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


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NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given RelTRS could be disproven:

(0) RelTRS
(1) RelTRSLoopFinderProof [COMPLETE, 6678 ms]
(2) NO


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(0)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

   top(ok(U(x, y))) -> top(check(D(x, y)))
   D(x, B) -> U(x, B)
   F(x, U(O(y), z)) -> U(x, F(y, z))
   F(x, U(N(y), z)) -> U(x, F(y, z))
   D(O(x), F(y, z)) -> F(x, D(y, z))
   D(N(x), F(y, z)) -> F(x, D(y, z))
   F(x, U(E, y)) -> U(x, F(E, y))
   D(E, F(x, y)) -> F(E, D(x, y))

The relative TRS consists of the following S rules:

   U(O(x), y) -> U(x, y)
   U(N(x), y) -> U(x, y)
   D(O(x), y) -> D(x, y)
   D(N(x), y) -> D(x, y)
   E -> N(E)
   check(O(x)) -> ok(O(x))
   check(U(x, y)) -> U(check(x), y)
   check(U(x, y)) -> U(x, check(y))
   check(D(x, y)) -> D(check(x), y)
   check(D(x, y)) -> D(x, check(y))
   check(F(x, y)) -> F(check(x), y)
   check(F(x, y)) -> F(x, check(y))
   check(O(x)) -> O(check(x))
   check(N(x)) -> N(check(x))
   U(ok(x), y) -> ok(U(x, y))
   U(x, ok(y)) -> ok(U(x, y))
   D(ok(x), y) -> ok(D(x, y))
   D(x, ok(y)) -> ok(D(x, y))
   F(ok(x), y) -> ok(F(x, y))
   F(x, ok(y)) -> ok(F(x, y))
   O(ok(x)) -> ok(O(x))
   N(ok(x)) -> ok(N(x))


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(1) RelTRSLoopFinderProof (COMPLETE)
The following loop was found:

---------- Loop: ----------

top(check(D(O(x), B))) -> top(D(check(O(x)), B)) with rule check(D(x', y)) -> D(check(x'), y) at position [0] and matcher [x' / O(x), y / B]

top(D(check(O(x)), B)) -> top(D(ok(O(x)), B)) with rule check(O(x')) -> ok(O(x')) at position [0,0] and matcher [x' / x]

top(D(ok(O(x)), B)) -> top(U(ok(O(x)), B)) with rule D(x', B) -> U(x', B) at position [0] and matcher [x' / ok(O(x))]

top(U(ok(O(x)), B)) -> top(ok(U(O(x), B))) with rule U(ok(x'), y') -> ok(U(x', y')) at position [0] and matcher [x' / O(x), y' / B]

top(ok(U(O(x), B))) -> top(check(D(O(x), B))) with rule top(ok(U(x', y))) -> top(check(D(x', y))) at position [] and matcher [x' / O(x), y / B]

Now an instance of the first term with Matcher [ ] occurs in the last term at position [].

Context: []

Therefore, the relative TRS problem does not terminate.
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(2)
NO