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

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

(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)) = x1 + x2   
POL(hd(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   
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))
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(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) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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.

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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.

(8) 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.

(9) 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.

(10) 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

(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(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.

(13) 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.

(14) 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.

(15) 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

(16) TRUE

(17) 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.

(18) 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.

(19) 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.

(20) 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

(21) TRUE

(22) 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.

(23) 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.

(24) 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.

(25) 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

(26) TRUE

(27) 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.

(28) 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.

(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)

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

(30) 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

(31) TRUE

(32) 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.

(33) 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.

(34) 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.

(35) 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

(36) TRUE

(37) 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.

(38) 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)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → MARK(X)


Used ordering: 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)) = 1 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(s(x1)) = x1   
POL(tl(x1)) = 1 + x1   
POL(zeros) = 0   

(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(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.

(40) Narrowing (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(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))

(41) 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.

(42) DependencyGraphProof (EQUIVALENT transformation)

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

(43) 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.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(x0, active(x1))) → ACTIVE(cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(adx(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\10/
·x2

POL(s(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(hd(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(active(x1)) =
/1\
\0/
 +
/11\
\00/
·x1

POL(nats) =
/0\
\0/

POL(zeros) =
/0\
\0/

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(45) 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))

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) 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: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(adx(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/1\
\1/
 +
/11\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\00/
·x2

POL(s(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(hd(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(tl(x1)) =
/1\
\0/
 +
/01\
\00/
·x1

POL(active(x1)) =
/0\
\0/
 +
/11\
\11/
·x1

POL(nats) =
/1\
\1/

POL(zeros) =
/1\
\1/

POL(0) =
/1\
\1/

The following usable rules [FROCOS05] were oriented:

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

(47) 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)))
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))

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) 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: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(adx(x1)) =
/1\
\0/
 +
/01\
\01/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/1\
\0/
 +
/11\
\01/
·x1

POL(incr(x1)) =
/0\
\1/
 +
/10\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\10/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/1\
\1/
 +
/00\
\00/
·x1

POL(hd(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(active(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(nats) =
/0\
\1/

POL(zeros) =
/0\
\0/

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(49) 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)
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))

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].


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: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\10/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(s(x1)) =
/0\
\1/
 +
/00\
\10/
·x1

POL(mark(x1)) =
/1\
\0/
 +
/11\
\00/
·x1

POL(adx(x1)) =
/1\
\0/
 +
/00\
\10/
·x1

POL(hd(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(active(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(nats) =
/0\
\1/

POL(zeros) =
/0\
\1/

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(51) 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)
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))

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].


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: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\11/
·x1 +
/00\
\01/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/01\
\11/
·x1

POL(hd(x1)) =
/1\
\1/
 +
/00\
\00/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(active(x1)) =
/1\
\0/
 +
/01\
\11/
·x1

POL(nats) =
/0\
\0/

POL(zeros) =
/0\
\1/

POL(0) =
/0\
\0/

POL(adx(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

The following usable rules [FROCOS05] were oriented:

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

(53) 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)
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))

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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(active(x0), x1)) → ACTIVE(cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/11\
\10/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\1/
 +
/10\
\00/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/11\
\01/
·x1

POL(tl(x1)) =
/0\
\1/
 +
/11\
\10/
·x1

POL(active(x1)) =
/1\
\1/
 +
/10\
\01/
·x1

POL(nats) =
/0\
\0/

POL(hd(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(zeros) =
/0\
\1/

POL(0) =
/1\
\1/

POL(adx(x1)) =
/1\
\0/
 +
/11\
\00/
·x1

The following usable rules [FROCOS05] were oriented:

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

(55) 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)
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))

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.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

POL(incr(x1)) =
/1\
\0/
 +
/11\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\11/
·x1

POL(s(x1)) =
/1\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\1/
 +
/00\
\11/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(nats) =
/1\
\0/

POL(active(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(hd(x1)) =
/1\
\0/
 +
/00\
\00/
·x1

POL(zeros) =
/1\
\1/

POL(0) =
/1\
\0/

POL(adx(x1)) =
/1\
\1/
 +
/00\
\11/
·x1

The following usable rules [FROCOS05] were oriented:

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

(57) Obligation:

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

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

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.

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(x0, mark(x1))) → ACTIVE(cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(incr(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\11/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/11\
\11/
·x1

POL(s(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\1/
 +
/01\
\10/
·x1

POL(nats) =
/0\
\0/

POL(active(x1)) =
/0\
\0/
 +
/10\
\11/
·x1

POL(hd(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

POL(zeros) =
/1\
\0/

POL(0) =
/0\
\0/

POL(adx(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

The following usable rules [FROCOS05] were oriented:

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

(59) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), 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.

(60) 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)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(incr(x1)) =
/1\
\1/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

POL(tl(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

POL(nats) =
/1\
\1/

POL(active(x1)) =
/1\
\1/
 +
/00\
\11/
·x1

POL(hd(x1)) =
/0\
\1/
 +
/10\
\10/
·x1

POL(zeros) =
/0\
\1/

POL(0) =
/0\
\0/

POL(adx(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

The following usable rules [FROCOS05] were oriented:

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

(61) Obligation:

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

MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(cons(mark(x0), 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.

(62) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(63) TRUE