Termination w.r.t. Q proof of TRS/TRCSREx4_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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( S₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = 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(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.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = x₁

POL( ok₁(x₁) ) = x₁ + 1

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( PROPER₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = x₁ + 1

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( ACTIVE₁(x₁) ) = x₁ + 1

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( p₁(x₁) ) = x₁ + 1

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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.