Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(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:

ACTIVE(__(__(X, Y), Z)) → MARK(__(X, __(Y, Z)))
ACTIVE(__(__(X, Y), Z)) → __1(X, __(Y, Z))
ACTIVE(__(__(X, Y), Z)) → __1(Y, Z)
ACTIVE(__(X, nil)) → MARK(X)
ACTIVE(__(nil, X)) → MARK(X)
ACTIVE(and(tt, X)) → MARK(X)
ACTIVE(isNePal(__(I, __(P, I)))) → MARK(tt)
MARK(__(X1, X2)) → ACTIVE(__(mark(X1), mark(X2)))
MARK(__(X1, X2)) → __1(mark(X1), mark(X2))
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(isNePal(X)) → ACTIVE(isNePal(mark(X)))
MARK(isNePal(X)) → ISNEPAL(mark(X))
MARK(isNePal(X)) → MARK(X)
__1(mark(X1), X2) → __1(X1, X2)
__1(X1, mark(X2)) → __1(X1, X2)
__1(active(X1), X2) → __1(X1, X2)
__1(X1, active(X2)) → __1(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
ISNEPAL(mark(X)) → ISNEPAL(X)
ISNEPAL(active(X)) → ISNEPAL(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(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 4 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ISNEPAL(active(X)) → ISNEPAL(X)
ISNEPAL(mark(X)) → ISNEPAL(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(X)) → isNePal(X)

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

(6) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(X)) → isNePal(X)

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

(7) Obligation:

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

__1(X1, mark(X2)) → __1(X1, X2)
__1(mark(X1), X2) → __1(X1, X2)
__1(active(X1), X2) → __1(X1, X2)
__1(X1, active(X2)) → __1(X1, X2)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(X)) → isNePal(X)

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

(8) Obligation:

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

MARK(__(X1, X2)) → ACTIVE(__(mark(X1), mark(X2)))
ACTIVE(__(__(X, Y), Z)) → MARK(__(X, __(Y, Z)))
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(__(X, nil)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(isNePal(X)) → ACTIVE(isNePal(mark(X)))
ACTIVE(__(nil, X)) → MARK(X)
MARK(isNePal(X)) → MARK(X)
ACTIVE(and(tt, X)) → MARK(X)

The TRS R consists of the following rules:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(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(__(X1, X2)) → ACTIVE(__(mark(X1), mark(X2)))
ACTIVE(__(__(X, Y), Z)) → MARK(__(X, __(Y, Z)))
MARK(__(X1, X2)) → MARK(X1)
MARK(__(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(__(X, nil)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(isNePal(X)) → ACTIVE(isNePal(mark(X)))
ACTIVE(__(nil, X)) → MARK(X)
MARK(isNePal(X)) → MARK(X)
ACTIVE(and(tt, 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)
__(x1, x2)  =  __(x1, x2)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
nil  =  nil
isNePal(x1)  =  isNePal(x1)
tt  =  tt
active(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
_2 > [MARK1, and2, nil, tt]
isNePal1 > [MARK1, and2, nil, tt]

Status:
MARK1: [1]
_2: [1,2]
and2: [2,1]
nil: []
isNePal1: [1]
tt: []


The following usable rules [FROCOS05] were oriented:

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(X)) → isNePal(X)

(10) Obligation:

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

active(__(__(X, Y), Z)) → mark(__(X, __(Y, Z)))
active(__(X, nil)) → mark(X)
active(__(nil, X)) → mark(X)
active(and(tt, X)) → mark(X)
active(isNePal(__(I, __(P, I)))) → mark(tt)
mark(__(X1, X2)) → active(__(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(isNePal(X)) → active(isNePal(mark(X)))
__(mark(X1), X2) → __(X1, X2)
__(X1, mark(X2)) → __(X1, X2)
__(active(X1), X2) → __(X1, X2)
__(X1, active(X2)) → __(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
isNePal(mark(X)) → isNePal(X)
isNePal(active(X)) → isNePal(X)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE