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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP

(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: Lexicographic path order with status [LPO].
Precedence:
free1 > OLD1

Status:
OLD1: [1]
free1: [1]

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: Lexicographic path order with status [LPO].
Precedence:
free1 > NEW1

Status:
NEW1: [1]
free1: [1]

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(new(x)) → CHECK(x)
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)  =  new(x1)
free(x1)  =  free(x1)
old(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
new1: [1]
free1: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

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.

(18) 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: Lexicographic path order with status [LPO].
Precedence:
old1 > CHECK1

Status:
CHECK1: [1]
old1: [1]

The following usable rules [FROCOS05] were oriented: none

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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.