Termination w.r.t. Q proof of /tmp/tmpafsTZ8/PALINDROME_nosorts_noand

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

↳ QTRS

  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(U12(X)) → MARK(X)
A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))
MARK(isNePal(X)) → MARK(X)
A____(__(X, Y), Z) → MARK(Z)
A____(__(X, Y), Z) → A____(mark(Y), mark(Z))
MARK(__(X1, X2)) → MARK(X1)
A____(__(X, Y), Z) → MARK(X)
A____(X, nil) → MARK(X)
A____(nil, X) → MARK(X)
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
A____(__(X, Y), Z) → MARK(Y)
MARK(__(X1, X2)) → MARK(X2)
MARK(U11(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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(U12(X)) → MARK(X)
MARK(isNePal(X)) → MARK(X)
MARK(U11(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(a____(x₁, x₂)) = x_1 + x_2
POL(A____(x₁, x₂)) = (4)x_1 + (4)x_2
POL(__(x₁, x₂)) = x_1 + x_2
POL(a__isNePal(x₁)) = 4 + x_1
POL(mark(x₁)) = x_1
POL(isNePal(x₁)) = 4 + x_1
POL(a__U12(x₁)) = 4 + (4)x_1
POL(MARK(x₁)) = (4)x_1
POL(tt) = 0
POL(a__U11(x₁)) = 4 + x_1
POL(U12(x₁)) = 4 + (4)x_1
POL(U11(x₁)) = 4 + x_1
POL(nil) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A____(__(X, Y), Z) → A____(mark(X), a____(mark(Y), mark(Z)))
MARK(__(X1, X2)) → MARK(X1)
A____(__(X, Y), Z) → MARK(X)
A____(__(X, Y), Z) → MARK(Z)
A____(__(X, Y), Z) → A____(mark(Y), mark(Z))
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
A____(__(X, Y), Z) → MARK(Y)
MARK(__(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.

A____(nil, X) → MARK(X)
A____(X, nil) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(a____(x₁, x₂)) = 4 + (4)x_1 + x_2
POL(A____(x₁, x₂)) = (4)x_1 + x_2
POL(__(x₁, x₂)) = 4 + (4)x_1 + x_2
POL(a__isNePal(x₁)) = 0
POL(mark(x₁)) = x_1
POL(isNePal(x₁)) = 0
POL(a__U12(x₁)) = 0
POL(MARK(x₁)) = x_1
POL(tt) = 0
POL(a__U11(x₁)) = 0
POL(U12(x₁)) = 0
POL(U11(x₁)) = 0
POL(nil) = 0

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.