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

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

(0) Obligation:

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

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
INF(X) → S(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
LENGTH(nil) → 01
LENGTH(cons(X, L)) → S(n__length(activate(L)))
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__0) → 01
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__inf(X)) → INF(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
ACTIVATE(n__length(X)) → LENGTH(X)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(X)
LENGTH(cons(X, L)) → ACTIVATE(L)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(6) 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.

(7) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(X)
LENGTH(cons(X, L)) → ACTIVATE(L)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)

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

(8) 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:

  • LENGTH(cons(X, L)) → ACTIVATE(L)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
    The graph contains the following edges 1 > 1, 1 > 2

  • ACTIVATE(n__length(X)) → LENGTH(X)
    The graph contains the following edges 1 > 1

  • TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
    The graph contains the following edges 2 > 1

  • TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
    The graph contains the following edges 1 > 1

  • TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
    The graph contains the following edges 2 > 1

(9) TRUE

(10) Obligation:

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

EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(11) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y)) at position [0] we obtained the following new rules [LPAR04]:

EQ(n__s(n__0), n__s(y1)) → EQ(0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(s(x0), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1))
EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))

(12) Obligation:

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

EQ(n__s(n__0), n__s(y1)) → EQ(0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(s(x0), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1))
EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(13) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__0), n__s(y1)) → EQ(0, activate(y1)) at position [0] we obtained the following new rules [LPAR04]:

EQ(n__s(n__0), n__s(y0)) → EQ(n__0, activate(y0))

(14) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(s(x0), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1))
EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__0), n__s(y0)) → EQ(n__0, activate(y0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(s(x0), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1))
EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(17) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__s(x0)), n__s(y1)) → EQ(s(x0), activate(y1)) at position [0] we obtained the following new rules [LPAR04]:

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))

(18) Obligation:

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

EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1))
EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(inf(x0), activate(y1)) at position [0] we obtained the following new rules [LPAR04]:

EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(cons(x0, n__inf(s(x0))), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(n__inf(x0), activate(y1))

(20) Obligation:

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

EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(cons(x0, n__inf(s(x0))), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(n__inf(x0), activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1))
EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__take(x0, x1)), n__s(y1)) → EQ(take(x0, x1), activate(y1)) at position [0] we obtained the following new rules [LPAR04]:

EQ(n__s(n__take(0, x0)), n__s(y2)) → EQ(nil, activate(y2))
EQ(n__s(n__take(s(x0), cons(x1, x2))), n__s(y2)) → EQ(cons(activate(x1), n__take(activate(x0), activate(x2))), activate(y2))
EQ(n__s(n__take(x0, x1)), n__s(y2)) → EQ(n__take(x0, x1), activate(y2))

(24) Obligation:

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

EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__take(0, x0)), n__s(y2)) → EQ(nil, activate(y2))
EQ(n__s(n__take(s(x0), cons(x1, x2))), n__s(y2)) → EQ(cons(activate(x1), n__take(activate(x0), activate(x2))), activate(y2))
EQ(n__s(n__take(x0, x1)), n__s(y2)) → EQ(n__take(x0, x1), activate(y2))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(26) Obligation:

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

EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1))
EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(27) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__length(x0)), n__s(y1)) → EQ(length(x0), activate(y1)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), 0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)

(28) Obligation:

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

EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1))
EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), 0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(29) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(x0), n__s(y1)) → EQ(x0, activate(y1)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)

(30) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), 0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(31) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), 0) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), n__0)

(32) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__0)) → EQ(length(y0), n__0)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(33) DependencyGraphProof (EQUIVALENT transformation)

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

(34) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(35) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), s(x0)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

(36) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(37) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), inf(x0)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), cons(x0, n__inf(s(x0))))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), n__inf(x0))

(38) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), cons(x0, n__inf(s(x0))))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), n__inf(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(41) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), take(x0, x1)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(n__length(y0)), n__s(n__take(0, x0))) → EQ(length(y0), nil)
EQ(n__s(n__length(y0)), n__s(n__take(s(x0), cons(x1, x2)))) → EQ(length(y0), cons(activate(x1), n__take(activate(x0), activate(x2))))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), n__take(x0, x1))

(42) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(n__length(y0)), n__s(n__take(0, x0))) → EQ(length(y0), nil)
EQ(n__s(n__length(y0)), n__s(n__take(s(x0), cons(x1, x2)))) → EQ(length(y0), cons(activate(x1), n__take(activate(x0), activate(x2))))
EQ(n__s(n__length(y0)), n__s(n__take(x0, x1))) → EQ(length(y0), n__take(x0, x1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(43) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(44) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(45) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(y0), n__s(n__0)) → EQ(y0, 0) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(y0), n__s(n__0)) → EQ(y0, n__0)

(46) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__0)) → EQ(y0, n__0)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(49) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(x0)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))

(50) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(51) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, inf(x0)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, cons(x0, n__inf(s(x0))))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, n__inf(x0))

(52) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, cons(x0, n__inf(s(x0))))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, n__inf(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(53) DependencyGraphProof (EQUIVALENT transformation)

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

(54) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(55) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, take(x0, x1)) at position [1] we obtained the following new rules [LPAR04]:

EQ(n__s(y0), n__s(n__take(0, x0))) → EQ(y0, nil)
EQ(n__s(y0), n__s(n__take(s(x0), cons(x1, x2)))) → EQ(y0, cons(activate(x1), n__take(activate(x0), activate(x2))))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, n__take(x0, x1))

(56) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))
EQ(n__s(y0), n__s(n__take(0, x0))) → EQ(y0, nil)
EQ(n__s(y0), n__s(n__take(s(x0), cons(x1, x2)))) → EQ(y0, cons(activate(x1), n__take(activate(x0), activate(x2))))
EQ(n__s(y0), n__s(n__take(x0, x1))) → EQ(y0, n__take(x0, x1))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(57) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(58) Obligation:

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

EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(n__s(n__s(x0)), n__s(y1)) → EQ(n__s(x0), activate(y1))
EQ(n__s(y0), n__s(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, x0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, n__s(x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(EQ(x1, x2)) = x1   
POL(activate(x1)) = 1 + x1   
POL(cons(x1, x2)) = 0   
POL(inf(x1)) = 0   
POL(length(x1)) = 1   
POL(n__0) = 0   
POL(n__inf(x1)) = 0   
POL(n__length(x1)) = 0   
POL(n__s(x1)) = 1 + x1   
POL(n__take(x1, x2)) = x2   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   
POL(take(x1, x2)) = x2   

The following usable rules [FROCOS05] were oriented:

activate(n__length(X)) → length(X)
activate(n__take(X1, X2)) → take(X1, X2)
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(X) → n__length(X)
take(X1, X2) → n__take(X1, X2)
0 → n__0
s(X) → n__s(X)
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))

(60) Obligation:

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

EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))
EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(n__s(n__length(y0)), n__s(x0)) → EQ(length(y0), x0)
EQ(n__s(n__length(y0)), n__s(n__s(x0))) → EQ(length(y0), n__s(x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(EQ(x1, x2)) = x2   
POL(activate(x1)) = 1   
POL(cons(x1, x2)) = 0   
POL(inf(x1)) = 0   
POL(length(x1)) = 1   
POL(n__0) = 0   
POL(n__inf(x1)) = 0   
POL(n__length(x1)) = 0   
POL(n__s(x1)) = 1 + x1   
POL(n__take(x1, x2)) = 0   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   
POL(take(x1, x2)) = 0   

The following usable rules [FROCOS05] were oriented:

activate(n__length(X)) → length(X)
activate(n__take(X1, X2)) → take(X1, X2)
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(X) → n__length(X)
take(X1, X2) → n__take(X1, X2)
0 → n__0
s(X) → n__s(X)
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))

(62) Obligation:

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

EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(63) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = EQ(length(activate(n__inf(X))), length(activate(n__inf(X')))) evaluates to t =EQ(length(activate(n__inf(s(X)))), length(activate(n__inf(s(X')))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [X / s(X), X' / s(X')]
  • Semiunifier: [ ]



Rewriting sequence

EQ(length(activate(n__inf(X))), length(activate(n__inf(X')))) → EQ(length(activate(n__inf(X))), length(inf(X')))
with rule activate(n__inf(X'')) → inf(X'') at position [1,0] and matcher [X'' / X']

EQ(length(activate(n__inf(X))), length(inf(X'))) → EQ(length(activate(n__inf(X))), length(cons(X', n__inf(s(X')))))
with rule inf(X'') → cons(X'', n__inf(s(X''))) at position [1,0] and matcher [X'' / X']

EQ(length(activate(n__inf(X))), length(cons(X', n__inf(s(X'))))) → EQ(length(activate(n__inf(X))), s(n__length(activate(n__inf(s(X'))))))
with rule length(cons(X'', L)) → s(n__length(activate(L))) at position [1] and matcher [X'' / X', L / n__inf(s(X'))]

EQ(length(activate(n__inf(X))), s(n__length(activate(n__inf(s(X')))))) → EQ(length(activate(n__inf(X))), n__s(n__length(activate(n__inf(s(X'))))))
with rule s(X'') → n__s(X'') at position [1] and matcher [X'' / n__length(activate(n__inf(s(X'))))]

EQ(length(activate(n__inf(X))), n__s(n__length(activate(n__inf(s(X')))))) → EQ(length(inf(X)), n__s(n__length(activate(n__inf(s(X'))))))
with rule activate(n__inf(X'')) → inf(X'') at position [0,0] and matcher [X'' / X]

EQ(length(inf(X)), n__s(n__length(activate(n__inf(s(X')))))) → EQ(length(cons(X, n__inf(s(X)))), n__s(n__length(activate(n__inf(s(X'))))))
with rule inf(X'') → cons(X'', n__inf(s(X''))) at position [0,0] and matcher [X'' / X]

EQ(length(cons(X, n__inf(s(X)))), n__s(n__length(activate(n__inf(s(X')))))) → EQ(s(n__length(activate(n__inf(s(X))))), n__s(n__length(activate(n__inf(s(X'))))))
with rule length(cons(X'', L)) → s(n__length(activate(L))) at position [0] and matcher [X'' / X, L / n__inf(s(X))]

EQ(s(n__length(activate(n__inf(s(X))))), n__s(n__length(activate(n__inf(s(X')))))) → EQ(n__s(n__length(activate(n__inf(s(X))))), n__s(n__length(activate(n__inf(s(X'))))))
with rule s(X'') → n__s(X'') at position [0] and matcher [X'' / n__length(activate(n__inf(s(X))))]

EQ(n__s(n__length(activate(n__inf(s(X))))), n__s(n__length(activate(n__inf(s(X')))))) → EQ(length(activate(n__inf(s(X)))), length(activate(n__inf(s(X')))))
with rule EQ(n__s(n__length(y0)), n__s(n__length(x0))) → EQ(length(y0), length(x0))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(64) FALSE