Z) → MARK(X)
A____(__(X, Y), Z) → A____(mark(Y), mark(Z))
A____(__(X, Y), Z) → MARK(Y)
A____(__(X, Y), Z) → MARK(Z)
A____(X, nil) → MARK(X)
A____(nil, X) → MARK(X)
A__U11(tt) → A__U12(tt)
A__ISNEPAL(__(I, __(P, I))) → A__U11(tt)
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(U11(X)) → A__U11(mark(X))
MARK(U11(X)) → MARK(X)
MARK(U12(X)) → A__U12(mark(X))
MARK(U12(X)) → MARK(X)
MARK(isNePal(X)) → A__ISNEPAL(mark(X))
MARK(isNePal(X)) → MARK(X)

The TRS R consists of the following rules:

a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(X, nil) → mark(X)
a____(nil, X) → mark(X)
a__U11(tt) → a__U12(tt)
a__U12(tt) → tt
a__isNePal(__(I, __(P, I))) → a__U11(tt)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(X) → U11(X)
a__U12(X) → U12(X)
a__isNePal(X) → isNePal(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A____(__(X, Y), Z) → MARK(X)
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))
A____(__(X, Y), Z) → A____(mark(Y), mark(Z))
A____(__(X, Y), Z) → MARK(Y)
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(U11(X)) → MARK(X)
MARK(U12(X)) → MARK(X)
MARK(isNePal(X)) → MARK(X)
A____(__(X, Y), Z) → MARK(Z)
A____(X, nil) → MARK(X)
A____(nil, X) → MARK(X)

The TRS R consists of the following rules:

a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(X, nil) → mark(X)
a____(nil, X) → mark(X)
a__U11(tt) → a__U12(tt)
a__U12(tt) → tt
a__isNePal(__(I, __(P, I))) → a__U11(tt)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(X) → U11(X)
a__U12(X) → U12(X)
a__isNePal(X) → isNePal(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A____(__(X, Y), Z) → MARK(X)
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
A____(__(X, Y), Z) → A____(mark(Y), mark(Z))
A____(__(X, Y), Z) → MARK(Y)
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(U11(X)) → MARK(X)
MARK(U12(X)) → MARK(X)
MARK(isNePal(X)) → MARK(X)
A____(__(X, Y), Z) → MARK(Z)
A____(X, nil) → MARK(X)
A____(nil, X) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A____(x₁, x₂)) = x₁ + x₂   
POL(MARK(x₁)) = x₁   
POL(U11(x₁)) = 1 + x₁   
POL(U12(x₁)) = 1 + x₁   
POL(__(x₁, x₂)) = 1 + x₁ + x₂   
POL(a__U11(x₁)) = 1 + x₁   
POL(a__U12(x₁)) = 1 + x₁   
POL(a____(x₁, x₂)) = 1 + x₁ + x₂   
POL(a__isNePal(x₁)) = 1 + x₁   
POL(isNePal(x₁)) = 1 + x₁   
POL(mark(x₁)) = x₁   
POL(nil) = 1   
POL(tt) = 1   

The following usable rules [FROCOS05] were oriented:

a____(X, nil) → mark(X)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(nil, X) → mark(X)
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(tt) → a__U12(tt)
a__U11(X) → U11(X)
a__U12(tt) → tt
a__U12(X) → U12(X)
a__isNePal(__(I, __(P, I))) → a__U11(tt)
a__isNePal(X) → isNePal(X)

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))

The TRS R consists of the following rules:

a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(X, nil) → mark(X)
a____(nil, X) → mark(X)
a__U11(tt) → a__U12(tt)
a__U12(tt) → tt
a__isNePal(__(I, __(P, I))) → a__U11(tt)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(X) → U11(X)
a__U12(X) → U12(X)
a__isNePal(X) → isNePal(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A____(x₁, x₂)) = x₁   
POL(U11(x₁)) = 1   
POL(U12(x₁)) = x₁   
POL(__(x₁, x₂)) = 1 + x₁ + x₂   
POL(a__U11(x₁)) = 1   
POL(a__U12(x₁)) = x₁   
POL(a____(x₁, x₂)) = 1 + x₁ + x₂   
POL(a__isNePal(x₁)) = 1   
POL(isNePal(x₁)) = 1   
POL(mark(x₁)) = x₁   
POL(nil) = 0   
POL(tt) = 0   

The following usable rules [FROCOS05] were oriented:

a____(X, nil) → mark(X)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(nil, X) → mark(X)
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(tt) → a__U12(tt)
a__U11(X) → U11(X)
a__U12(tt) → tt
a__U12(X) → U12(X)
a__isNePal(__(I, __(P, I))) → a__U11(tt)
a__isNePal(X) → isNePal(X)

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a____(__(X, Y), Z) → a____(mark(X), a____(mark(Y), mark(Z)))
a____(X, nil) → mark(X)
a____(nil, X) → mark(X)
a__U11(tt) → a__U12(tt)
a__U12(tt) → tt
a__isNePal(__(I, __(P, I))) → a__U11(tt)
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(U11(X)) → a__U11(mark(X))
mark(U12(X)) → a__U12(mark(X))
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__U11(X) → U11(X)
a__U12(X) → U12(X)
a__isNePal(X) → isNePal(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE