(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)
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)
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: Combined order from the following AFS and order.
A____(x1, x2)  =  A____(x1, x2)
__(x1, x2)  =  __(x1, x2)
MARK(x1)  =  x1
mark(x1)  =  x1
a____(x1, x2)  =  a____(x1, x2)
U11(x1)  =  x1
U12(x1)  =  x1
isNePal(x1)  =  x1
nil  =  nil
a__U11(x1)  =  x1
tt  =  tt
a__U12(x1)  =  x1
a__isNePal(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
nil > [A2, 2, a2, tt]

Status:
A2: [1,2]
_2: [1,2]
a2: [1,2]
nil: multiset
tt: multiset


The following usable rules [FROCOS05] were oriented:

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)

(6) Obligation:

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

MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
MARK(U11(X)) → MARK(X)
MARK(U12(X)) → 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.

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

MARK(U12(X)) → MARK(X)
MARK(U11(X)) → 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.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(U12(X)) → MARK(X)
MARK(U11(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
U12(x1)  =  U12(x1)
U11(x1)  =  U11(x1)
isNePal(x1)  =  x1
a____(x1, x2)  =  a____(x1, x2)
__(x1, x2)  =  __(x1, x2)
mark(x1)  =  x1
nil  =  nil
a__U11(x1)  =  a__U11(x1)
tt  =  tt
a__U12(x1)  =  a__U12(x1)
a__isNePal(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
MARK1 > tt
[U111, a2, 2, aU111] > [U121, aU121] > tt
nil > tt

Status:
MARK1: multiset
U121: multiset
U111: [1]
a2: [1,2]
_2: [1,2]
nil: multiset
aU111: [1]
tt: multiset
aU121: multiset


The following usable rules [FROCOS05] were oriented:

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)

(10) Obligation:

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

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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(isNePal(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
isNePal(x1)  =  isNePal(x1)
a____(x1, x2)  =  a____(x1, x2)
__(x1, x2)  =  __(x1, x2)
mark(x1)  =  x1
nil  =  nil
a__U11(x1)  =  a__U11(x1)
tt  =  tt
a__U12(x1)  =  x1
a__isNePal(x1)  =  a__isNePal(x1)
U11(x1)  =  U11(x1)
U12(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[isNePal1, aisNePal1] > [aU111, tt, U111]
[a2, 2] > [aU111, tt, U111]
nil > [aU111, tt, U111]

Status:
isNePal1: multiset
a2: [1,2]
_2: [1,2]
nil: multiset
aU111: multiset
tt: multiset
aisNePal1: multiset
U111: multiset


The following usable rules [FROCOS05] were oriented:

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)

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE