Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AProVESLASHLiveness8.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 strictly oriented and are deleted.

F₁(ok₁(x)) -> F₁(x)

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
F₁(x1) = F₁(x1)
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
found₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

[F₁, ok_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)
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 strictly oriented and are deleted.

F₁(found₁(x)) -> F₁(x)

The remaining pairs can at least by weakly be oriented.

F₁(mark₁(x)) -> F₁(x)

Used ordering: Combined order from the following AFS and order.
F₁(x1) = F₁(x1)
mark₁(x1) = x1
found₁(x1) = found₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 strictly oriented and are deleted.

F₁(mark₁(x)) -> F₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₁(x1) = F₁(x1)
mark₁(x1) = mark₁(x1)

Lexicographic Path Order [19].
Precedence:

mark₁ > F₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 strictly oriented and are deleted.

ACTIVE₁(f₁(x)) -> ACTIVE₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE₁(x1) = ACTIVE₁(x1)
f₁(x1) = f₁(x1)

Lexicographic Path Order [19].
Precedence:

f₁ > ACTIVE₁

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 strictly oriented and are deleted.

PROPER₁(f₁(x)) -> PROPER₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER₁(x1) = PROPER₁(x1)
f₁(x1) = f₁(x1)

Lexicographic Path Order [19].
Precedence:

f₁ > PROPER₁

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 strictly oriented and are deleted.

MATCH₂(f₁(x), f₁(y)) -> MATCH₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MATCH₂(x1, x2) = MATCH₁(x1)
f₁(x1) = f₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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 strictly oriented and are deleted.

CHECK₁(f₁(x)) -> CHECK₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CHECK₁(x1) = CHECK₁(x1)
f₁(x1) = f₁(x1)

Lexicographic Path Order [19].
Precedence:

f₁ > CHECK₁

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 strictly oriented and are deleted.

TOP₁(active₁(c)) -> TOP₁(mark₁(c))

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
TOP₁(x1) = TOP₁(x1)
found₁(x1) = x1
active₁(x1) = x1
c = c
mark₁(x1) = mark
check₁(x1) = check
f₁(x1) = f
ok₁(x1) = x1
start₁(x1) = x1
match₂(x1, x2) = x1
X = X
proper₁(x1) = proper

Lexicographic Path Order [19].
Precedence:

[TOP₁, c, X, proper] > [mark, check, f]

The following usable rules [14] were oriented:

active₁(f₁(x)) -> mark₁(x)
active₁(f₁(x)) -> f₁(active₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
f₁(found₁(x)) -> found₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
start₁(ok₁(x)) -> found₁(x)
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(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 strictly oriented and are deleted.

TOP₁(mark₁(x)) -> TOP₁(check₁(x))

The remaining pairs can at least by weakly be oriented.

TOP₁(found₁(x)) -> TOP₁(active₁(x))

Used ordering: Combined order from the following AFS and order.
TOP₁(x1) = TOP₁(x1)
found₁(x1) = x1
active₁(x1) = x1
mark₁(x1) = mark₁(x1)
check₁(x1) = x1
f₁(x1) = f₁(x1)
ok₁(x1) = x1
start₁(x1) = x1
match₂(x1, x2) = x2
X = X
proper₁(x1) = x1
c = c

Lexicographic Path Order [19].
Precedence:

TOP₁ > X
[mark₁, f_1] > X
c > X

The following usable rules [14] were oriented:

active₁(f₁(x)) -> mark₁(x)
active₁(f₁(x)) -> f₁(active₁(x))
f₁(ok₁(x)) -> ok₁(f₁(x))
f₁(found₁(x)) -> found₁(f₁(x))
f₁(mark₁(x)) -> mark₁(f₁(x))
check₁(f₁(x)) -> f₁(check₁(x))
check₁(x) -> start₁(match₂(f₁(X), x))
match₂(f₁(x), f₁(y)) -> f₁(match₂(x, y))
start₁(ok₁(x)) -> found₁(x)
match₂(X, x) -> proper₁(x)
proper₁(c) -> ok₁(c)
proper₁(f₁(x)) -> f₁(proper₁(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 strictly oriented and are deleted.

TOP₁(found₁(x)) -> TOP₁(active₁(x))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TOP₁(x1) = x1
found₁(x1) = found₁(x1)
active₁(x1) = x1
f₁(x1) = f₁(x1)
mark₁(x1) = x1
ok₁(x1) = ok

Lexicographic Path Order [19].
Precedence:

[f₁, ok] > found₁

The following usable rules [14] were oriented:

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