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:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:

ACTIVE1(g1(X)) -> H1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(h1(d)) -> G1(c)
PROPER1(h1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
PROPER1(h1(X)) -> H1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
H1(ok1(X)) -> H1(X)

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:

ACTIVE1(g1(X)) -> H1(X)
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(h1(d)) -> G1(c)
PROPER1(h1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
PROPER1(h1(X)) -> H1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
H1(ok1(X)) -> H1(X)

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

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

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

H1(ok1(X)) -> H1(X)

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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 oriented strictly and are deleted.


H1(ok1(X)) -> H1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( H1(x1) ) = 2x1 + 3


POL( ok1(x1) ) = 3x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

G1(ok1(X)) -> G1(X)

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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 oriented strictly and are deleted.


G1(ok1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G1(x1) ) = 2x1 + 3


POL( ok1(x1) ) = 3x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(h1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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 oriented strictly and are deleted.


PROPER1(g1(X)) -> PROPER1(X)
PROPER1(h1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( h1(x1) ) = 2x1 + 2


POL( g1(x1) ) = 3x1 + 1


POL( PROPER1(x1) ) = 2x1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(g1(X)) -> mark1(h1(X))
active1(c) -> mark1(d)
active1(h1(d)) -> mark1(g1(c))
proper1(g1(X)) -> g1(proper1(X))
proper1(h1(X)) -> h1(proper1(X))
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
g1(ok1(X)) -> ok1(g1(X))
h1(ok1(X)) -> ok1(h1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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