Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(adx1(cons2(X, Y))) -> CONS2(X, adx1(Y))
ACTIVE1(tl1(X)) -> TL1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> INCR1(Y)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
TL1(ok1(X)) -> TL1(X)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(adx1(cons2(X, Y))) -> ADX1(Y)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
ACTIVE1(incr1(cons2(X, Y))) -> S1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(nats) -> ADX1(zeros)
PROPER1(hd1(X)) -> HD1(proper1(X))
INCR1(ok1(X)) -> INCR1(X)
PROPER1(hd1(X)) -> PROPER1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(adx1(cons2(X, Y))) -> INCR1(cons2(X, adx1(Y)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> CONS2(s1(X), incr1(Y))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
PROPER1(tl1(X)) -> TL1(proper1(X))
HD1(mark1(X)) -> HD1(X)
ADX1(ok1(X)) -> ADX1(X)
ACTIVE1(hd1(X)) -> HD1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
HD1(ok1(X)) -> HD1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(adx1(X)) -> ADX1(active1(X))
TL1(mark1(X)) -> TL1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(adx1(cons2(X, Y))) -> CONS2(X, adx1(Y))
ACTIVE1(tl1(X)) -> TL1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> INCR1(Y)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
TL1(ok1(X)) -> TL1(X)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(adx1(cons2(X, Y))) -> ADX1(Y)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
ACTIVE1(incr1(cons2(X, Y))) -> S1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(nats) -> ADX1(zeros)
PROPER1(hd1(X)) -> HD1(proper1(X))
INCR1(ok1(X)) -> INCR1(X)
PROPER1(hd1(X)) -> PROPER1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(adx1(cons2(X, Y))) -> INCR1(cons2(X, adx1(Y)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> CONS2(s1(X), incr1(Y))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
PROPER1(tl1(X)) -> TL1(proper1(X))
HD1(mark1(X)) -> HD1(X)
ADX1(ok1(X)) -> ADX1(X)
ACTIVE1(hd1(X)) -> HD1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
HD1(ok1(X)) -> HD1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(adx1(X)) -> ADX1(active1(X))
TL1(mark1(X)) -> TL1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 9 SCCs with 20 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S1(ok1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


S1(ok1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S1(x1)) = 2·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS2(x1, x2)) = 2·x1 + 2·x2   
POL(ok1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TL1(ok1(X)) -> TL1(X)
TL1(mark1(X)) -> TL1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TL1(ok1(X)) -> TL1(X)
The remaining pairs can at least be oriented weakly.

TL1(mark1(X)) -> TL1(X)
Used ordering: Polynomial interpretation [21]:

POL(TL1(x1)) = x1   
POL(mark1(x1)) = 3·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TL1(mark1(X)) -> TL1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TL1(mark1(X)) -> TL1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TL1(x1)) = 2·x1   
POL(mark1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

HD1(mark1(X)) -> HD1(X)
HD1(ok1(X)) -> HD1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


HD1(mark1(X)) -> HD1(X)
The remaining pairs can at least be oriented weakly.

HD1(ok1(X)) -> HD1(X)
Used ordering: Polynomial interpretation [21]:

POL(HD1(x1)) = x1   
POL(mark1(x1)) = 1 + 2·x1   
POL(ok1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

HD1(ok1(X)) -> HD1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


HD1(ok1(X)) -> HD1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HD1(x1)) = 2·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

INCR1(ok1(X)) -> INCR1(X)
INCR1(mark1(X)) -> INCR1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INCR1(ok1(X)) -> INCR1(X)
The remaining pairs can at least be oriented weakly.

INCR1(mark1(X)) -> INCR1(X)
Used ordering: Polynomial interpretation [21]:

POL(INCR1(x1)) = x1   
POL(mark1(x1)) = 3·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

INCR1(mark1(X)) -> INCR1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INCR1(mark1(X)) -> INCR1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INCR1(x1)) = 2·x1   
POL(mark1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ADX1(ok1(X)) -> ADX1(X)
ADX1(mark1(X)) -> ADX1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADX1(ok1(X)) -> ADX1(X)
The remaining pairs can at least be oriented weakly.

ADX1(mark1(X)) -> ADX1(X)
Used ordering: Polynomial interpretation [21]:

POL(ADX1(x1)) = x1   
POL(mark1(x1)) = 3·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ADX1(mark1(X)) -> ADX1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADX1(mark1(X)) -> ADX1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADX1(x1)) = 2·x1   
POL(mark1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(hd1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(hd1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(cons2(x1, x2)) = 3·x1 + 3·x2   
POL(hd1(x1)) = 1 + 2·x1   
POL(incr1(x1)) = 3·x1   
POL(s1(x1)) = 3·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(s1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(cons2(x1, x2)) = 3·x1 + 3·x2   
POL(incr1(x1)) = 3·x1   
POL(s1(x1)) = 1 + 2·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
The remaining pairs can at least be oriented weakly.

PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(cons2(x1, x2)) = 1 + 2·x1 + 3·x2   
POL(incr1(x1)) = 3·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(incr1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(incr1(x1)) = 1 + 2·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(adx1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(tl1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(adx1(x1)) = 1 + 2·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(tl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(tl1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(tl1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(hd1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(tl1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(hd1(x1)) = 1 + 2·x1   
POL(incr1(x1)) = 3·x1   
POL(tl1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(tl1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(tl1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(adx1(x1)) = 3·x1   
POL(incr1(x1)) = 3·x1   
POL(tl1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(adx1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(incr1(X)) -> ACTIVE1(X)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(adx1(x1)) = 1 + 2·x1   
POL(incr1(x1)) = 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(incr1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(incr1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(incr1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof
          ↳ QDP

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

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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