Termination w.r.t. Q proof of /tmp/tmpKCLgH1/Ex1_GL02a

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 YES

    ↳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 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) 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__length(y0)), n__s(x0)) → EQ(length(y0), 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 remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( EQ(x₁, x₂) ) = max{0, 2x₂ - 2}

POL( cons(x₁, x₂) ) = max{0, -2}

POL( n__take(x₁, x₂) ) = x₂ + 2

POL( n__length(x₁) ) = max{0, -2}

POL( s(x₁) ) = 2x₁ + 2

POL( activate(x₁) ) = x₁ + 2

POL( n__0 ) = 0

POL( 0 ) = 0

POL( n__s(x₁) ) = 2x₁ + 2

POL( n__inf(x₁) ) = max{0, -1}

POL( inf(x₁) ) = max{0, -2}

POL( take(x₁, x₂) ) = x₂ + 2

POL( length(x₁) ) = 2

POL( nil ) = 0

The following usable rules [FROCOS05] 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__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(y0), n__s(n__length(x0))) → EQ(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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

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

POL( cons(x₁, x₂) ) = max{0, 2x₁ - 2}

POL( n__take(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( n__length(x₁) ) = max{0, -1}

POL( s(x₁) ) = x₁ + 2

POL( activate(x₁) ) = x₁ + 2

POL( n__0 ) = 0

POL( 0 ) = 0

POL( n__s(x₁) ) = x₁ + 2

POL( n__inf(x₁) ) = 2x₁

POL( inf(x₁) ) = 2x₁ + 2

POL( take(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( length(x₁) ) = 2

POL( nil ) = 2

The following usable rules [FROCOS05] 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

(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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPSizeChangeProof (EQUIVALENT transformation)

(9) YES

(10) Obligation:

(11) Narrowing (EQUIVALENT transformation)

(12) Obligation:

(13) Narrowing (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) Narrowing (EQUIVALENT transformation)

(18) Obligation:

(19) Narrowing (EQUIVALENT transformation)

(20) Obligation:

(21) DependencyGraphProof (EQUIVALENT transformation)

(22) Obligation:

(23) Narrowing (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) Obligation:

(27) Narrowing (EQUIVALENT transformation)

(28) Obligation:

(29) Narrowing (EQUIVALENT transformation)

(30) Obligation:

(31) Narrowing (EQUIVALENT transformation)

(32) Obligation:

(33) DependencyGraphProof (EQUIVALENT transformation)

(34) Obligation:

(35) Narrowing (EQUIVALENT transformation)

(36) Obligation:

(37) Narrowing (EQUIVALENT transformation)

(38) Obligation:

(39) DependencyGraphProof (EQUIVALENT transformation)

(40) Obligation:

(41) Narrowing (EQUIVALENT transformation)

(42) Obligation:

(43) DependencyGraphProof (EQUIVALENT transformation)

(44) Obligation:

(45) Narrowing (EQUIVALENT transformation)

(46) Obligation:

(47) DependencyGraphProof (EQUIVALENT transformation)

(48) Obligation:

(49) Narrowing (EQUIVALENT transformation)

(50) Obligation:

(51) Narrowing (EQUIVALENT transformation)

(52) Obligation:

(53) DependencyGraphProof (EQUIVALENT transformation)

(54) Obligation:

(55) Narrowing (EQUIVALENT transformation)

(56) Obligation:

(57) DependencyGraphProof (EQUIVALENT transformation)

(58) Obligation:

(59) QDPOrderProof (EQUIVALENT transformation)

(60) Obligation:

(61) QDPOrderProof (EQUIVALENT transformation)

(62) Obligation:

(63) NonTerminationProof (EQUIVALENT transformation)

(64) NO