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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSToCSRProof [EQUIVALENT, 0 ms]
(2) CSR
(3) CSRInnermostProof [EQUIVALENT, 0 ms]
(4) CSR
(5) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(6) QCSDP
(7) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QCSDP
        (10) QCSDPSubtermProof [EQUIVALENT, 5 ms]
        (11) QCSDP
        (12) PIsEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QCSDP
        (15) QCSDPSubtermProof [EQUIVALENT, 4 ms]
        (16) QCSDP
        (17) PIsEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(head(cons(X, XS))) -> mark(X)
   active(2nd(cons(X, XS))) -> mark(head(XS))
   active(take(0, XS)) -> mark(nil)
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(sel(0, cons(X, XS))) -> mark(X)
   active(sel(s(N), cons(X, XS))) -> mark(sel(N, XS))
   active(from(X)) -> from(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(s(X)) -> s(active(X))
   active(head(X)) -> head(active(X))
   active(2nd(X)) -> 2nd(active(X))
   active(take(X1, X2)) -> take(active(X1), X2)
   active(take(X1, X2)) -> take(X1, active(X2))
   active(sel(X1, X2)) -> sel(active(X1), X2)
   active(sel(X1, X2)) -> sel(X1, active(X2))
   from(mark(X)) -> mark(from(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   s(mark(X)) -> mark(s(X))
   head(mark(X)) -> mark(head(X))
   2nd(mark(X)) -> mark(2nd(X))
   take(mark(X1), X2) -> mark(take(X1, X2))
   take(X1, mark(X2)) -> mark(take(X1, X2))
   sel(mark(X1), X2) -> mark(sel(X1, X2))
   sel(X1, mark(X2)) -> mark(sel(X1, X2))
   proper(from(X)) -> from(proper(X))
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(s(X)) -> s(proper(X))
   proper(head(X)) -> head(proper(X))
   proper(2nd(X)) -> 2nd(proper(X))
   proper(take(X1, X2)) -> take(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(nil) -> ok(nil)
   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
   from(ok(X)) -> ok(from(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   s(ok(X)) -> ok(s(X))
   head(ok(X)) -> ok(head(X))
   2nd(ok(X)) -> ok(2nd(X))
   take(ok(X1), ok(X2)) -> ok(take(X1, X2))
   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.

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

(1) QTRSToCSRProof (EQUIVALENT)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(head(cons(X, XS))) -> mark(X)
   active(2nd(cons(X, XS))) -> mark(head(XS))
   active(take(0, XS)) -> mark(nil)
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(sel(0, cons(X, XS))) -> mark(X)
   active(sel(s(N), cons(X, XS))) -> mark(sel(N, XS))
   active(from(X)) -> from(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(s(X)) -> s(active(X))
   active(head(X)) -> head(active(X))
   active(2nd(X)) -> 2nd(active(X))
   active(take(X1, X2)) -> take(active(X1), X2)
   active(take(X1, X2)) -> take(X1, active(X2))
   active(sel(X1, X2)) -> sel(active(X1), X2)
   active(sel(X1, X2)) -> sel(X1, active(X2))
   from(mark(X)) -> mark(from(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   s(mark(X)) -> mark(s(X))
   head(mark(X)) -> mark(head(X))
   2nd(mark(X)) -> mark(2nd(X))
   take(mark(X1), X2) -> mark(take(X1, X2))
   take(X1, mark(X2)) -> mark(take(X1, X2))
   sel(mark(X1), X2) -> mark(sel(X1, X2))
   sel(X1, mark(X2)) -> mark(sel(X1, X2))
   proper(from(X)) -> from(proper(X))
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(s(X)) -> s(proper(X))
   proper(head(X)) -> head(proper(X))
   proper(2nd(X)) -> 2nd(proper(X))
   proper(take(X1, X2)) -> take(proper(X1), proper(X2))
   proper(0) -> ok(0)
   proper(nil) -> ok(nil)
   proper(sel(X1, X2)) -> sel(proper(X1), proper(X2))
   from(ok(X)) -> ok(from(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   s(ok(X)) -> ok(s(X))
   head(ok(X)) -> ok(head(X))
   2nd(ok(X)) -> ok(2nd(X))
   take(ok(X1), ok(X2)) -> ok(take(X1, X2))
   sel(ok(X1), ok(X2)) -> ok(sel(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
The replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
head: {1}
2nd: {1}
take: {1, 2}
0: empty set
nil: empty set
sel: {1, 2}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
----------------------------------------

(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
head: {1}
2nd: {1}
take: {1, 2}
0: empty set
nil: empty set
sel: {1, 2}

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(3) CSRInnermostProof (EQUIVALENT)
The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
----------------------------------------

(4)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
head: {1}
2nd: {1}
take: {1, 2}
0: empty set
nil: empty set
sel: {1, 2}


Innermost Strategy.

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(5) CSDependencyPairsProof (EQUIVALENT)
Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
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(6)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2, HEAD_1, 2ND_1, SEL_2, FROM_1, TAKE_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.
The symbols in {U_1} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:
   2ND(cons(X, XS)) -> HEAD(XS)
   SEL(s(N), cons(X, XS)) -> SEL(N, XS)

The collapsing dependency pairs are DP_c:
   2ND(cons(X, XS)) -> XS
   SEL(s(N), cons(X, XS)) -> XS


The hidden terms of R are:

   from(s(x0))
   take(x0, x1)

Every hiding context is built from:
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@1e55042e
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@12caaa75
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@3d8f544e

Hence, the new unhiding pairs DP_u are :
   2ND(cons(X, XS)) -> U(XS)
   SEL(s(N), cons(X, XS)) -> U(XS)
   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_1)
   U(from(s(x0))) -> FROM(s(x0))
   U(take(x0, x1)) -> TAKE(x0, x1)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The set Q consists of the following terms:

   from(x0)
   head(cons(x0, x1))
   2nd(cons(x0, x1))
   take(0, x0)
   take(s(x0), cons(x1, x2))
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))


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(7) QCSDependencyGraphProof (EQUIVALENT)
The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 2 SCCs with 4 less nodes.

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(8)
Complex Obligation (AND)

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(9)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.
The symbols in {U_1} are not replacing on any position.

The TRS P consists of the following rules:

   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_1)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The set Q consists of the following terms:

   from(x0)
   head(cons(x0, x1))
   2nd(cons(x0, x1))
   take(0, x0)
   take(s(x0), cons(x1, x2))
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))


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(10) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_0)
   U(take(x_0, x_1)) -> U(x_1)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
U(x1)  =  x1


Subterm Order

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(11)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The set Q consists of the following terms:

   from(x0)
   head(cons(x0, x1))
   2nd(cons(x0, x1))
   take(0, x0)
   take(s(x0), cons(x1, x2))
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))


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(12) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(13)
YES

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(14)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2, SEL_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:

   SEL(s(N), cons(X, XS)) -> SEL(N, XS)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The set Q consists of the following terms:

   from(x0)
   head(cons(x0, x1))
   2nd(cons(x0, x1))
   take(0, x0)
   take(s(x0), cons(x1, x2))
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))


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

(15) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   SEL(s(N), cons(X, XS)) -> SEL(N, XS)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
SEL(x1, x2)  =  x1


Subterm Order

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(16)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {from_1, s_1, head_1, 2nd_1, take_2, sel_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   head(cons(X, XS)) -> X
   2nd(cons(X, XS)) -> head(XS)
   take(0, XS) -> nil
   take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)

The set Q consists of the following terms:

   from(x0)
   head(cons(x0, x1))
   2nd(cons(x0, x1))
   take(0, x0)
   take(s(x0), cons(x1, x2))
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))


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(17) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(18)
YES