NO Termination w.r.t. Q proof of Transformed_CSR_04_Ex1_GL02a_Z.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 TransformationProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 TransformationProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 DependencyGraphProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 TransformationProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 TransformationProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 TransformationProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 TransformationProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 TransformationProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 TransformationProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 DependencyGraphProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 TransformationProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 TransformationProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 TransformationProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 DependencyGraphProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 TransformationProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 DependencyGraphProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 TransformationProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 TransformationProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 DependencyGraphProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 TransformationProof (⇔, 0 ms)
        ↳56 QDP
        ↳57 DependencyGraphProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 QDPOrderProof (⇔, 37 ms)
        ↳60 QDP
        ↳61 QDPOrderProof (⇔, 13 ms)
        ↳62 QDP
        ↳63 NonTerminationLoopProof (⇒, 1296 ms)
        ↳64 NO

(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) YES

(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) TransformationProof (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__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__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__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__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(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(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) TransformationProof (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)) → 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) TransformationProof (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)) → 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) TransformationProof (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(cons(x0, n__inf(s(x0))), activate(y1))
EQ(n__s(n__inf(x0)), n__s(y1)) → EQ(n__inf(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) TransformationProof (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(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(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)) → 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) TransformationProof (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__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__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__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__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(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(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) TransformationProof (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__0)) → EQ(y0, 0)
EQ(n__s(y0), n__s(n__s(x0))) → EQ(y0, s(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__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__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(n__length(x0))) → EQ(y0, length(x0))
EQ(n__s(y0), n__s(x0)) → EQ(y0, 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) TransformationProof (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) → 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) TransformationProof (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)) → 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) TransformationProof (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), cons(x0, n__inf(s(x0))))
EQ(n__s(n__length(y0)), n__s(n__inf(x0))) → EQ(length(y0), n__inf(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) TransformationProof (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(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(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)) → 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) TransformationProof (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) → 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) TransformationProof (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)) → 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) TransformationProof (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, cons(x0, n__inf(s(x0))))
EQ(n__s(y0), n__s(n__inf(x0))) → EQ(y0, n__inf(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) TransformationProof (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(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(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)) → 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,JAR06].


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 Order [NEGPOLO,POLO] with Interpretation:
POL( EQ(x1, x2) ) = x1 + 1

POL( cons(x1, x2) ) = 2x2 + 1

POL( n__take(x1, x2) ) = 0

POL( n__length(x1) ) = 2x1

POL( s(x1) ) = x1 + 1

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

POL( n__0 ) = 0

POL( 0 ) = 0

POL( n__s(x1) ) = x1 + 1

POL( n__inf(x1) ) = 0

POL( inf(x1) ) = 1

POL( take(x1, x2) ) = 1

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

POL( nil ) = 0


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

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

(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,JAR06].


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 Order [NEGPOLO,POLO] with Interpretation:
POL( EQ(x1, x2) ) = x1 + 2x2 + 1

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

POL( nil ) = 0

POL( 0 ) = 0

POL( cons(x1, x2) ) = 2x2 + 2

POL( s(x1) ) = x1 + 2

POL( n__length(x1) ) = max{0, 2x1 - 1}

POL( activate(x1) ) = x1 + 2

POL( n__take(x1, x2) ) = max{0, x2 - 2}

POL( take(x1, x2) ) = x2

POL( n__0 ) = 0

POL( n__s(x1) ) = x1 + 2

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

POL( inf(x1) ) = 2


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

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

(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) NonTerminationLoopProof (COMPLETE 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) NO