YES Termination w.r.t. Q proof of AProVE_07_kabasci01.ari

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

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 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPSizeChangeProof (⇔, 0 ms)
        ↳34 YES
    ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QDPSizeChangeProof (⇔, 0 ms)
        ↳39 YES
    ↳40 QDP
        ↳41 UsableRulesProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 QDPSizeChangeProof (⇔, 0 ms)
        ↳44 YES
    ↳45 QDP
        ↳46 UsableRulesProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 QDPSizeChangeProof (⇔, 0 ms)
        ↳49 YES
    ↳50 QDP
        ↳51 UsableRulesReductionPairsProof (⇔, 4 ms)
        ↳52 QDP
        ↳53 QDPOrderProof (⇔, 97 ms)
        ↳54 QDP
        ↳55 UsableRulesReductionPairsProof (⇔, 16 ms)
        ↳56 QDP
        ↳57 UsableRulesProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 MRRProof (⇔, 0 ms)
        ↳60 QDP
        ↳61 PisEmptyProof (⇔, 0 ms)
        ↳62 YES

(0) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

ACTIVE(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
ACTIVE(inf(X)) → INF(s(X))
ACTIVE(inf(X)) → S(X)
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(length(cons(X, L))) → S(length(L))
ACTIVE(length(cons(X, L))) → LENGTH(L)
ACTIVE(inf(X)) → INF(active(X))
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
INF(mark(X)) → INF(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
PROPER(eq(X1, X2)) → EQ(proper(X1), proper(X2))
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(inf(X)) → INF(proper(X))
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → CONS(any(any(proper(X1))), any(proper(X2)))
PROPER(cons(any(X1), X2)) → ANY(any(proper(X1)))
PROPER(cons(any(X1), X2)) → ANY(proper(X1))
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → ANY(proper(X2))
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
EQ(ok(X1), ok(X2)) → EQ(X1, X2)
S(ok(X)) → S(X)
INF(ok(X)) → INF(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
LENGTH(ok(X)) → LENGTH(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)
ANY(X) → S(X)
ANY(proper(X)) → ANY(any(any(X)))
ANY(proper(X)) → ANY(any(X))
ANY(proper(X)) → ANY(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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 10 SCCs with 26 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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:

  • CONS(ok(X1), ok(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 > 2

(9) YES

(10) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(11) 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.

(12) Obligation:

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

S(ok(X)) → S(X)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • S(ok(X)) → S(X)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

ANY(proper(X)) → ANY(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(16) 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.

(17) Obligation:

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

ANY(proper(X)) → ANY(X)

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

(18) 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:

  • ANY(proper(X)) → ANY(X)
    The graph contains the following edges 1 > 1

(19) YES

(20) Obligation:

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

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(21) 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.

(22) Obligation:

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

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

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

(23) 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:

  • EQ(ok(X1), ok(X2)) → EQ(X1, X2)
    The graph contains the following edges 1 > 1, 2 > 2

(24) YES

(25) Obligation:

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

LENGTH(ok(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(26) 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.

(27) Obligation:

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

LENGTH(ok(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

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

(28) 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(ok(X)) → LENGTH(X)
    The graph contains the following edges 1 > 1

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

(29) YES

(30) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(31) 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.

(32) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

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

(33) 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:

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

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

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

(34) YES

(35) Obligation:

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

INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(36) 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.

(37) Obligation:

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

INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

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

(38) 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:

  • INF(ok(X)) → INF(X)
    The graph contains the following edges 1 > 1

  • INF(mark(X)) → INF(X)
    The graph contains the following edges 1 > 1

(39) YES

(40) Obligation:

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

PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(41) 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.

(42) Obligation:

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

PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)

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

(43) 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:

  • PROPER(eq(X1, X2)) → PROPER(X2)
    The graph contains the following edges 1 > 1

  • PROPER(eq(X1, X2)) → PROPER(X1)
    The graph contains the following edges 1 > 1

  • PROPER(s(X)) → PROPER(X)
    The graph contains the following edges 1 > 1

  • PROPER(inf(X)) → PROPER(X)
    The graph contains the following edges 1 > 1

  • PROPER(cons(any(X1), X2)) → PROPER(X1)
    The graph contains the following edges 1 > 1

  • PROPER(cons(any(X1), X2)) → PROPER(X2)
    The graph contains the following edges 1 > 1

  • PROPER(take(X1, X2)) → PROPER(X1)
    The graph contains the following edges 1 > 1

  • PROPER(take(X1, X2)) → PROPER(X2)
    The graph contains the following edges 1 > 1

  • PROPER(length(X)) → PROPER(X)
    The graph contains the following edges 1 > 1

(44) YES

(45) Obligation:

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

ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(46) 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.

(47) Obligation:

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

ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(length(X)) → ACTIVE(X)

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

(48) 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:

  • ACTIVE(take(X1, X2)) → ACTIVE(X1)
    The graph contains the following edges 1 > 1

  • ACTIVE(inf(X)) → ACTIVE(X)
    The graph contains the following edges 1 > 1

  • ACTIVE(take(X1, X2)) → ACTIVE(X2)
    The graph contains the following edges 1 > 1

  • ACTIVE(length(X)) → ACTIVE(X)
    The graph contains the following edges 1 > 1

(49) YES

(50) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(51) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 2·x1   
POL(any(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(eq(x1, x2)) = 2·x1 + 2·x2   
POL(false) = 0   
POL(inf(x1)) = 2·x1   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = 2·x1   
POL(proper(x1)) = x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(true) = 0   

(52) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))

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

(53) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(TOP(x1)) = 0 +
[1,1]
·x1

POL(ok(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(active(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(mark(x1)) =
/1\
\0/
 +
/00\
\11/
·x1

POL(proper(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(eq(x1, x2)) =
/1\
\0/
 +
/00\
\10/
·x1 +
/00\
\00/
·x2

POL(0) =
/1\
\0/

POL(true) =
/0\
\0/

POL(s(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

POL(false) =
/0\
\0/

POL(inf(x1)) =
/1\
\1/
 +
/00\
\11/
·x1

POL(cons(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(take(x1, x2)) =
/1\
\1/
 +
/00\
\11/
·x1 +
/00\
\11/
·x2

POL(nil) =
/1\
\1/

POL(length(x1)) =
/1\
\1/
 +
/00\
\11/
·x1

POL(any(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
s(ok(X)) → ok(s(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))

(54) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))

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

(55) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 1   
POL(TOP(x1)) = 2·x1   
POL(active(x1)) = 2 + 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(eq(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(false) = 1   
POL(inf(x1)) = x1   
POL(length(x1)) = 2 + 2·x1   
POL(mark(x1)) = 1 + x1   
POL(nil) = 0   
POL(ok(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(true) = 1   

(56) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(inf(X)) → inf(active(X))
active(length(X)) → length(active(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
s(ok(X)) → ok(s(X))

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

(57) 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.

(58) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(inf(X)) → inf(active(X))
active(length(X)) → length(active(X))
length(ok(X)) → ok(length(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))

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

(59) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

TOP(ok(X)) → TOP(active(X))

Strictly oriented rules of the TRS R:

active(inf(X)) → inf(active(X))
active(length(X)) → length(active(X))
length(ok(X)) → ok(length(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))

Used ordering: Knuth-Bendix order [KBO] with precedence:
active1 > TOP1 > inf1 > mark1 > length1 > ok1

and weight map:

active_1=1
inf_1=3
length_1=3
ok_1=2
mark_1=1
TOP_1=1

The variable weight is 1

(60) Obligation:

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

(61) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(62) YES