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

(0) Obligation:

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

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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(a))) → C(f(g(f(a))))
ACTIVE(f(f(a))) → F(g(f(a)))
ACTIVE(f(f(a))) → G(f(a))
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
F(mark(X)) → F(X)
G(mark(X)) → G(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)
F(ok(X)) → F(X)
C(ok(X)) → C(X)
G(ok(X)) → G(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(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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 6 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) 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(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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.


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)  =  x1
a  =  a
mark(x1)  =  mark(x1)
c(x1)  =  x1
g(x1)  =  g(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[active1, top] > [ok1, mark1, g1]
a > [ok1, mark1, g1]
proper1 > [ok1, mark1, g1]


The following usable rules [FROCOS05] were oriented:

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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)  =  G(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
f(x1)  =  f(x1)
a  =  a
c(x1)  =  c(x1)
g(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
top > [ok1, active1, f1, c1, proper1] > a


The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
G1 > ok
[active1, top] > mark1 > ok
[active1, top] > a > ok
[active1, top] > c1 > ok


The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

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

Lexicographic Path Order [LPO].
Precedence:
top > [ok1, active1, f1, c1, proper1] > a


The following usable rules [FROCOS05] were oriented:

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

(19) Obligation:

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  F(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  x1
a  =  a
c(x1)  =  c(x1)
g(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
F1 > ok
[active1, top] > mark1 > ok
[active1, top] > a > ok
[active1, top] > c1 > ok


The following usable rules [FROCOS05] were oriented:

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

(21) Obligation:

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

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

(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:

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

The TRS R consists of the following rules:

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[proper1, top] > ok > [f1, active1] > a


The following usable rules [FROCOS05] were oriented:

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

(26) Obligation:

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

PROPER(c(X)) → PROPER(X)
PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

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

(27) 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)
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)
g(x1)  =  x1
active(x1)  =  active(x1)
f(x1)  =  f
a  =  a
mark(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[c1, active1, f, proper1, top] > a


The following usable rules [FROCOS05] were oriented:

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

(28) Obligation:

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

PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

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

(29) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
g(x1)  =  g(x1)
active(x1)  =  active(x1)
f(x1)  =  x1
a  =  a
mark(x1)  =  x1
c(x1)  =  c
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
PROPER1 > [g1, ok]
active1 > a > [g1, ok]
active1 > c > [g1, ok]
proper1 > a > [g1, ok]
proper1 > c > [g1, ok]
top > [g1, ok]


The following usable rules [FROCOS05] were oriented:

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

(30) Obligation:

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

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  ACTIVE(x1)
g(x1)  =  x1
f(x1)  =  f(x1)
active(x1)  =  active(x1)
a  =  a
mark(x1)  =  mark(x1)
c(x1)  =  c(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[active1, proper1, top] > [f1, mark1] > a
[active1, proper1, top] > [f1, mark1] > c1


The following usable rules [FROCOS05] were oriented:

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

(35) Obligation:

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

ACTIVE(g(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
ACTIVE1 > [g1, ok]
active1 > a > [g1, ok]
active1 > c > [g1, ok]
proper1 > a > [g1, ok]
proper1 > c > [g1, ok]
top > [g1, ok]


The following usable rules [FROCOS05] were oriented:

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

(37) Obligation:

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

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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.

(41) 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
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  proper(x1)
f(x1)  =  x1
a  =  a
c(x1)  =  c
g(x1)  =  x1
top(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
a > mark1 > proper1
a > c > proper1


The following usable rules [FROCOS05] were oriented:

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

(42) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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)  =  x1
ok(x1)  =  ok
active(x1)  =  active
f(x1)  =  x1
a  =  a
mark(x1)  =  x1
c(x1)  =  c(x1)
g(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[ok, proper1] > [active, top] > a > c1


The following usable rules [FROCOS05] were oriented:

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

(44) Obligation:

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

active(f(f(a))) → mark(c(f(g(f(a)))))
active(f(X)) → f(active(X))
active(g(X)) → g(active(X))
f(mark(X)) → mark(f(X))
g(mark(X)) → mark(g(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(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) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE