YES Termination w.r.t. Q proof of Transformed_CSR_04_ExIntrod_GM04_iGM.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔, 77 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔, 0 ms)
        ↳33 YES
    ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔, 0 ms)
        ↳38 YES
    ↳39 QDP
        ↳40 MRRProof (⇔, 23 ms)
        ↳41 QDP
        ↳42 MRRProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 MRRProof (⇔, 17 ms)
        ↳45 QDP
        ↳46 TransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 DependencyGraphProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 QDPOrderProof (⇔, 127 ms)
        ↳51 QDP
        ↳52 QDPOrderProof (⇔, 0 ms)
        ↳53 QDP
        ↳54 PisEmptyProof (⇔, 0 ms)
        ↳55 YES

(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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(hd(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 2   
POL(s(x1)) = x1   
POL(tl(x1)) = 2 + 2·x1   
POL(zeros) = 2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)


(2) 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))))
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.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(hd(x1)) = 2 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 2   
POL(s(x1)) = x1   
POL(tl(x1)) = 2 + x1   
POL(zeros) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

active(nats) → mark(adx(zeros))


(4) Obligation:

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

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))))
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.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

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)
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(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))))
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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 13 less nodes.

(8) Complex Obligation (AND)

(9) 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(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))))
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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

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

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • TL(active(X)) → TL(X)
    The graph contains the following edges 1 > 1

  • TL(mark(X)) → TL(X)
    The graph contains the following edges 1 > 1

(13) YES

(14) 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(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))))
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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

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

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

(17) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • HD(active(X)) → HD(X)
    The graph contains the following edges 1 > 1

  • HD(mark(X)) → HD(X)
    The graph contains the following edges 1 > 1

(18) YES

(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(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))))
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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(21) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

(22) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(23) YES

(24) 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(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))))
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.

(25) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(26) Obligation:

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

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

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

(27) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • INCR(active(X)) → INCR(X)
    The graph contains the following edges 1 > 1

  • INCR(mark(X)) → INCR(X)
    The graph contains the following edges 1 > 1

(28) YES

(29) 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(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))))
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.

(30) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(31) 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)

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

(32) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • CONS(X1, mark(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS(mark(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(active(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(X1, active(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(33) YES

(34) 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(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))))
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.

(35) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(36) Obligation:

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

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

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

(37) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ADX(active(X)) → ADX(X)
    The graph contains the following edges 1 > 1

  • ADX(mark(X)) → ADX(X)
    The graph contains the following edges 1 > 1

(38) YES

(39) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
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(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
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(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))))
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) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(tl(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 2·x1   
POL(MARK(x1)) = 2·x1   
POL(active(x1)) = x1   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(hd(x1)) = 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   

(41) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
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(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(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(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))))
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) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(adx(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 2 + 2·x1   
POL(MARK(x1)) = 2 + 2·x1   
POL(active(x1)) = x1   
POL(adx(x1)) = 2 + 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(hd(x1)) = x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = x1   
POL(tl(x1)) = 2 + x1   
POL(zeros) = 2   

(43) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
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(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(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(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))))
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.

(44) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(hd(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 2 + 2·x1   
POL(MARK(x1)) = 2 + 2·x1   
POL(active(x1)) = x1   
POL(adx(x1)) = 1 + 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(hd(x1)) = 1 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = x1   
POL(tl(x1)) = 2 + x1   
POL(zeros) = 2   

(45) Obligation:

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

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

The TRS R consists of the following rules:

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))))
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.

(46) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2)) at position [0] we obtained the following new rules [LPAR04]:

MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1)) → MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1)) → MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1)) → MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1)) → MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))

(47) 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(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))

The TRS R consists of the following rules:

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))))
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.

(48) DependencyGraphProof (EQUIVALENT transformation)

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

(49) 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(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))

The TRS R consists of the following rules:

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))))
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.

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


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)))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ACTIVE(x1) ) = x1 + 1

POL( adx(x1) ) = 2

POL( hd(x1) ) = 2x1

POL( incr(x1) ) = 2x1 + 1

POL( tl(x1) ) = max{0, -2}

POL( mark(x1) ) = 2x1

POL( nats ) = 1

POL( active(x1) ) = x1

POL( cons(x1, x2) ) = max{0, -1}

POL( zeros ) = 2

POL( 0 ) = 2

POL( s(x1) ) = 0

POL( MARK(x1) ) = 2x1 + 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(X1, X2))
active(zeros) → mark(cons(0, zeros))
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(zeros) → active(zeros)
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)))
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)

(51) Obligation:

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

ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))

The TRS R consists of the following rules:

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))))
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,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE
MARK(x1)  =  x1
incr(x1)  =  incr
s(x1)  =  s
hd(x1)  =  hd
tl(x1)  =  tl
cons(x1, x2)  =  cons

Knuth-Bendix order [KBO] with precedence:
s > ACTIVE > incr

and weight map:

s=1
hd=2
ACTIVE=1
cons=2
tl=2
incr=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)

(53) Obligation:

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

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))))
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) PisEmptyProof (EQUIVALENT transformation)

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

(55) YES