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

(0) Obligation:

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

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(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__APP(nil, YS) → MARK(YS)
A__APP(cons(X, XS), YS) → MARK(X)
A__FROM(X) → MARK(X)
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → A__PREFIX(mark(X))
MARK(prefix(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(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:

MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
A__APP(nil, YS) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
A__APP(cons(X, XS), YS) → MARK(X)
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(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.

MARK(app(X1, X2)) → A__APP(mark(X1), mark(X2))
A__APP(nil, YS) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(zWadr(X1, X2)) → A__ZWADR(mark(X1), mark(X2))
A__ZWADR(cons(X, XS), cons(Y, YS)) → A__APP(mark(Y), cons(mark(X), nil))
A__APP(cons(X, XS), YS) → MARK(X)
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
A__ZWADR(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__ZWADR(cons(X, XS), cons(Y, YS)) → 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)
app(x1, x2)  =  app(x1, x2)
A__APP(x1, x2)  =  A__APP(x1, x2)
mark(x1)  =  x1
nil  =  nil
from(x1)  =  x1
A__FROM(x1)  =  A__FROM(x1)
zWadr(x1, x2)  =  zWadr(x1, x2)
A__ZWADR(x1, x2)  =  A__ZWADR(x1, x2)
cons(x1, x2)  =  x1
prefix(x1)  =  x1
s(x1)  =  x1
a__from(x1)  =  x1
a__zWadr(x1, x2)  =  a__zWadr(x1, x2)
a__app(x1, x2)  =  a__app(x1, x2)
a__prefix(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[zWadr₂, A_ZWADR2, a_zWadr2] > [app₂, A_APP2, a_app2] > [MARK₁, A_FROM1] > nil

Status:

a_app2: [1,2]
MARK₁: [1]
A_APP2: [1,2]
A_FROM1: [1]
a_zWadr2: [2,1]
A_ZWADR2: [2,1]
app₂: [1,2]
zWadr₂: [2,1]
nil: []

The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
a__app(nil, YS) → mark(YS)
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__prefix(X) → prefix(X)
a__zWadr(X1, X2) → zWadr(X1, X2)

(6) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(prefix(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(prefix(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  x1
prefix(x1)  =  prefix(x1)
cons(x1, x2)  =  cons(x1)
s(x1)  =  x1
a__from(x1)  =  a__from(x1)
a__zWadr(x1, x2)  =  a__zWadr(x1, x2)
nil  =  nil
a__app(x1, x2)  =  a__app(x1, x2)
app(x1, x2)  =  app(x1, x2)
a__prefix(x1)  =  a__prefix(x1)
zWadr(x1, x2)  =  zWadr(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[from₁, a_from1] > [MARK₁, A_FROM1, prefix₁, nil, a_prefix1] > cons₁
[a_zWadr2, a_app2, app₂, zWadr₂] > [MARK₁, A_FROM1, prefix₁, nil, a_prefix1] > cons₁

Status:

from₁: [1]
cons₁: [1]
a_from1: [1]
prefix₁: [1]
a_app2: [2,1]
MARK₁: [1]
A_FROM1: [1]
a_prefix1: [1]
a_zWadr2: [1,2]
app₂: [2,1]
zWadr₂: [1,2]
nil: []

The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
a__app(nil, YS) → mark(YS)
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__prefix(X) → prefix(X)
a__zWadr(X1, X2) → zWadr(X1, X2)

(8) Obligation:

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

A__FROM(X) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(X)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(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(s(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
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

a__app(nil, YS) → mark(YS)
a__app(cons(X, XS), YS) → cons(mark(X), app(XS, YS))
a__from(X) → cons(mark(X), from(s(X)))
a__zWadr(nil, YS) → nil
a__zWadr(XS, nil) → nil
a__zWadr(cons(X, XS), cons(Y, YS)) → cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS))
a__prefix(L) → cons(nil, zWadr(L, prefix(L)))
mark(app(X1, X2)) → a__app(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(zWadr(X1, X2)) → a__zWadr(mark(X1), mark(X2))
mark(prefix(X)) → a__prefix(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__app(X1, X2) → app(X1, X2)
a__from(X) → from(X)
a__zWadr(X1, X2) → zWadr(X1, X2)
a__prefix(X) → prefix(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