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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 UsableRulesProof (⇔)
    ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 UsableRulesProof (⇔)
    ↳32 QDP
    ↳33 QDPSizeChangeProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 QDPOrderProof (⇔)
    ↳45 QDP
    ↳46 QDPOrderProof (⇔)
    ↳47 QDP
    ↳48 DependencyGraphProof (⇔)
    ↳49 TRUE

(0) Obligation:

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

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

ACTIVE(app(nil, YS)) → MARK(YS)
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
ACTIVE(app(cons(X, XS), YS)) → CONS(X, app(XS, YS))
ACTIVE(app(cons(X, XS), YS)) → APP(XS, YS)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(zWadr(nil, YS)) → MARK(nil)
ACTIVE(zWadr(XS, nil)) → MARK(nil)
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → CONS(app(Y, cons(X, nil)), zWadr(XS, YS))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → APP(Y, cons(X, nil))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → CONS(X, nil)
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → ZWADR(XS, YS)
ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
ACTIVE(prefix(L)) → CONS(nil, zWadr(L, prefix(L)))
ACTIVE(prefix(L)) → ZWADR(L, prefix(L))
MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(app(X1, X2)) → APP(mark(X1), mark(X2))
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(zWadr(X1, X2)) → ZWADR(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
MARK(prefix(X)) → PREFIX(mark(X))
MARK(prefix(X)) → MARK(X)
APP(mark(X1), X2) → APP(X1, X2)
APP(X1, mark(X2)) → APP(X1, X2)
APP(active(X1), X2) → APP(X1, X2)
APP(X1, active(X2)) → APP(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
ZWADR(mark(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, mark(X2)) → ZWADR(X1, X2)
ZWADR(active(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, active(X2)) → ZWADR(X1, X2)
PREFIX(mark(X)) → PREFIX(X)
PREFIX(active(X)) → PREFIX(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(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 7 SCCs with 20 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PREFIX(active(X)) → PREFIX(X)
PREFIX(mark(X)) → PREFIX(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(active(X)) → PREFIX(X)
PREFIX(mark(X)) → PREFIX(X)

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

(8) 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:

  • PREFIX(active(X)) → PREFIX(X)
    The graph contains the following edges 1 > 1

  • PREFIX(mark(X)) → PREFIX(X)
    The graph contains the following edges 1 > 1

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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:

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

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

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

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

(14) TRUE

(15) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(16) 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.

(17) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

(18) 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:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(19) TRUE

(20) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(21) 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.

(22) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

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

(23) 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:

  • FROM(active(X)) → FROM(X)
    The graph contains the following edges 1 > 1

  • FROM(mark(X)) → FROM(X)
    The graph contains the following edges 1 > 1

(24) TRUE

(25) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(26) 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.

(27) Obligation:

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

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

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

(28) 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:

  • CONS(X1, mark(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS(mark(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(active(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(X1, active(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(29) TRUE

(30) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(31) 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.

(32) Obligation:

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

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

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

(33) 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(X1, mark(X2)) → APP(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

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

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

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

(34) TRUE

(35) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
ACTIVE(app(nil, YS)) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
MARK(prefix(X)) → MARK(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(prefix(x1)) = x1   
POL(s(x1)) = x1   
POL(zWadr(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(37) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
ACTIVE(app(nil, YS)) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
MARK(prefix(X)) → MARK(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(38) 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: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(prefix(x1)) = x1   
POL(s(x1)) = 1 + x1   
POL(zWadr(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(39) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
ACTIVE(app(nil, YS)) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
MARK(prefix(X)) → MARK(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
MARK(prefix(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(prefix(x1)) = 1 + x1   
POL(s(x1)) = 0   
POL(zWadr(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(41) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(prefix(x1)) = 0   
POL(s(x1)) = 0   
POL(zWadr(x1, x2)) = 1 + x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(43) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
ACTIVE(app(nil, YS)) → MARK(YS)
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(app(nil, YS)) → MARK(YS)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(prefix(x1)) = 1   
POL(s(x1)) = 0   
POL(zWadr(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(45) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 1 + x1   
POL(active(x1)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 1 + x1   
POL(from(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(prefix(x1)) = 1 + x1   
POL(s(x1)) = 0   
POL(zWadr(x1, x2)) = 1 + x1 + x2   

The following usable rules [FROCOS05] were oriented:

app(active(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
app(mark(X1), X2) → app(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)
mark(nil) → active(nil)
active(from(X)) → mark(cons(X, from(s(X))))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(prefix(X)) → active(prefix(mark(X)))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
mark(s(X)) → active(s(mark(X)))
active(app(nil, YS)) → mark(YS)
mark(from(X)) → active(from(mark(X)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)

(47) Obligation:

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

ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(48) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(49) TRUE