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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 111 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(c(c(a, y, a), b(x, z), a)) -> b(y, f(c(f(a), z, z)))
   f(b(b(x, f(y)), z)) -> c(z, x, f(b(b(f(a), y), y)))
   c(b(a, a), b(y, z), x) -> b(a, b(z, z))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
[f_1, c_3] > [a, b_2]


Status:
f_1: [1]
c_3: [1,3,2]
a: multiset status
b_2: [2,1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(c(c(a, y, a), b(x, z), a)) -> b(y, f(c(f(a), z, z)))
   f(b(b(x, f(y)), z)) -> c(z, x, f(b(b(f(a), y), y)))
   c(b(a, a), b(y, z), x) -> b(a, b(z, z))




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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