YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 76 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 2 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   active(f(X1, X2)) -> f(active(X1), X2)
   active(g(X)) -> g(active(X))
   f(mark(X1), X2) -> mark(f(X1, X2))
   g(mark(X)) -> mark(g(X))
   proper(f(X1, X2)) -> f(proper(X1), proper(X2))
   proper(g(X)) -> g(proper(X))
   f(ok(X1), ok(X2)) -> ok(f(X1, X2))
   g(ok(X)) -> ok(g(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(f(x0, x1))
   active(g(x0))
   f(mark(x0), x1)
   g(mark(x0))
   proper(f(x0, x1))
   proper(g(x0))
   f(ok(x0), ok(x1))
   g(ok(x0))
   top(mark(x0))
   top(ok(x0))


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

   POL(active(x_1)) = 1 + 2*x_1
   POL(f(x_1, x_2)) = x_1 + x_2
   POL(g(x_1)) = x_1
   POL(mark(x_1)) = x_1
   POL(ok(x_1)) = 1 + 2*x_1
   POL(proper(x_1)) = x_1
   POL(top(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   f(ok(X1), ok(X2)) -> ok(f(X1, X2))




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

   active(f(X1, X2)) -> f(active(X1), X2)
   active(g(X)) -> g(active(X))
   f(mark(X1), X2) -> mark(f(X1, X2))
   g(mark(X)) -> mark(g(X))
   proper(f(X1, X2)) -> f(proper(X1), proper(X2))
   proper(g(X)) -> g(proper(X))
   g(ok(X)) -> ok(g(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(f(x0, x1))
   active(g(x0))
   f(mark(x0), x1)
   g(mark(x0))
   proper(f(x0, x1))
   proper(g(x0))
   f(ok(x0), ok(x1))
   g(ok(x0))
   top(mark(x0))
   top(ok(x0))


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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:proper_1 > top_1 > active_1 > g_1 > ok_1 > f_2 > mark_1

and weight map:

   active_1=2
   g_1=4
   mark_1=1
   proper_1=0
   ok_1=3
   top_1=1
   f_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:

   active(f(X1, X2)) -> f(active(X1), X2)
   active(g(X)) -> g(active(X))
   f(mark(X1), X2) -> mark(f(X1, X2))
   g(mark(X)) -> mark(g(X))
   proper(f(X1, X2)) -> f(proper(X1), proper(X2))
   proper(g(X)) -> g(proper(X))
   g(ok(X)) -> ok(g(X))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   active(f(x0, x1))
   active(g(x0))
   f(mark(x0), x1)
   g(mark(x0))
   proper(f(x0, x1))
   proper(g(x0))
   f(ok(x0), ok(x1))
   g(ok(x0))
   top(mark(x0))
   top(ok(x0))


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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES