Termination w.r.t. Q proof of TRS/AProVELiveness8.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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(found₁(x)) -> TOP₁(active₁(x))
CHECK₁(f₁(x)) -> CHECK₁(x)
MATCH₂(X, x) -> PROPER₁(x)
ACTIVE₁(f₁(x)) -> ACTIVE₁(x)
TOP₁(mark₁(x)) -> CHECK₁(x)
ACTIVE₁(f₁(x)) -> F₁(active₁(x))
CHECK₁(x) -> MATCH₂(f₁(X), x)
TOP₁(mark₁(x)) -> TOP₁(check₁(x))
CHECK₁(f₁(x)) -> F₁(check₁(x))
F₁(mark₁(x)) -> F₁(x)
CHECK₁(x) -> F₁(X)
MATCH₂(f₁(x), f₁(y)) -> F₁(match₂(x, y))
PROPER₁(f₁(x)) -> PROPER₁(x)
TOP₁(found₁(x)) -> ACTIVE₁(x)
F₁(ok₁(x)) -> F₁(x)
CHECK₁(x) -> START₁(match₂(f₁(X), x))
TOP₁(active₁(c)) -> TOP₁(mark₁(c))
MATCH₂(f₁(x), f₁(y)) -> MATCH₂(x, y)
F₁(found₁(x)) -> F₁(x)
PROPER₁(f₁(x)) -> F₁(proper₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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(found₁(x)) -> TOP₁(active₁(x))
CHECK₁(f₁(x)) -> CHECK₁(x)
MATCH₂(X, x) -> PROPER₁(x)
ACTIVE₁(f₁(x)) -> ACTIVE₁(x)
TOP₁(mark₁(x)) -> CHECK₁(x)
ACTIVE₁(f₁(x)) -> F₁(active₁(x))
CHECK₁(x) -> MATCH₂(f₁(X), x)
TOP₁(mark₁(x)) -> TOP₁(check₁(x))
CHECK₁(f₁(x)) -> F₁(check₁(x))
F₁(mark₁(x)) -> F₁(x)
CHECK₁(x) -> F₁(X)
MATCH₂(f₁(x), f₁(y)) -> F₁(match₂(x, y))
PROPER₁(f₁(x)) -> PROPER₁(x)
TOP₁(found₁(x)) -> ACTIVE₁(x)
F₁(ok₁(x)) -> F₁(x)
CHECK₁(x) -> START₁(match₂(f₁(X), x))
TOP₁(active₁(c)) -> TOP₁(mark₁(c))
MATCH₂(f₁(x), f₁(y)) -> MATCH₂(x, y)
F₁(found₁(x)) -> F₁(x)
PROPER₁(f₁(x)) -> F₁(proper₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₁(mark₁(x)) -> F₁(x)
F₁(ok₁(x)) -> F₁(x)
F₁(found₁(x)) -> F₁(x)

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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)
F₁(found₁(x)) -> F₁(x)

The remaining pairs can at least be oriented weakly.

F₁(mark₁(x)) -> F₁(x)

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

POL( found₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁

POL( ok₁(x₁) ) = x₁ + 1

POL( F₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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 Order [17,21] with Interpretation:

POL( F₁(x₁) ) = x₁

POL( mark₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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

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(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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 Order [17,21] with Interpretation:

POL( ACTIVE₁(x₁) ) = x₁

POL( f₁(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

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

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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

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

PROPER₁(f₁(x)) -> PROPER₁(x)

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( PROPER₁(x₁) ) = x₁

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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

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

MATCH₂(f₁(x), f₁(y)) -> MATCH₂(x, y)

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.

MATCH₂(f₁(x), f₁(y)) -> MATCH₂(x, y)

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

POL( MATCH₂(x₁, x₂) ) = x₁

POL( f₁(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

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

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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

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

CHECK₁(f₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.

CHECK₁(f₁(x)) -> CHECK₁(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( CHECK₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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

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

TOP₁(found₁(x)) -> TOP₁(active₁(x))
TOP₁(active₁(c)) -> TOP₁(mark₁(c))
TOP₁(mark₁(x)) -> TOP₁(check₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.

TOP₁(active₁(c)) -> TOP₁(mark₁(c))

The remaining pairs can at least be oriented weakly.

TOP₁(found₁(x)) -> TOP₁(active₁(x))
TOP₁(mark₁(x)) -> TOP₁(check₁(x))

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

POL( found₁(x₁) ) = x₁

POL( mark₁(x₁) ) = 0

POL( active₁(x₁) ) = x₁

POL( X ) = 0

POL( f₁(x₁) ) = max{0, -1}

POL( start₁(x₁) ) = x₁

POL( c ) = 1

POL( match₂(x₁, x₂) ) = max{0, -1}

POL( TOP₁(x₁) ) = x₁

POL( ok₁(x₁) ) = x₁

POL( check₁(x₁) ) = 0

POL( proper₁(x₁) ) = 1

The following usable rules [14] were oriented:

f₁(mark₁(x)) -> mark₁(f₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
f₁(found₁(x)) -> found₁(f₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
start₁(ok₁(x)) -> found₁(x)
active₁(f₁(x)) -> mark₁(x)
check₁(x) -> start₁(match₂(f₁(X), x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

TOP₁(found₁(x)) -> TOP₁(active₁(x))
TOP₁(mark₁(x)) -> TOP₁(check₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.

TOP₁(mark₁(x)) -> TOP₁(check₁(x))

The remaining pairs can at least be oriented weakly.

TOP₁(found₁(x)) -> TOP₁(active₁(x))

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

POL( found₁(x₁) ) = x₁

POL( mark₁(x₁) ) = x₁ + 1

POL( active₁(x₁) ) = x₁

POL( X ) = 1

POL( f₁(x₁) ) = x₁ + 1

POL( start₁(x₁) ) = x₁

POL( c ) = max{0, -1}

POL( match₂(x₁, x₂) ) = x₂

POL( TOP₁(x₁) ) = x₁

POL( ok₁(x₁) ) = x₁

POL( check₁(x₁) ) = x₁

POL( proper₁(x₁) ) = x₁

The following usable rules [14] were oriented:

proper₁(c) -> ok₁(c)
f₁(mark₁(x)) -> mark₁(f₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(found₁(x)) -> found₁(f₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
start₁(ok₁(x)) -> found₁(x)
active₁(f₁(x)) -> mark₁(x)
match₂(X, x) -> proper₁(x)
check₁(x) -> start₁(match₂(f₁(X), x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

TOP₁(found₁(x)) -> TOP₁(active₁(x))

The TRS R consists of the following rules:

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.

TOP₁(found₁(x)) -> TOP₁(active₁(x))

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

POL( found₁(x₁) ) = x₁ + 1

POL( TOP₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = max{0, -1}

POL( ok₁(x₁) ) = 1

POL( active₁(x₁) ) = x₁

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented:

f₁(mark₁(x)) -> mark₁(f₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
f₁(found₁(x)) -> found₁(f₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
active₁(f₁(x)) -> mark₁(x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

active₁(f₁(x)) -> mark₁(x)
top₁(active₁(c)) -> top₁(mark₁(c))
top₁(mark₁(x)) -> top₁(check₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
start₁(ok₁(x)) -> found₁(x)
f₁(found₁(x)) -> found₁(f₁(x))
top₁(found₁(x)) -> top₁(active₁(x))
active₁(f₁(x)) -> f₁(active₁(x))
f₁(mark₁(x)) -> mark₁(f₁(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.