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(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(a, b, X)) → F(X, X, X)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
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(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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), ok(X3)) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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), ok(X3)) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)

Recursive Path Order [RPO].
Precedence:

[F₁, ok₁, mark₁]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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:

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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.

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)

The remaining pairs can at least be oriented weakly.
Used ordering: 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:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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:

ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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.

ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)

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, x3)  =  f(x1, x2, x3)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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:

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

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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.