Termination w.r.t. Q proof of /tmp/tmp4D5Y7z/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

            ↳ UsableRulesProof

          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

F(mark(x)) → F(x)
The graph contains the following edges 1 > 1
F(found(x)) → F(x)
The graph contains the following edges 1 > 1
F(ok(x)) → F(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ACTIVE(f(x)) → ACTIVE(x)

From the DPs we obtained the following set of size-change graphs:

ACTIVE(f(x)) → ACTIVE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PROPER(f(x)) → PROPER(x)

From the DPs we obtained the following set of size-change graphs:

PROPER(f(x)) → PROPER(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

MATCH(f(x), f(y)) → MATCH(x, y)

From the DPs we obtained the following set of size-change graphs:

MATCH(f(x), f(y)) → MATCH(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

CHECK(f(x)) → CHECK(x)

From the DPs we obtained the following set of size-change graphs:

CHECK(f(x)) → CHECK(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

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

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ RuleRemovalProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

proper(c) → ok(c)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

TOP(found(x)) → TOP(active(x))

Strictly oriented rules of the TRS R:

f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
start(ok(x)) → found(x)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

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)
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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(f(x)) → f(proper(x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

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:

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)
f(mark(x)) → mark(f(x))
proper(f(x)) → f(proper(x))

TOP(mark(x)) → TOP(check(x))

Strictly oriented rules of the TRS R:

match(f(x), f(y)) → f(match(x, y))
f(mark(x)) → mark(f(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(TOP(x₁)) = x₁
POL(X) = 0
POL(check(x₁)) = 1 + 2·x₁
POL(f(x₁)) = 1 + 2·x₁
POL(mark(x₁)) = 2 + 2·x₁
POL(match(x₁, x₂)) = x₁ + x₂
POL(proper(x₁)) = x₁
POL(start(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

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

check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(X, x) → proper(x)
proper(f(x)) → f(proper(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.