Termination w.r.t. Q proof of /tmp/tmpbDBTjT/PALINDROME_nosorts

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

(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(isNePal(X)) → ACTIVE(isNePal(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, x2) = __
ACTIVE(x1) = x1
mark(x1) = mark
and(x1, x2) = and
nil = nil
isNePal(x1) = isNePal
tt = tt
active(x1) = active

Recursive Path Order [RPO].
Precedence:

nil > [MARK, , and, tt] > [mark, isNePal, active]

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:
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)
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.