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:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(X, X)) → F(a, b)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
F(mark(X1), X2) → F(X1, X2)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(f(X1, X2)) → PROPER(X2)
F(ok(X1), ok(X2)) → F(X1, X2)
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(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(ok(X1), ok(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

F(ok(X1), ok(X2)) → F(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x2
ok(x1)  =  ok(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(mark(X1), X2) → F(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
f(x1, x2)  =  f(x1, x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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:

ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

ACTIVE(f(X1, X2)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
f(x1, x2)  =  f(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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:

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

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.