Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx4_7_15_Bor03

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.

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 contains 7 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P₁(ok₁(X)) -> P₁(X)

Used argument filtering: P₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P₁(mark₁(X)) -> P₁(X)

Used argument filtering: P₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: CONS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₁(ok₁(X)) -> F₁(X)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₁(mark₁(X)) -> F₁(X)

Used argument filtering: F₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
f₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
p₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(f₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(p₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
f₁(x1) = x1
s₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(f₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
f₁(x1) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = x1
cons₂(x1, x2) = x1
s₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = x1
cons₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(f₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
f₁(x1) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ 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.