/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 64 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


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

   fst(0, Z) -> nil
   fst(s, cons(Y)) -> cons(Y)
   from(X) -> cons(X)
   add(0, X) -> X
   add(s, Y) -> s
   len(nil) -> 0
   len(cons(X)) -> s

The set Q consists of the following terms:

   fst(0, x0)
   fst(s, cons(x0))
   from(x0)
   add(0, x0)
   add(s, x0)
   len(nil)
   len(cons(x0))


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

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

   fst(0, Z) -> nil
   fst(s, cons(Y)) -> cons(Y)
   from(X) -> cons(X)
   add(0, X) -> X
   add(s, Y) -> s
   len(nil) -> 0
   len(cons(X)) -> s




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   fst(0, x0)
   fst(s, cons(x0))
   from(x0)
   add(0, x0)
   add(s, x0)
   len(nil)
   len(cons(x0))


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(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(4)
YES