Termination w.r.t. Q of the given QTRS could not be shown:

(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) 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.

(7) 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.

(8) 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.

(9) 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.

(10) 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.

(11) 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.