Termination w.r.t. Q proof of /tmp/tmpiFIjzF/ExIntrod_GM01

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

0 QTRS

↳1 QTRSRRRProof (⇔)

↳2 QTRS

↳3 QTRSRRRProof (⇔)

↳4 QTRS

↳5 DependencyPairsProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 AND

  ↳9 QDP

    ↳10 UsableRulesProof (⇔)

    ↳11 QDP

    ↳12 QDPSizeChangeProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 UsableRulesProof (⇔)

    ↳16 QDP

    ↳17 QDPSizeChangeProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 UsableRulesProof (⇔)

    ↳21 QDP

    ↳22 QDPSizeChangeProof (⇔)

    ↳23 TRUE

  ↳24 QDP

    ↳25 UsableRulesProof (⇔)

    ↳26 QDP

    ↳27 QDPSizeChangeProof (⇔)

    ↳28 TRUE

  ↳29 QDP

    ↳30 UsableRulesProof (⇔)

    ↳31 QDP

    ↳32 QDPSizeChangeProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 UsableRulesProof (⇔)

    ↳36 QDP

    ↳37 QDPSizeChangeProof (⇔)

    ↳38 TRUE

  ↳39 QDP

    ↳40 MRRProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 DependencyGraphProof (⇔)

    ↳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 DependencyGraphProof (⇔)

    ↳61 TRUE

(0) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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 + 2·x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 1
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·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))

(2) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(head(x₁)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(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(adx(nil)) → mark(nil)
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)

(4) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(incr(nil)) → MARK(nil)
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
ACTIVE(incr(cons(X, L))) → S(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
ACTIVE(adx(cons(X, L))) → ADX(L)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(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)
S(mark(X)) → S(X)
S(active(X)) → S(X)
ADX(mark(X)) → ADX(X)
ADX(active(X)) → ADX(X)
HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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 17 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(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:

TAIL(active(X)) → TAIL(X)
The graph contains the following edges 1 > 1
TAIL(mark(X)) → TAIL(X)
The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

HEAD(active(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

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

HEAD(active(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(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:

HEAD(active(X)) → HEAD(X)
The graph contains the following edges 1 > 1
HEAD(mark(X)) → HEAD(X)
The graph contains the following edges 1 > 1

(18) TRUE

(19) 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(20) UsableRulesProof (EQUIVALENT transformation)

(21) 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.

(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:

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

(23) TRUE

(24) 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(25) UsableRulesProof (EQUIVALENT transformation)

(26) 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.

(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:

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

(28) TRUE

(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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(30) UsableRulesProof (EQUIVALENT transformation)

(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) TRUE

(34) 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(35) UsableRulesProof (EQUIVALENT transformation)

(36) 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.

(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:

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

(38) TRUE

(39) Obligation:

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

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(adx(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(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(head(x₁)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁
POL(zeros) = 0

(41) Obligation:

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

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(42) 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(mark(X)))

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

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(MARK(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(zeros) =
/ 0 \

\ 1 /

POL(0) =
/ 0 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(tail(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(nil) =
/ 1 \

\ 0 /

POL(active(x₁)) =
/ 1 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(nats) =
/ 0 \

\ 1 /

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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

(43) Obligation:

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

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(zeros) → ACTIVE(zeros)

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

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 1 0 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(zeros) =
/ 0 \

\ 1 /

POL(0) =
/ 0 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(tail(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(nil) =
/ 0 \

\ 0 /

POL(active(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 1 0 /

· x₁

POL(nats) =
/ 0 \

\ 0 /

The following usable rules [FROCOS05] were oriented:

mark(nil) → active(nil)
mark(0) → active(0)
mark(nats) → active(nats)
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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(incr(nil)) → mark(nil)
active(zeros) → mark(cons(0, zeros))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
mark(zeros) → active(zeros)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(tail(X)) → active(tail(mark(X)))

(45) Obligation:

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

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(46) DependencyGraphProof (EQUIVALENT transformation)

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

(47) Obligation:

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

MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))

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

POL(MARK(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(head(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(tail(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(nil) =
/ 0 \

\ 0 /

POL(active(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(nats) =
/ 0 \

\ 0 /

POL(zeros) =
/ 0 \

\ 0 /

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)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

(49) Obligation:

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

MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → MARK(X)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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.

MARK(cons(X1, X2)) → MARK(X1)

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

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 1 /

+
/ 1 1 \

\ 1 0 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 1 \

\ 1 /

+
/ 1 1 \

\ 1 1 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(nil) =
/ 0 \

\ 0 /

POL(active(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 1 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(nats) =
/ 0 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(tail(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(zeros) =
/ 1 \

\ 1 /

The following usable rules [FROCOS05] were oriented:

mark(nil) → active(nil)
mark(0) → active(0)
mark(nats) → active(nats)
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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
adx(active(X)) → adx(X)
adx(mark(X)) → adx(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(incr(nil)) → mark(nil)
active(zeros) → mark(cons(0, zeros))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
mark(zeros) → active(zeros)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(tail(X)) → active(tail(mark(X)))

(51) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(s(X)) → MARK(X)

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

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 1 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(active(x₁)) =
/ 0 \

\ 1 /

+
/ 0 1 \

\ 0 1 /

· x₁

POL(nil) =
/ 1 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(tail(x₁)) =
/ 1 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(nats) =
/ 0 \

\ 0 /

POL(zeros) =
/ 0 \

\ 0 /

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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)

(53) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(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 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 1 1 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 1 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 1 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(active(x₁)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 1 1 /

· x₁

POL(nil) =
/ 0 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(tail(x₁)) =
/ 1 \

\ 1 /

+
/ 0 1 \

\ 0 0 /

· x₁

POL(nats) =
/ 0 \

\ 0 /

POL(zeros) =
/ 1 \

\ 1 /

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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)

(55) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))

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

POL(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 1 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(cons(x₁, x₂)) =
/ 1 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁ +
/ 0 1 \

\ 0 0 /

· x₂

POL(s(x₁)) =
/ 1 \

\ 1 /

+
/ 1 1 \

\ 1 0 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 1 /

· x₁

POL(active(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(nil) =
/ 0 \

\ 0 /

POL(head(x₁)) =
/ 0 \

\ 1 /

+
/ 1 1 \

\ 0 1 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(tail(x₁)) =
/ 1 \

\ 1 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(nats) =
/ 0 \

\ 0 /

POL(zeros) =
/ 0 \

\ 1 /

The following usable rules [FROCOS05] were oriented:

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

(57) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(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(MARK(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(incr(x₁)) =
/ 1 \

\ 0 /

+
/ 1 0 \

\ 0 0 /

· x₁

POL(ACTIVE(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(mark(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 1 0 /

· x₁

POL(cons(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 1 0 /

· x₁ +
/ 0 0 \

\ 1 0 /

· x₂

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(active(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁

POL(adx(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(nil) =
/ 0 \

\ 1 /

POL(head(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 1 1 /

· x₁

POL(0) =
/ 0 \

\ 1 /

POL(tail(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 1 1 /

· x₁

POL(nats) =
/ 0 \

\ 1 /

POL(zeros) =
/ 1 \

\ 1 /

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
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, L))) → MARK(cons(s(X), incr(L)))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(60) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPSizeChangeProof (EQUIVALENT transformation)

(13) TRUE

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPSizeChangeProof (EQUIVALENT transformation)

(18) TRUE

(19) Obligation:

(20) UsableRulesProof (EQUIVALENT transformation)

(21) Obligation:

(22) QDPSizeChangeProof (EQUIVALENT transformation)

(23) TRUE

(24) Obligation:

(25) UsableRulesProof (EQUIVALENT transformation)

(26) Obligation:

(27) QDPSizeChangeProof (EQUIVALENT transformation)

(28) TRUE

(29) Obligation:

(30) UsableRulesProof (EQUIVALENT transformation)

(31) Obligation:

(32) QDPSizeChangeProof (EQUIVALENT transformation)

(33) TRUE

(34) Obligation:

(35) UsableRulesProof (EQUIVALENT transformation)

(36) Obligation:

(37) QDPSizeChangeProof (EQUIVALENT transformation)

(38) TRUE

(39) Obligation:

(40) MRRProof (EQUIVALENT transformation)

(41) Obligation:

(42) QDPOrderProof (EQUIVALENT transformation)

(43) Obligation:

(44) QDPOrderProof (EQUIVALENT transformation)

(45) Obligation:

(46) DependencyGraphProof (EQUIVALENT transformation)

(47) Obligation:

(48) QDPOrderProof (EQUIVALENT transformation)

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) QDPOrderProof (EQUIVALENT transformation)

(59) Obligation:

(60) DependencyGraphProof (EQUIVALENT transformation)

(61) TRUE