Termination w.r.t. Q of the given QTRS could be proven:

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 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 PisEmptyProof (⇔)
    ↳36 TRUE
  ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 PisEmptyProof (⇔)
    ↳41 TRUE
  ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 PisEmptyProof (⇔)
    ↳48 TRUE

(0) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(f(f(X))) → C(f(g(f(X))))
ACTIVE(f(f(X))) → F(g(f(X)))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(c(X)) → D(X)
ACTIVE(h(X)) → C(d(X))
ACTIVE(h(X)) → D(X)
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(h(X)) → H(active(X))
ACTIVE(h(X)) → ACTIVE(X)
F(mark(X)) → F(X)
H(mark(X)) → H(X)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(c(X)) → C(proper(X))
PROPER(c(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → D(proper(X))
PROPER(d(X)) → PROPER(X)
PROPER(h(X)) → H(proper(X))
PROPER(h(X)) → PROPER(X)
F(ok(X)) → F(X)
C(ok(X)) → C(X)
G(ok(X)) → G(X)
D(ok(X)) → D(X)
H(ok(X)) → H(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

D(ok(X)) → D(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


D(ok(X)) → D(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
mark(x1)  =  mark
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[ok1, active1, f1, h1] > [mark, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(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:

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


G(ok(X)) → G(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
mark(x1)  =  mark
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[ok1, active1, f1, h1] > [mark, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(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:

C(ok(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


C(ok(X)) → C(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
C(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
mark(x1)  =  mark
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[ok1, active1, f1, h1] > [mark, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
H(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
proper(x1)  =  x1
top(x1)  =  top(x1)

Recursive Path Order [RPO].
Precedence:
top1 > [ok1, active1, f1, h1] > mark1


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

F(ok(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(ok(X)) → F(X)
F(mark(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
proper(x1)  =  x1
top(x1)  =  top(x1)

Recursive Path Order [RPO].
Precedence:
top1 > [ok1, active1, f1, h1] > mark1


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(27) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

PROPER(c(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(c(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
c(x1)  =  c(x1)
f(x1)  =  f(x1)
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)
active(x1)  =  x1
mark(x1)  =  mark
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
f1 > [c1, mark, top]
h1 > [c1, mark, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(32) Obligation:

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

PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
g(x1)  =  g(x1)
d(x1)  =  d(x1)
active(x1)  =  active(x1)
f(x1)  =  f
mark(x1)  =  mark
c(x1)  =  c(x1)
h(x1)  =  h(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[f, ok] > [active1, h1] > [g1, d1, mark, c1, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(34) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
h(x1)  =  h(x1)
f(x1)  =  f(x1)
active(x1)  =  x1
mark(x1)  =  mark(x1)
c(x1)  =  c(x1)
g(x1)  =  g
d(x1)  =  d
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Recursive Path Order [RPO].
Precedence:
h1 > [mark1, c1, g, d]
f1 > [mark1, c1, g, d]
top1 > [mark1, c1, g, d]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  ok
active(x1)  =  active
mark(x1)  =  mark
proper(x1)  =  proper
f(x1)  =  x1
c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  x1
top(x1)  =  top(x1)

Recursive Path Order [RPO].
Precedence:
ok > active > [TOP1, mark, proper, top1]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
active(x1)  =  active(x1)
f(x1)  =  f(x1)
c(x1)  =  c(x1)
g(x1)  =  g
d(x1)  =  x1
h(x1)  =  h(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > f1 > [mark1, c1, g, top]
active1 > h1 > [mark1, c1, g, top]


The following usable rules [FROCOS05] were oriented:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

(47) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE