(0) Obligation:

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

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
niln__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → 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:

APP(cons(X, XS), YS) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
PREFIX(L) → NIL
PREFIX(L) → PREFIX(L)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
niln__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → 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 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PREFIX(L) → PREFIX(L)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
niln__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

PREFIX(L) → PREFIX(L)

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

(8) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = PREFIX(L) evaluates to t =PREFIX(L)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from PREFIX(L) to PREFIX(L).



(9) FALSE

(10) Obligation:

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

ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
niln__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP(cons(X, XS), YS) → ACTIVATE(XS)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
    The graph contains the following edges 1 > 1, 1 > 2

  • ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
    The graph contains the following edges 1 > 1, 1 > 2

  • ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
    The graph contains the following edges 2 > 1

  • ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
    The graph contains the following edges 1 > 1

  • ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
    The graph contains the following edges 2 > 1

(14) TRUE