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

Recursive Path Order [RPO].
Precedence:
[from1, afrom1] > [AAPP2, nil, AFROM1, cons1, s1]
[zWadr2, azWadr2] > [app2, aapp2] > [AAPP2, nil, AFROM1, cons1, s1]
[zWadr2, azWadr2] > AZWADR2 > [AAPP2, nil, AFROM1, cons1, s1]
[prefix1, aprefix1] > [AAPP2, nil, AFROM1, cons1, s1]


The following usable rules [FROCOS05] were oriented:

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)

(6) 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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE