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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSToCSRProof [SOUND, 0 ms]
(2) CSR
(3) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(4) QCSDP
(5) QCSDPForwardInstantiationProcessor [EQUIVALENT, 0 ms]
(6) QCSDP
(7) PIsEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
   active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
   f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
   f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
   proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
   proper(a) -> ok(a)
   proper(b) -> ok(b)
   proper(c) -> ok(c)
   f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(c)
   active(f(x0, x1, x2))
   f(mark(x0), x1, x2)
   f(x0, x1, mark(x2))
   proper(f(x0, x1, x2))
   proper(a)
   proper(b)
   proper(c)
   f(ok(x0), ok(x1), ok(x2))
   top(mark(x0))
   top(ok(x0))


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(1) QTRSToCSRProof (SOUND)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   active(f(X1, X2, X3)) -> f(active(X1), X2, X3)
   active(f(X1, X2, X3)) -> f(X1, X2, active(X3))
   f(mark(X1), X2, X3) -> mark(f(X1, X2, X3))
   f(X1, X2, mark(X3)) -> mark(f(X1, X2, X3))
   proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
   proper(a) -> ok(a)
   proper(b) -> ok(b)
   proper(c) -> ok(c)
   f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

The set Q consists of the following terms:

   active(c)
   active(f(x0, x1, x2))
   f(mark(x0), x1, x2)
   f(x0, x1, mark(x2))
   proper(f(x0, x1, x2))
   proper(a)
   proper(b)
   proper(c)
   f(ok(x0), ok(x1), ok(x2))
   top(mark(x0))
   top(ok(x0))

Special symbols used for the transformation (see [GM04]):
top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
The replacement map contains the following entries:

f: {1, 3}
a: empty set
b: empty set
c: empty set
The QTRS contained just a subset of rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is sound, but not necessarily complete.
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(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   f(a, b, X) -> f(X, X, X)
   c -> a
   c -> b

The replacement map contains the following entries:

f: {1, 3}
a: empty set
b: empty set
c: empty set

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(3) CSDependencyPairsProof (EQUIVALENT)
Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
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(4)
Obligation:
Q-restricted context-sensitive dependency pair problem:
For all symbols f in {f_3, F_3} we have mu(f) = {1, 3}.

The ordinary context-sensitive dependency pairs DP_o are:
   F(a, b, X) -> F(X, X, X)

The TRS R consists of the following rules:

   f(a, b, X) -> f(X, X, X)
   c -> a
   c -> b

Q is empty.

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(5) QCSDPForwardInstantiationProcessor (EQUIVALENT)
Using the Context-Sensitive Forward Instantiation[DA_EMMES] Processor

the pair F(a, b, X) -> F(X, X, X)

was transformed to the following new pairs: 
   F(a, b, b) -> F(b, b, b)



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(6)
Obligation:
Q-restricted context-sensitive dependency pair problem:
For all symbols f in {f_3} we have mu(f) = {1, 3}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   f(a, b, X) -> f(X, X, X)
   c -> a
   c -> b

Q is empty.

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(7) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
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(8)
YES