Termination w.r.t. Q proof of /tmp/tmp5A5HfT/Liveness8.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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(mark(x)) → F(x)
TOP(mark(x)) → TOP(check(x))
CHECK(x) → MATCH(f(X), x)
TOP(active(c)) → TOP(mark(c))
MATCH(f(x), f(y)) → MATCH(x, y)
ACTIVE(f(x)) → ACTIVE(x)
ACTIVE(f(x)) → F(active(x))
CHECK(x) → START(match(f(X), x))
TOP(found(x)) → ACTIVE(x)
TOP(found(x)) → TOP(active(x))
F(found(x)) → F(x)
PROPER(f(x)) → F(proper(x))
CHECK(f(x)) → CHECK(x)
CHECK(x) → F(X)
TOP(mark(x)) → CHECK(x)
F(ok(x)) → F(x)
CHECK(f(x)) → F(check(x))
MATCH(X, x) → PROPER(x)
PROPER(f(x)) → PROPER(x)
MATCH(f(x), f(y)) → F(match(x, y))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

F(mark(x)) → F(x)
TOP(mark(x)) → TOP(check(x))
CHECK(x) → MATCH(f(X), x)
TOP(active(c)) → TOP(mark(c))
MATCH(f(x), f(y)) → MATCH(x, y)
ACTIVE(f(x)) → ACTIVE(x)
ACTIVE(f(x)) → F(active(x))
CHECK(x) → START(match(f(X), x))
TOP(found(x)) → ACTIVE(x)
TOP(found(x)) → TOP(active(x))
F(found(x)) → F(x)
PROPER(f(x)) → F(proper(x))
CHECK(f(x)) → CHECK(x)
CHECK(x) → F(X)
TOP(mark(x)) → CHECK(x)
F(ok(x)) → F(x)
CHECK(f(x)) → F(check(x))
MATCH(X, x) → PROPER(x)
PROPER(f(x)) → PROPER(x)
MATCH(f(x), f(y)) → F(match(x, y))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

F(mark(x)) → F(x)
F(found(x)) → F(x)
F(ok(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(mark(x)) → F(x)
F(found(x)) → F(x)
F(ok(x)) → F(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(found(x₁)) = 4 + (3)x_1
POL(mark(x₁)) = 4 + (4)x_1
POL(ok(x₁)) = 4 + (4)x_1
POL(F(x₁)) = (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

ACTIVE(f(x)) → ACTIVE(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVE(f(x)) → ACTIVE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁)) = 1 + (4)x_1
POL(ACTIVE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

PROPER(f(x)) → PROPER(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

PROPER(f(x)) → PROPER(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PROPER(x₁)) = (4)x_1
POL(f(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

MATCH(f(x), f(y)) → MATCH(x, y)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MATCH(f(x), f(y)) → MATCH(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁)) = 4 + (2)x_1
POL(MATCH(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] 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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

CHECK(f(x)) → CHECK(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

CHECK(f(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁)) = 1 + (4)x_1
POL(CHECK(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:

TOP(found(x)) → TOP(active(x))
TOP(mark(x)) → TOP(check(x))
TOP(active(c)) → TOP(mark(c))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(active(c)) → TOP(mark(c))

The remaining pairs can at least be oriented weakly.

TOP(found(x)) → TOP(active(x))
TOP(mark(x)) → TOP(check(x))

Used ordering: Polynomial interpretation [25,35]:

POL(found(x₁)) = (4)x_1
POL(active(x₁)) = (4)x_1
POL(c) = 1
POL(f(x₁)) = 0
POL(match(x₁, x₂)) = 0
POL(X) = 0
POL(check(x₁)) = 0
POL(start(x₁)) = (2)x_1
POL(mark(x₁)) = 0
POL(ok(x₁)) = (4)x_1
POL(TOP(x₁)) = (4)x_1
POL(proper(x₁)) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
f(found(x)) → found(f(x))
active(f(x)) → mark(x)
check(x) → start(match(f(X), x))
check(f(x)) → f(check(x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)
f(ok(x)) → ok(f(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:

TOP(found(x)) → TOP(active(x))
TOP(mark(x)) → TOP(check(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(found(x)) → TOP(active(x))

The remaining pairs can at least be oriented weakly.

TOP(mark(x)) → TOP(check(x))

Used ordering: Polynomial interpretation [25,35]:

POL(found(x₁)) = 2 + x_1
POL(active(x₁)) = 1 + x_1
POL(c) = 2
POL(f(x₁)) = 3 + x_1
POL(match(x₁, x₂)) = x_2
POL(X) = 0
POL(check(x₁)) = 4 + x_1
POL(start(x₁)) = 2 + x_1
POL(mark(x₁)) = 4 + x_1
POL(ok(x₁)) = x_1
POL(TOP(x₁)) = (4)x_1
POL(proper(x₁)) = x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
f(found(x)) → found(f(x))
active(f(x)) → mark(x)
check(x) → start(match(f(X), x))
check(f(x)) → f(check(x))
match(X, x) → proper(x)
match(f(x), f(y)) → f(match(x, y))
proper(f(x)) → f(proper(x))
proper(c) → ok(c)
start(ok(x)) → found(x)
f(ok(x)) → ok(f(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:

TOP(mark(x)) → TOP(check(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(mark(x)) → TOP(check(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(found(x₁)) = 0
POL(c) = 0
POL(f(x₁)) = x_1
POL(match(x₁, x₂)) = 4 + (3)x_2
POL(X) = 3
POL(check(x₁)) = 3 + (3)x_1
POL(start(x₁)) = 3
POL(mark(x₁)) = 4 + (4)x_1
POL(ok(x₁)) = 0
POL(TOP(x₁)) = (2)x_1
POL(proper(x₁)) = 3 + x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

f(mark(x)) → mark(f(x))
f(found(x)) → found(f(x))
check(x) → start(match(f(X), x))
check(f(x)) → f(check(x))
start(ok(x)) → found(x)
f(ok(x)) → ok(f(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:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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.