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:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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:

TOP2(x, free(y)) → TOP1(x, check(new(y)))
TOP2(x, free(y)) → CHECK(new(y))
TOP1(free(x), y) → CHECK(new(x))
TOP2(x, free(y)) → TOP1(new(x), check(y))
TOP2(x, free(y)) → TOP1(check(x), new(y))
TOP2(x, free(y)) → CHECK(x)
TOP2(x, free(y)) → TOP1(check(new(x)), y)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → OLD(check(x))
TOP1(free(x), y) → CHECK(new(y))
TOP1(free(x), y) → CHECK(x)
TOP1(free(x), y) → NEW(x)
CHECK(new(x)) → CHECK(x)
TOP2(x, free(y)) → NEW(x)
TOP1(free(x), y) → TOP2(check(x), new(y))
TOP1(free(x), y) → TOP2(new(x), check(y))
TOP2(x, free(y)) → CHECK(new(x))
TOP1(free(x), y) → NEW(y)
TOP1(free(x), y) → CHECK(y)
TOP1(free(x), y) → TOP2(check(new(x)), y)
OLD(free(x)) → OLD(x)
CHECK(old(x)) → CHECK(x)
CHECK(new(x)) → NEW(check(x))
TOP2(x, free(y)) → NEW(y)
TOP2(x, free(y)) → CHECK(y)
TOP1(free(x), y) → TOP2(x, check(new(y)))
NEW(free(x)) → NEW(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

TOP2(x, free(y)) → TOP1(x, check(new(y)))
TOP2(x, free(y)) → CHECK(new(y))
TOP1(free(x), y) → CHECK(new(x))
TOP2(x, free(y)) → TOP1(new(x), check(y))
TOP2(x, free(y)) → TOP1(check(x), new(y))
TOP2(x, free(y)) → CHECK(x)
TOP2(x, free(y)) → TOP1(check(new(x)), y)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → OLD(check(x))
TOP1(free(x), y) → CHECK(new(y))
TOP1(free(x), y) → CHECK(x)
TOP1(free(x), y) → NEW(x)
CHECK(new(x)) → CHECK(x)
TOP2(x, free(y)) → NEW(x)
TOP1(free(x), y) → TOP2(check(x), new(y))
TOP1(free(x), y) → TOP2(new(x), check(y))
TOP2(x, free(y)) → CHECK(new(x))
TOP1(free(x), y) → NEW(y)
TOP1(free(x), y) → CHECK(y)
TOP1(free(x), y) → TOP2(check(new(x)), y)
OLD(free(x)) → OLD(x)
CHECK(old(x)) → CHECK(x)
CHECK(new(x)) → NEW(check(x))
TOP2(x, free(y)) → NEW(y)
TOP2(x, free(y)) → CHECK(y)
TOP1(free(x), y) → TOP2(x, check(new(y)))
NEW(free(x)) → NEW(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

TOP2(x, free(y)) → TOP1(x, check(new(y)))
TOP2(x, free(y)) → CHECK(new(y))
TOP1(free(x), y) → CHECK(new(x))
TOP2(x, free(y)) → CHECK(x)
TOP2(x, free(y)) → TOP1(check(x), new(y))
TOP2(x, free(y)) → TOP1(new(x), check(y))
CHECK(free(x)) → CHECK(x)
TOP2(x, free(y)) → TOP1(check(new(x)), y)
TOP1(free(x), y) → CHECK(new(y))
CHECK(old(x)) → OLD(check(x))
TOP1(free(x), y) → CHECK(x)
CHECK(new(x)) → CHECK(x)
TOP1(free(x), y) → NEW(x)
TOP2(x, free(y)) → NEW(x)
TOP1(free(x), y) → TOP2(new(x), check(y))
TOP1(free(x), y) → TOP2(check(x), new(y))
TOP2(x, free(y)) → CHECK(new(x))
TOP1(free(x), y) → CHECK(y)
TOP1(free(x), y) → NEW(y)
TOP1(free(x), y) → TOP2(check(new(x)), y)
OLD(free(x)) → OLD(x)
CHECK(new(x)) → NEW(check(x))
CHECK(old(x)) → CHECK(x)
TOP2(x, free(y)) → NEW(y)
TOP2(x, free(y)) → CHECK(y)
TOP1(free(x), y) → TOP2(x, check(new(y)))
NEW(free(x)) → NEW(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 14 less nodes.

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

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

OLD(free(x)) → OLD(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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.


OLD(free(x)) → OLD(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
OLD(x1)  =  x1
free(x1)  =  free(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

NEW(free(x)) → NEW(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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.


NEW(free(x)) → NEW(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NEW(x1)  =  x1
free(x1)  =  free(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

CHECK(new(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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.


CHECK(free(x)) → CHECK(x)
The remaining pairs can at least be oriented weakly.

CHECK(new(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  x1
old(x1)  =  x1
free(x1)  =  free(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

CHECK(new(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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.


CHECK(old(x)) → CHECK(x)
The remaining pairs can at least be oriented weakly.

CHECK(new(x)) → CHECK(x)
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  x1
old(x1)  =  old(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

CHECK(new(x)) → CHECK(x)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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.


CHECK(new(x)) → CHECK(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  new(x1)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

TOP2(x, free(y)) → TOP1(x, check(new(y)))
TOP1(free(x), y) → TOP2(check(new(x)), y)
TOP1(free(x), y) → TOP2(x, check(new(y)))
TOP1(free(x), y) → TOP2(check(x), new(y))
TOP1(free(x), y) → TOP2(new(x), check(y))
TOP2(x, free(y)) → TOP1(check(x), new(y))
TOP2(x, free(y)) → TOP1(new(x), check(y))
TOP2(x, free(y)) → TOP1(check(new(x)), y)

The TRS R consists of the following rules:

top1(free(x), y) → top2(check(new(x)), y)
top1(free(x), y) → top2(new(x), check(y))
top1(free(x), y) → top2(check(x), new(y))
top1(free(x), y) → top2(x, check(new(y)))
top2(x, free(y)) → top1(check(new(x)), y)
top2(x, free(y)) → top1(new(x), check(y))
top2(x, free(y)) → top1(check(x), new(y))
top2(x, free(y)) → top1(x, check(new(y)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

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