Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

OLD(free(x)) → OLD(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
OLD(x1)  =  x1
free(x1)  =  free(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

NEW(free(x)) → NEW(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
NEW(x1)  =  x1
free(x1)  =  free(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

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

CHECK(new(x)) → CHECK(x)
CHECK(free(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CHECK(old(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  x1
free(x1)  =  x1
old(x1)  =  old(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

CHECK(new(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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CHECK(free(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  x1
free(x1)  =  free(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CHECK(new(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
new(x1)  =  new(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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.

(22) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(23) TRUE

(24) Obligation:

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

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