(0) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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(nats) → MARK(adx(zeros))
ACTIVE(nats) → ADX(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
ACTIVE(incr(cons(X, Y))) → S(X)
ACTIVE(incr(cons(X, Y))) → INCR(Y)
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(adx(cons(X, Y))) → ADX(Y)
ACTIVE(hd(cons(X, Y))) → MARK(X)
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(nats) → ACTIVE(nats)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(0) → ACTIVE(0)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → HD(mark(X))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → TL(mark(X))
MARK(tl(X)) → MARK(X)
ADX(mark(X)) → ADX(X)
ADX(active(X)) → ADX(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
HD(mark(X)) → HD(X)
HD(active(X)) → HD(X)
TL(mark(X)) → TL(X)
TL(active(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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 7 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TL(active(X)) → TL(X)
TL(mark(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
TL 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(TL(x1)) = 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:

TL(mark(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


TL(mark(X)) → TL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
TL(x0, x1)  =  TL(x1)

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

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

POL(TL(x1)) = 0   
POL(mark(x1)) = 1 + x1   

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(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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:

HD(active(X)) → HD(X)
HD(mark(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
HD 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(HD(x1)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

HD(mark(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HD(mark(X)) → HD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
HD(x0, x1)  =  HD(x1)

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

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

POL(HD(x1)) = 0   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
S 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(S(x1)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
S(x0, x1)  =  S(x1)

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

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

POL(S(x1)) = 0   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
INCR 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(INCR(x1)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


INCR(mark(X)) → INCR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
INCR(x0, x1)  =  INCR(x1)

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

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

POL(INCR(x1)) = 0   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

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

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

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

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
CONS has argument tags [1,2,2] and root tag 0

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

POL(CONS(x1, x2)) = 1 + x1   
POL(active(x1)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

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

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

POL(CONS(x1, x2)) = 1   
POL(active(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

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

POL(CONS(x1, x2)) = 0   
POL(active(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

ADX(active(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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.


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

Tags:
ADX 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(ADX(x1)) = 1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADX(mark(X)) → ADX(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ADX(x0, x1)  =  ADX(x1)

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

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

POL(ADX(x1)) = 0   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(adx(X)) → MARK(X)
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)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(adx(x1)) = 1 + x1   
POL(cons(x1, x2)) = x1 + x2   
POL(hd(x1)) = x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = 0   
POL(tl(x1)) = x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented:

mark(adx(X)) → active(adx(mark(X)))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
mark(cons(X1, X2)) → active(cons(X1, X2))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(incr(X)) → active(incr(mark(X)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
mark(s(X)) → active(s(X))
active(hd(cons(X, Y))) → mark(X)
mark(hd(X)) → active(hd(mark(X)))
active(tl(cons(X, Y))) → mark(Y)
mark(tl(X)) → active(tl(mark(X)))
mark(0) → active(0)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
hd(active(X)) → hd(X)
hd(mark(X)) → hd(X)
tl(active(X)) → tl(X)
tl(mark(X)) → tl(X)

(53) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → ACTIVE(s(X))
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(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(adx(x1)) = 1   
POL(cons(x1, x2)) = 1   
POL(hd(x1)) = 1   
POL(incr(x1)) = 1   
POL(mark(x1)) = 1   
POL(nats) = 1   
POL(s(x1)) = 0   
POL(tl(x1)) = 1   
POL(zeros) = 1   

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(tl(X)) → MARK(X)
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)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(hd(x1)) = x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented:

mark(adx(X)) → active(adx(mark(X)))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
mark(cons(X1, X2)) → active(cons(X1, X2))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(incr(X)) → active(incr(mark(X)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
mark(s(X)) → active(s(X))
active(hd(cons(X, Y))) → mark(X)
mark(hd(X)) → active(hd(mark(X)))
active(tl(cons(X, Y))) → mark(Y)
mark(tl(X)) → active(tl(mark(X)))
mark(0) → active(0)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
hd(active(X)) → hd(X)
hd(mark(X)) → hd(X)
tl(active(X)) → tl(X)
tl(mark(X)) → tl(X)

(57) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(hd(X)) → MARK(X)
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)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 0   
POL(active(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(hd(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = 0   
POL(tl(x1)) = x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented:

mark(adx(X)) → active(adx(mark(X)))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
mark(cons(X1, X2)) → active(cons(X1, X2))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(incr(X)) → active(incr(mark(X)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
mark(s(X)) → active(s(X))
active(hd(cons(X, Y))) → mark(X)
mark(hd(X)) → active(hd(mark(X)))
active(tl(cons(X, Y))) → mark(Y)
mark(tl(X)) → active(tl(mark(X)))
mark(0) → active(0)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
hd(active(X)) → hd(X)
hd(mark(X)) → hd(X)
tl(active(X)) → tl(X)
tl(mark(X)) → tl(X)

(59) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(nats) → MARK(adx(zeros))
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(x1)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(hd(x1)) = x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = 0   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(nats) → ACTIVE(nats)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(62) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(63) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Tags:
ACTIVE has argument tags [1,0] and root tag 0
MARK 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(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = 1   
POL(adx(x1)) = 1   
POL(cons(x1, x2)) = 1   
POL(hd(x1)) = 1   
POL(incr(x1)) = 1   
POL(mark(x1)) = 0   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = 0   
POL(zeros) = 1   

The following usable rules [FROCOS05] were oriented: none

(65) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(hd(X)) → ACTIVE(hd(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(adx(x1)) = 1   
POL(cons(x1, x2)) = 0   
POL(hd(x1)) = 1   
POL(incr(x1)) = 1   
POL(mark(x1)) = 0   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = x1   
POL(zeros) = 1   

The following usable rules [FROCOS05] were oriented: none

(67) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(hd(X)) → ACTIVE(hd(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(68) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(69) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → ACTIVE(hd(mark(X)))

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(hd(X)) → ACTIVE(hd(mark(X)))
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)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 1 + x1   
POL(active(x1)) = x1   
POL(adx(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(hd(x1)) = 1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = 0   
POL(tl(x1)) = 1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented:

incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)

(71) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(adx(X)) → ACTIVE(adx(mark(X)))
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, x1)
ACTIVE(x0, x1)  =  ACTIVE(x0)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 0   
POL(active(x1)) = 1 + x1   
POL(adx(x1)) = 1   
POL(cons(x1, x2)) = 0   
POL(hd(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = 1   
POL(zeros) = 1   

The following usable rules [FROCOS05] were oriented: none

(73) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x1)
MARK(x0, x1)  =  MARK(x0)

Tags:
ACTIVE has argument tags [3,0] and root tag 0
MARK 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(0) = 1   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = 0   
POL(active(x1)) = 1   
POL(adx(x1)) = 1   
POL(cons(x1, x2)) = x1   
POL(hd(x1)) = x1   
POL(incr(x1)) = 0   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = 1 + x1   
POL(tl(x1)) = x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented: none

(75) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(76) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVE(x0, x1)  =  ACTIVE(x0)
MARK(x0, x1)  =  MARK(x0, x1)

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

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

POL(0) = 0   
POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(adx(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(hd(x1)) = x1   
POL(incr(x1)) = 1 + x1   
POL(mark(x1)) = 1   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = x1   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented: none

(77) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(78) PisEmptyProof (EQUIVALENT transformation)

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

(79) TRUE