/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES
    (30) QDP
        (31) UsableRulesProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (39) YES
    (40) QDP
        (41) UsableRulesProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (44) YES
    (45) QDP
        (46) UsableRulesProof [EQUIVALENT, 0 ms]
        (47) QDP
        (48) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (49) YES
    (50) QDP
        (51) UsableRulesReductionPairsProof [EQUIVALENT, 66 ms]
        (52) QDP
        (53) QDPOrderProof [EQUIVALENT, 91 ms]
        (54) QDP
        (55) UsableRulesReductionPairsProof [EQUIVALENT, 21 ms]
        (56) QDP
        (57) MRRProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) MRRProof [EQUIVALENT, 14 ms]
        (62) QDP
        (63) PisEmptyProof [EQUIVALENT, 0 ms]
        (64) 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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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(x_1)) = x_1
   POL(active(x_1)) = 2*x_1
   POL(any(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(false) = 0
   POL(inf(x_1)) = 2*x_1
   POL(length(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(ok(x_1)) = 2*x_1
   POL(proper(x_1)) = x_1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = 2*x_1 + x_2
   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)
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(x_1)) =  	[[0]] 	 +  	[[1, 0]] 	* 	x_1
>>>

   <<<
 POL(ok(x_1)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1
>>>

   <<<
 POL(active(x_1)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(proper(x_1)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1
>>>

   <<<
 POL(eq(x_1, x_2)) =  	[[1], [0]] 	 +  	[[0, 1], [0, 0]] 	* 	x_1 	 +  	[[0, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[0], [0]]
>>>

   <<<
 POL(true) =  	[[0], [0]]
>>>

   <<<
 POL(s(x_1)) =  	[[0], [1]] 	 +  	[[0, 0], [0, 1]] 	* 	x_1
>>>

   <<<
 POL(false) =  	[[0], [0]]
>>>

   <<<
 POL(inf(x_1)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1 	 +  	[[0, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(take(x_1, x_2)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1 	 +  	[[1, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(nil) =  	[[0], [0]]
>>>

   <<<
 POL(length(x_1)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(any(x_1)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1
>>>


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)
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))
   any(X) -> s(X)
   any(proper(X)) -> any(any(any(X)))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(TOP(x_1)) = x_1
   POL(active(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(eq(x_1, x_2)) = x_1 + 2*x_2
   POL(false) = 0
   POL(inf(x_1)) = 2*x_1
   POL(length(x_1)) = x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(ok(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = x_1 + 2*x_2
   POL(true) = 0


----------------------------------------

(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(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))
   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))
   inf(mark(X)) -> mark(inf(X))
   inf(ok(X)) -> ok(inf(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   eq(ok(X1), ok(X2)) -> ok(eq(X1, X2))

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

(57) MRRProof (EQUIVALENT)
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 rules of the TRS R:

   active(eq(0, 0)) -> mark(true)
   active(length(nil)) -> mark(0)

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(TOP(x_1)) = 2*x_1
   POL(active(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = 2*x_1 + x_2
   POL(false) = 0
   POL(inf(x_1)) = x_1
   POL(length(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 2
   POL(ok(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = x_1 + 2*x_2
   POL(true) = 1


----------------------------------------

(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(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(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))
   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))
   inf(mark(X)) -> mark(inf(X))
   inf(ok(X)) -> ok(inf(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   eq(ok(X1), ok(X2)) -> ok(eq(X1, X2))

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

(59) UsableRulesReductionPairsProof (EQUIVALENT)
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:

   active(take(0, X)) -> mark(nil)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(TOP(x_1)) = x_1
   POL(active(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = x_1 + x_2
   POL(false) = 0
   POL(inf(x_1)) = 2*x_1
   POL(length(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 2
   POL(ok(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = 2*x_1 + x_2


----------------------------------------

(60)
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(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(s(X), cons(Y, L))) -> mark(cons(Y, take(X, L)))
   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))
   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))
   inf(mark(X)) -> mark(inf(X))
   inf(ok(X)) -> ok(inf(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   eq(ok(X1), ok(X2)) -> ok(eq(X1, X2))

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

(61) MRRProof (EQUIVALENT)
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(eq(s(X), s(Y))) -> mark(eq(X, Y))
   active(eq(X, Y)) -> mark(false)
   active(take(s(X), cons(Y, L))) -> mark(cons(Y, take(X, L)))
   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))
   active(length(X)) -> length(active(X))
   take(ok(X1), ok(X2)) -> ok(take(X1, X2))
   inf(ok(X)) -> ok(inf(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   eq(ok(X1), ok(X2)) -> ok(eq(X1, X2))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP(x_1)) = x_1
   POL(active(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(false) = 1
   POL(inf(x_1)) = 2*x_1
   POL(length(x_1)) = 2 + 2*x_1
   POL(mark(x_1)) = x_1
   POL(ok(x_1)) = 2 + 2*x_1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = 1 + 2*x_1 + 2*x_2


----------------------------------------

(62)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   active(inf(X)) -> mark(cons(X, inf(s(X))))
   active(inf(X)) -> inf(active(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))
   inf(mark(X)) -> mark(inf(X))
   s(ok(X)) -> ok(s(X))

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

(63) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(64)
YES