/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 85 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 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:

   p(s(x)) -> x
   s(p(x)) -> x
   +(0, y) -> y
   +(s(x), y) -> s(+(x, y))
   +(p(x), y) -> p(+(x, y))
   minus(0) -> 0
   minus(s(x)) -> p(minus(x))
   minus(p(x)) -> s(minus(x))
   *(0, y) -> 0
   *(s(x), y) -> +(*(x, y), y)
   *(p(x), y) -> +(*(x, y), minus(y))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
p/1(YES)
s/1(YES)
+/2(YES,YES)
0/0)
minus/1)YES(
*/2(YES,YES)

Quasi precedence:
*_2 > +_2 > [p_1, s_1] > 0


Status:
p_1: multiset status
s_1: multiset status
+_2: multiset status
0: multiset status
*_2: multiset status

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

   p(s(x)) -> x
   s(p(x)) -> x
   +(0, y) -> y
   +(s(x), y) -> s(+(x, y))
   +(p(x), y) -> p(+(x, y))
   *(0, y) -> 0
   *(s(x), y) -> +(*(x, y), y)
   *(p(x), y) -> +(*(x, y), minus(y))




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

   minus(0) -> 0
   minus(s(x)) -> p(minus(x))
   minus(p(x)) -> s(minus(x))

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:minus_1 > p_1 > 0 > s_1

and weight map:

   0=1
   minus_1=2
   s_1=1
   p_1=1

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   minus(0) -> 0
   minus(s(x)) -> p(minus(x))
   minus(p(x)) -> s(minus(x))




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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