/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 67 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 12 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) RisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   first(X1, X2) -> n__first(X1, X2)
   from(X) -> n__from(X)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(X) -> X

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(activate(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(first(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(from(x_1)) = 2 + 2*x_1
   POL(n__first(x_1, x_2)) = x_1 + x_2
   POL(n__from(x_1)) = 1 + x_1
   POL(nil) = 0
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   from(X) -> n__from(X)




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   first(0, X) -> nil
   from(X) -> cons(X, n__from(s(X)))
   first(X1, X2) -> n__first(X1, X2)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(X) -> X

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(activate(x_1)) = 1 + 2*x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(first(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
   POL(from(x_1)) = 1 + 2*x_1
   POL(n__first(x_1, x_2)) = 2 + 2*x_1 + x_2
   POL(n__from(x_1)) = x_1
   POL(nil) = 1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   first(0, X) -> nil
   from(X) -> cons(X, n__from(s(X)))
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(X) -> X




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   first(X1, X2) -> n__first(X1, X2)
   activate(n__from(X)) -> from(X)

Q is empty.

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(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:activate_1 > from_1 > n__from_1 > first_2 > n__first_2

and weight map:

   activate_1=1
   n__from_1=1
   from_1=2
   first_2=0
   n__first_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   first(X1, X2) -> n__first(X1, X2)
   activate(n__from(X)) -> from(X)




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(6)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(7) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(8)
YES