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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(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__and(tt, X) → mark(X)
a__isNePal(__(I, __(P, I))) → tt
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__and(X1, X2) → and(X1, X2)
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__AND(tt, X) → MARK(X)
MARK(__(X1, X2)) → A____(mark(X1), mark(X2))
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
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__and(tt, X) → mark(X)
a__isNePal(__(I, __(P, I))) → tt
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__and(X1, X2) → and(X1, X2)
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 1 less node.

(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(and(X1, X2)) → A__AND(mark(X1), X2)
A__AND(tt, X) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
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__and(tt, X) → mark(X)
a__isNePal(__(I, __(P, I))) → tt
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__and(X1, X2) → and(X1, X2)
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(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(and(X1, X2)) → A__AND(mark(X1), X2)
A__AND(tt, X) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
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: Combined order from the following AFS and order.
A____(x1, x2)  =  A____(x1, x2)
__(x1, x2)  =  __(x1, x2)
MARK(x1)  =  MARK(x1)
mark(x1)  =  x1
a____(x1, x2)  =  a____(x1, x2)
and(x1, x2)  =  and(x1, x2)
A__AND(x1, x2)  =  A__AND(x1, x2)
tt  =  tt
isNePal(x1)  =  isNePal(x1)
nil  =  nil
a__and(x1, x2)  =  a__and(x1, x2)
a__isNePal(x1)  =  a__isNePal(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[and2, aand2] > [MARK1, AAND2, nil]
[isNePal1, aisNePal1] > [A2, 2, a2, tt] > [MARK1, AAND2, nil]

Status:
A2: [1,2]
_2: [1,2]
MARK1: [1]
a2: [1,2]
and2: [2,1]
AAND2: [2,1]
tt: []
isNePal1: [1]
nil: []
aand2: [2,1]
aisNePal1: [1]


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__and(tt, X) → mark(X)
a__isNePal(__(I, __(P, I))) → tt
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNePal(X) → isNePal(X)

(6) 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__and(tt, X) → mark(X)
a__isNePal(__(I, __(P, I))) → tt
mark(__(X1, X2)) → a____(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNePal(X)) → a__isNePal(mark(X))
mark(nil) → nil
mark(tt) → tt
a____(X1, X2) → __(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNePal(X) → isNePal(X)

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE