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
            ↳ QDPOrderProof
          ↳ 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 use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(mark(X1), X2) → F(X1, X2)
F(ok(X1), ok(X2)) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(mark(x1)) = 4 + (2)x_1   
POL(ok(x1)) = 1 + (4)x_1   
POL(F(x1, x2)) = (4)x_1 + x_2   
The value of delta used in the strict ordering is 5.
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(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 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:

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))

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.


G(mark(X1), X2) → G(X1, X2)
G(ok(X1), ok(X2)) → G(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(mark(x1)) = 4 + (4)x_1   
POL(ok(x1)) = 1 + (4)x_1   
POL(G(x1, x2)) = (4)x_1 + x_2   
The value of delta used in the strict ordering is 5.
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(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 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:

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))

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.


H(ok(X)) → H(X)
H(mark(X)) → H(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(H(x1)) = (4)x_1   
POL(ok(x1)) = 4 + (4)x_1   
POL(mark(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
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
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 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:

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))

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

POL(PROPER(x1)) = (4)x_1   
POL(g(x1, x2)) = 4 + (4)x_1 + (3)x_2   
POL(f(x1, x2)) = 4 + (2)x_1 + (4)x_2   
POL(h(x1)) = 4 + (2)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
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
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 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:

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))

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(g(X1, X2)) → ACTIVE(X1)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(g(x1, x2)) = 4 + (3)x_1   
POL(f(x1, x2)) = 4 + (4)x_1   
POL(h(x1)) = 4 + (4)x_1   
POL(ACTIVE(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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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.