YES Termination w.r.t. Q proof of Transformed_CSR_04_ExIntrod_GM01_iGM.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 79 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QDPSizeChangeProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 MRRProof (⇔, 3 ms)
        ↳39 QDP
        ↳40 QDPOrderProof (⇔, 43 ms)
        ↳41 QDP
        ↳42 DependencyGraphProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 QDPOrderProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 QDPOrderProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 QDPOrderProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 QDPOrderProof (⇔, 9 ms)
        ↳51 QDP
        ↳52 DependencyGraphProof (⇔, 0 ms)
        ↳53 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)) = 2·x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(head(x1)) = 1 + 2·x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + 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(adx(nil)) → mark(nil)
active(nats) → mark(adx(zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)


(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(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.

(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(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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 17 less nodes.

(6) Complex Obligation (AND)

(7) 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.

(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:

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.

(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:

  • 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

(11) YES

(12) 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.

(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:

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.

(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:

  • 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

(16) YES

(17) 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.

(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:

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.

(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:

  • 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

(21) YES

(22) 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.

(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:

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.

(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:

  • 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

(26) YES

(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(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.

(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) YES

(32) 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.

(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:

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.

(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:

  • 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

(36) YES

(37) 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.

(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(head(X)) → MARK(X)
MARK(tail(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

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

(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(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.

(40) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


ACTIVE(zeros) → MARK(cons(0, zeros))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
incr(x1)  =  x1
cons(x1, x2)  =  x1
MARK(x1)  =  x1
s(x1)  =  x1
mark(x1)  =  x1
adx(x1)  =  x1
zeros  =  zeros
0  =  0
head(x1)  =  head
tail(x1)  =  tail
active(x1)  =  x1
nil  =  nil
nats  =  nats

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

tail=1
nats=2
0=2
head=2
zeros=3
nil=2

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

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

(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)
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) DependencyGraphProof (EQUIVALENT transformation)

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

(43) 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)) → ACTIVE(s(mark(X)))
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.

(44) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
incr(x1)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  x1
adx(x1)  =  adx(x1)
head(x1)  =  head
tail(x1)  =  tail
active(x1)  =  x1
zeros  =  zeros
0  =  0
nil  =  nil
nats  =  nats

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

tail=1
nats=2
zeros=3
0=2
head=2
adx_1=1
nil=2

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

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

(45) 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))
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(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) QDPOrderProof (EQUIVALENT transformation)

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


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: Combined order from the following AFS and order.
MARK(x1)  =  x1
incr(x1)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1)
s(x1)  =  x1
adx(x1)  =  x1
head(x1)  =  head
tail(x1)  =  tail
active(x1)  =  x1
zeros  =  zeros
0  =  0
nil  =  nil
nats  =  nats

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

tail=1
nats=2
zeros=4
0=2
head=2
cons_1=1
nil=2

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

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

(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))
MARK(s(X)) → ACTIVE(s(mark(X)))
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,JAR06].


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: Combined order from the following AFS and order.
MARK(x1)  =  x1
incr(x1)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  cons
s(x1)  =  s(x1)
adx(x1)  =  adx
head(x1)  =  head
tail(x1)  =  tail
active(x1)  =  x1
zeros  =  zeros
nil  =  nil
nats  =  nats
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
tail=1
nats=2
zeros=3
head=2
0=2
cons=2
adx=3
nil=2

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

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

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

(50) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
incr(x1)  =  incr(x1)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
cons(x1, x2)  =  cons
s(x1)  =  s
adx(x1)  =  adx
head(x1)  =  head
tail(x1)  =  tail
active(x1)  =  x1
zeros  =  zeros
nil  =  nil
nats  =  nats
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s=1
tail=2
nats=2
zeros=3
head=2
0=2
cons=2
incr_1=1
adx=4
nil=2

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

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

(51) Obligation:

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

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

(52) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 6 less nodes.

(53) TRUE