Termination w.r.t. Q proof of /tmp/tmp3FkImQ/Ex14_Luc06

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:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(h(X)) → H(active(X))
H(ok(X)) → H(X)
ACTIVE(h(X)) → G(X, X)
G(mark(X1), X2) → G(X1, X2)
ACTIVE(g(X1, X2)) → G(active(X1), X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
F(ok(X1), ok(X2)) → F(X1, X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(h(X)) → H(proper(X))
ACTIVE(f(X, X)) → H(a)
PROPER(g(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X2)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → ACTIVE(X)
F(mark(X1), X2) → F(X1, X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
PROPER(g(X1, X2)) → PROPER(X1)
TOP(ok(X)) → TOP(active(X))
ACTIVE(g(X1, X2)) → ACTIVE(X1)
G(ok(X1), ok(X2)) → G(X1, X2)
PROPER(h(X)) → PROPER(X)
PROPER(g(X1, X2)) → G(proper(X1), proper(X2))
TOP(mark(X)) → TOP(proper(X))
H(mark(X)) → H(X)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(h(X)) → H(active(X))
H(ok(X)) → H(X)
ACTIVE(h(X)) → G(X, X)
G(mark(X1), X2) → G(X1, X2)
ACTIVE(g(X1, X2)) → G(active(X1), X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
F(ok(X1), ok(X2)) → F(X1, X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(h(X)) → H(proper(X))
ACTIVE(f(X, X)) → H(a)
PROPER(g(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X2)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → ACTIVE(X)
F(mark(X1), X2) → F(X1, X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
PROPER(g(X1, X2)) → PROPER(X1)
TOP(ok(X)) → TOP(active(X))
ACTIVE(g(X1, X2)) → ACTIVE(X1)
G(ok(X1), ok(X2)) → G(X1, X2)
PROPER(h(X)) → PROPER(X)
PROPER(g(X1, X2)) → G(proper(X1), proper(X2))
TOP(mark(X)) → TOP(proper(X))
H(mark(X)) → H(X)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 11 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(X1), X2) → F(X1, X2)
F(ok(X1), ok(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(X1), X2) → F(X1, X2)
F(ok(X1), ok(X2)) → F(X1, X2)

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(X1), X2) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
F(ok(X1), ok(X2)) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(mark(X1), X2) → G(X1, X2)
G(ok(X1), ok(X2)) → G(X1, X2)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

G(mark(X1), X2) → G(X1, X2)
G(ok(X1), ok(X2)) → G(X1, X2)

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

G(mark(X1), X2) → G(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
G(ok(X1), ok(X2)) → G(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

H(ok(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

H(ok(X)) → H(X)
H(mark(X)) → H(X)

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

H(ok(X)) → H(X)
The graph contains the following edges 1 > 1
H(mark(X)) → H(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:

PROPER(g(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X2)
PROPER(g(X1, X2)) → PROPER(X1)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(h(X)) → PROPER(X)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(g(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X2)
PROPER(g(X1, X2)) → PROPER(X1)
PROPER(h(X)) → PROPER(X)
PROPER(f(X1, X2)) → PROPER(X1)

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

PROPER(g(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(f(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(g(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(f(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(h(X)) → PROPER(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(g(X1, X2)) → ACTIVE(X1)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(g(X1, X2)) → ACTIVE(X1)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

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

ACTIVE(g(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(h(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(f(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(TOP(x₁)) = x₁
POL(a) = 0
POL(active(x₁)) = x₁
POL(b) = 0
POL(f(x₁, x₂)) = 2·x₁ + x₂
POL(g(x₁, x₂)) = x₁ + x₂
POL(h(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
h(mark(X)) → mark(h(X))
h(ok(X)) → ok(h(X))
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:

TOP(mark(f(x0, x1))) → TOP(f(proper(x0), proper(x1)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(g(x0, x1))) → TOP(g(proper(x0), proper(x1)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(a)) → TOP(ok(a))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

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

TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(g(x0, x1))) → TOP(g(proper(x0), proper(x1)))
TOP(mark(f(x0, x1))) → TOP(f(proper(x0), proper(x1)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(a)) → TOP(ok(a))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
h(mark(X)) → mark(h(X))
h(ok(X)) → ok(h(X))
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

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

TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(g(x0, x1))) → TOP(g(proper(x0), proper(x1)))
TOP(mark(f(x0, x1))) → TOP(f(proper(x0), proper(x1)))
TOP(mark(b)) → TOP(ok(b))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
h(mark(X)) → mark(h(X))
h(ok(X)) → ok(h(X))
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:

TOP(ok(h(x0))) → TOP(h(active(x0)))
TOP(ok(g(a, x0))) → TOP(mark(f(b, x0)))
TOP(ok(f(x0, x0))) → TOP(mark(h(a)))
TOP(ok(f(x0, x1))) → TOP(f(active(x0), x1))
TOP(ok(h(x0))) → TOP(mark(g(x0, x0)))
TOP(ok(a)) → TOP(mark(b))
TOP(ok(g(x0, x1))) → TOP(g(active(x0), x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

TOP(ok(h(x0))) → TOP(h(active(x0)))
TOP(ok(g(a, x0))) → TOP(mark(f(b, x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(g(x0, x1))) → TOP(g(proper(x0), proper(x1)))
TOP(ok(f(x0, x0))) → TOP(mark(h(a)))
TOP(ok(f(x0, x1))) → TOP(f(active(x0), x1))
TOP(mark(f(x0, x1))) → TOP(f(proper(x0), proper(x1)))
TOP(ok(h(x0))) → TOP(mark(g(x0, x0)))
TOP(ok(a)) → TOP(mark(b))
TOP(ok(g(x0, x1))) → TOP(g(active(x0), x1))
TOP(mark(b)) → TOP(ok(b))

The TRS R consists of the following rules:

proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
h(mark(X)) → mark(h(X))
h(ok(X)) → ok(h(X))
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

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

TOP(ok(h(x0))) → TOP(h(active(x0)))
TOP(ok(g(a, x0))) → TOP(mark(f(b, x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(g(x0, x1))) → TOP(g(proper(x0), proper(x1)))
TOP(ok(f(x0, x0))) → TOP(mark(h(a)))
TOP(ok(f(x0, x1))) → TOP(f(active(x0), x1))
TOP(mark(f(x0, x1))) → TOP(f(proper(x0), proper(x1)))
TOP(ok(h(x0))) → TOP(mark(g(x0, x0)))
TOP(ok(g(x0, x1))) → TOP(g(active(x0), x1))

The TRS R consists of the following rules:

proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(mark(X1), X2) → mark(g(X1, X2))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
h(mark(X)) → mark(h(X))
h(ok(X)) → ok(h(X))
active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.