YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 27 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:

   f(X) -> if(X, c, n__f(true))
   if(true, X, Y) -> X
   if(false, X, Y) -> activate(Y)
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:f_1 > if_3 > activate_1 > false > n__f_1 > true > c

and weight map:

   c=1
   true=1
   false=2
   f_1=3
   n__f_1=1
   activate_1=3
   if_3=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:

   f(X) -> if(X, c, n__f(true))
   if(true, X, Y) -> X
   if(false, X, Y) -> activate(Y)
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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