(0) Obligation:

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

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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, X, X)) → MARK(f(X, b, b))
ACTIVE(f(a, X, X)) → F(X, b, b)
ACTIVE(b) → MARK(a)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
MARK(f(X1, X2, X3)) → F(X1, mark(X2), X3)
MARK(f(X1, X2, X3)) → MARK(X2)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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(active(X1), X2, X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x1)

Tags:
F has argument tags [1,0,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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, X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x1)

Tags:
F has argument tags [0,3,2,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = x2 + x3   
POL(active(x1)) = 0   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(X1, active(X2), X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x0)

Tags:
F has argument tags [0,2,2,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = x1 + x2   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(X1, mark(X2), 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x1, x2, x3)

Tags:
F has argument tags [2,0,0,0] and root tag 0

Comparison: DMS
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = x2   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(X1, X2, active(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x3)

Tags:
F has argument tags [0,0,0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = 1 + x1 + x2   
POL(active(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
ACTIVE(f(a, X, X)) → MARK(f(X, b, b))
MARK(f(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X1, X2, X3)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0)
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)

Tags:
MARK has argument tags [0,0] and root tag 0
ACTIVE has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = x1   
POL(a) = 0   
POL(active(x1)) = x1   
POL(b) = 0   
POL(f(x1, x2, x3)) = 1 + x1 + x2   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

f(X1, mark(X2), X3) → f(X1, X2, X3)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

(20) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
ACTIVE(f(a, X, X)) → MARK(f(X, b, b))

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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.


MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
ACTIVE(f(a, X, X)) → MARK(f(X, b, b))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)
ACTIVE(x0, x1)  =  ACTIVE(x0, x1)

Tags:
MARK has argument tags [3,2] and root tag 1
ACTIVE has argument tags [0,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1 + x1   
POL(a) = 1   
POL(active(x1)) = 1 + x1   
POL(b) = 0   
POL(f(x1, x2, x3)) = x1 + x3   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

f(X1, mark(X2), X3) → f(X1, X2, X3)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

(22) Obligation:

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

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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