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


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 70 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

   a__first(0, X) -> nil
   a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
   a__from(X) -> cons(mark(X), from(s(X)))
   mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
   mark(from(X)) -> a__from(mark(X))
   mark(0) -> 0
   mark(nil) -> nil
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__first(X1, X2) -> first(X1, X2)
   a__from(X) -> from(X)

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__FIRST(s(X), cons(Y, Z)) -> MARK(Y)
   A__FROM(X) -> MARK(X)
   MARK(first(X1, X2)) -> A__FIRST(mark(X1), mark(X2))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)
   MARK(from(X)) -> A__FROM(mark(X))
   MARK(from(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__first(0, X) -> nil
   a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
   a__from(X) -> cons(mark(X), from(s(X)))
   mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
   mark(from(X)) -> a__from(mark(X))
   mark(0) -> 0
   mark(nil) -> nil
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__first(X1, X2) -> first(X1, X2)
   a__from(X) -> from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A__FIRST(s(X), cons(Y, Z)) -> MARK(Y)
   MARK(from(X)) -> A__FROM(mark(X))
   MARK(from(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__FIRST_2(x_1, x_2) ) = max{0, x_1 + x_2 - 1}
POL( A__FROM_1(x_1) ) = x_1
POL( mark_1(x_1) ) = 2x_1
POL( first_2(x_1, x_2) ) = 2x_1 + 2x_2
POL( a__first_2(x_1, x_2) ) = 2x_1 + 2x_2
POL( from_1(x_1) ) = 2x_1 + 2
POL( a__from_1(x_1) ) = 2x_1 + 2
POL( 0 ) = 1
POL( nil ) = 0
POL( s_1(x_1) ) = 2x_1 + 1
POL( cons_2(x_1, x_2) ) = x_1 + 1
POL( MARK_1(x_1) ) = x_1

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

   mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
   mark(from(X)) -> a__from(mark(X))
   mark(0) -> 0
   mark(nil) -> nil
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__from(X) -> from(X)
   a__first(0, X) -> nil
   a__first(X1, X2) -> first(X1, X2)
   a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
   a__from(X) -> cons(mark(X), from(s(X)))


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

   A__FROM(X) -> MARK(X)
   MARK(first(X1, X2)) -> A__FIRST(mark(X1), mark(X2))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   a__first(0, X) -> nil
   a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
   a__from(X) -> cons(mark(X), from(s(X)))
   mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
   mark(from(X)) -> a__from(mark(X))
   mark(0) -> 0
   mark(nil) -> nil
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__first(X1, X2) -> first(X1, X2)
   a__from(X) -> from(X)

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

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

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

   MARK(first(X1, X2)) -> MARK(X2)
   MARK(first(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__first(0, X) -> nil
   a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
   a__from(X) -> cons(mark(X), from(s(X)))
   mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
   mark(from(X)) -> a__from(mark(X))
   mark(0) -> 0
   mark(nil) -> nil
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__first(X1, X2) -> first(X1, X2)
   a__from(X) -> from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) 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.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(first(X1, X2)) -> MARK(X2)
   MARK(first(X1, X2)) -> MARK(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) 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:
*MARK(first(X1, X2)) -> MARK(X2)
The graph contains the following edges 1 > 1


*MARK(first(X1, X2)) -> MARK(X1)
The graph contains the following edges 1 > 1


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(10)
YES