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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 34 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:

   a__U11(tt, M, N) -> a__U12(tt, M, N)
   a__U12(tt, M, N) -> s(a__plus(mark(N), mark(M)))
   a__plus(N, 0) -> mark(N)
   a__plus(N, s(M)) -> a__U11(tt, M, N)
   mark(U11(X1, X2, X3)) -> a__U11(mark(X1), X2, X3)
   mark(U12(X1, X2, X3)) -> a__U12(mark(X1), X2, X3)
   mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
   mark(tt) -> tt
   mark(s(X)) -> s(mark(X))
   mark(0) -> 0
   a__U11(X1, X2, X3) -> U11(X1, X2, X3)
   a__U12(X1, X2, X3) -> U12(X1, X2, X3)
   a__plus(X1, X2) -> plus(X1, X2)

The set Q consists of the following terms:

   mark(U11(x0, x1, x2))
   mark(U12(x0, x1, x2))
   mark(plus(x0, x1))
   mark(tt)
   mark(s(x0))
   mark(0)
   a__U11(x0, x1, x2)
   a__U12(x0, x1, x2)
   a__plus(x0, x1)


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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:mark_1 > a__plus_2 > plus_2 > tt > a__U11_3 > a__U12_3 > U12_3 > U11_3 > s_1 > 0

and weight map:

   tt=2
   0=1
   s_1=2
   mark_1=0
   a__U11_3=0
   a__U12_3=0
   a__plus_2=0
   U11_3=0
   U12_3=0
   plus_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:

   a__U11(tt, M, N) -> a__U12(tt, M, N)
   a__U12(tt, M, N) -> s(a__plus(mark(N), mark(M)))
   a__plus(N, 0) -> mark(N)
   a__plus(N, s(M)) -> a__U11(tt, M, N)
   mark(U11(X1, X2, X3)) -> a__U11(mark(X1), X2, X3)
   mark(U12(X1, X2, X3)) -> a__U12(mark(X1), X2, X3)
   mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
   mark(tt) -> tt
   mark(s(X)) -> s(mark(X))
   mark(0) -> 0
   a__U11(X1, X2, X3) -> U11(X1, X2, X3)
   a__U12(X1, X2, X3) -> U12(X1, X2, X3)
   a__plus(X1, X2) -> plus(X1, X2)




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   mark(U11(x0, x1, x2))
   mark(U12(x0, x1, x2))
   mark(plus(x0, x1))
   mark(tt)
   mark(s(x0))
   mark(0)
   a__U11(x0, x1, x2)
   a__U12(x0, x1, x2)
   a__plus(x0, x1)


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