/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) QDP
(17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(18) YES


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

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

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> 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:

   U11^1(tt, N, X, XS) -> U12^1(splitAt(activate(N), activate(XS)), activate(X))
   U11^1(tt, N, X, XS) -> SPLITAT(activate(N), activate(XS))
   U11^1(tt, N, X, XS) -> ACTIVATE(N)
   U11^1(tt, N, X, XS) -> ACTIVATE(XS)
   U11^1(tt, N, X, XS) -> ACTIVATE(X)
   U12^1(pair(YS, ZS), X) -> ACTIVATE(X)
   AFTERNTH(N, XS) -> SND(splitAt(N, XS))
   AFTERNTH(N, XS) -> SPLITAT(N, XS)
   AND(tt, X) -> ACTIVATE(X)
   SEL(N, XS) -> HEAD(afterNth(N, XS))
   SEL(N, XS) -> AFTERNTH(N, XS)
   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   SPLITAT(s(N), cons(X, XS)) -> ACTIVATE(XS)
   TAIL(cons(N, XS)) -> ACTIVATE(XS)
   TAKE(N, XS) -> FST(splitAt(N, XS))
   TAKE(N, XS) -> SPLITAT(N, XS)
   ACTIVATE(n__natsFrom(X)) -> NATSFROM(X)

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> 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 1 SCC with 15 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U11^1(tt, N, X, XS) -> SPLITAT(activate(N), activate(XS))
   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(5) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule U11^1(tt, N, X, XS) -> SPLITAT(activate(N), activate(XS)) at position [0] we obtained the following new rules [LPAR04]:

   (U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(natsFrom(x0), activate(y2)),U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(natsFrom(x0), activate(y2)))
   (U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2)),U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2)))


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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(natsFrom(x0), activate(y2))
   U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(7) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(natsFrom(x0), activate(y2)) at position [0] we obtained the following new rules [LPAR04]:

   (U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(cons(x0, n__natsFrom(s(x0))), activate(y2)),U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(cons(x0, n__natsFrom(s(x0))), activate(y2)))
   (U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(n__natsFrom(x0), activate(y2)),U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(n__natsFrom(x0), activate(y2)))


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

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2))
   U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(cons(x0, n__natsFrom(s(x0))), activate(y2))
   U11^1(tt, n__natsFrom(x0), y1, y2) -> SPLITAT(n__natsFrom(x0), activate(y2))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2))
   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(11) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule U11^1(tt, x0, y1, y2) -> SPLITAT(x0, activate(y2)) at position [1] we obtained the following new rules [LPAR04]:

   (U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, natsFrom(x0)),U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, natsFrom(x0)))
   (U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0),U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0))


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, natsFrom(x0))
   U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0)

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(13) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, natsFrom(x0)) at position [1] we obtained the following new rules [LPAR04]:

   (U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, cons(x0, n__natsFrom(s(x0)))),U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, cons(x0, n__natsFrom(s(x0)))))
   (U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, n__natsFrom(x0)),U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, n__natsFrom(x0)))


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0)
   U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))
   U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, n__natsFrom(x0))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(15) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0)
   SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
   U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))

The TRS R consists of the following rules:

   U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X))
   U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS)
   afterNth(N, XS) -> snd(splitAt(N, XS))
   and(tt, X) -> activate(X)
   fst(pair(X, Y)) -> X
   head(cons(N, XS)) -> N
   natsFrom(N) -> cons(N, n__natsFrom(s(N)))
   sel(N, XS) -> head(afterNth(N, XS))
   snd(pair(X, Y)) -> Y
   splitAt(0, XS) -> pair(nil, XS)
   splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS))
   tail(cons(N, XS)) -> activate(XS)
   take(N, XS) -> fst(splitAt(N, XS))
   natsFrom(X) -> n__natsFrom(X)
   activate(n__natsFrom(X)) -> natsFrom(X)
   activate(X) -> X

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

(17) 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:
*SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS))
The graph contains the following edges 1 > 2, 2 > 3


*U11^1(tt, y0, y1, x0) -> SPLITAT(y0, x0)
The graph contains the following edges 2 >= 1, 4 >= 2


*U11^1(tt, y0, y1, n__natsFrom(x0)) -> SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))
The graph contains the following edges 2 >= 1


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

(18)
YES