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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(X)) -> ACTIVE1(X)
P1(mark1(X)) -> P1(X)
ACTIVE1(f1(0)) -> CONS2(0, f1(s1(0)))
S1(mark1(X)) -> S1(X)
ACTIVE1(p1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> PROPER1(X)
ACTIVE1(f1(s1(0))) -> F1(p1(s1(0)))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
PROPER1(p1(X)) -> P1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(f1(s1(0))) -> P1(s1(0))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(f1(X)) -> F1(active1(X))
P1(ok1(X)) -> P1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(p1(X)) -> P1(active1(X))
ACTIVE1(f1(0)) -> S1(0)
F1(mark1(X)) -> F1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(f1(0)) -> F1(s1(0))
PROPER1(f1(X)) -> F1(proper1(X))
PROPER1(p1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(X)) -> ACTIVE1(X)
P1(mark1(X)) -> P1(X)
ACTIVE1(f1(0)) -> CONS2(0, f1(s1(0)))
S1(mark1(X)) -> S1(X)
ACTIVE1(p1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> PROPER1(X)
ACTIVE1(f1(s1(0))) -> F1(p1(s1(0)))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
PROPER1(p1(X)) -> P1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(f1(s1(0))) -> P1(s1(0))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(f1(X)) -> F1(active1(X))
P1(ok1(X)) -> P1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(p1(X)) -> P1(active1(X))
ACTIVE1(f1(0)) -> S1(0)
F1(mark1(X)) -> F1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(f1(0)) -> F1(s1(0))
PROPER1(f1(X)) -> F1(proper1(X))
PROPER1(p1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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 7 SCCs with 15 less nodes.

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

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

P1(mark1(X)) -> P1(X)
P1(ok1(X)) -> P1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.


P1(ok1(X)) -> P1(X)
The remaining pairs can at least be oriented weakly.

P1(mark1(X)) -> P1(X)
Used ordering: Polynomial interpretation [21]:

POL(P1(x1)) = 2·x1   
POL(mark1(x1)) = 2·x1   
POL(ok1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

P1(mark1(X)) -> P1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.


P1(mark1(X)) -> P1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(P1(x1)) = 2·x1   
POL(mark1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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

S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(mark1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.

S1(ok1(X)) -> S1(X)
Used ordering: Polynomial interpretation [21]:

POL(S1(x1)) = 2·x1   
POL(mark1(x1)) = 2 + 2·x1   
POL(ok1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ 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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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
          ↳ 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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(CONS2(x1, x2)) = 2·x2   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 2 + 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

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(mark1(X1), 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   
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

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

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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

F1(mark1(X)) -> F1(X)
F1(ok1(X)) -> F1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.


F1(ok1(X)) -> F1(X)
The remaining pairs can at least be oriented weakly.

F1(mark1(X)) -> F1(X)
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = 2·x1   
POL(mark1(x1)) = 2·x1   
POL(ok1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

F1(mark1(X)) -> F1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.


F1(mark1(X)) -> F1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(p1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(p1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2·x1 + 2·x2   
POL(f1(x1)) = 2·x1   
POL(p1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 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

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

PROPER1(f1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(f1(x1)) = 2·x1   
POL(s1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



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

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(s1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(f1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(s1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(s1(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
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(p1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(p1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2·x1   
POL(f1(x1)) = 2·x1   
POL(p1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 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

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

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(s1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2·x1   
POL(f1(x1)) = 2·x1   
POL(s1(x1)) = 2 + 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
                    ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(cons2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least be oriented weakly.

ACTIVE1(f1(X)) -> ACTIVE1(X)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2 + 2·x1   
POL(f1(x1)) = 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
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(f1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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(f1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 2·x1   
POL(f1(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
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof
          ↳ QDP

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

active1(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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

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(f1(0)) -> mark1(cons2(0, f1(s1(0))))
active1(f1(s1(0))) -> mark1(f1(p1(s1(0))))
active1(p1(s1(0))) -> mark1(0)
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(p1(X)) -> p1(active1(X))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
p1(mark1(X)) -> mark1(p1(X))
proper1(f1(X)) -> f1(proper1(X))
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(p1(X)) -> p1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
p1(ok1(X)) -> ok1(p1(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.