/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 73 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 16 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QReductionProof [EQUIVALENT, 0 ms]
(14) QDP
(15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(16) YES


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

   max(L(x)) -> x
   max(N(L(0), L(y))) -> y
   max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 2
   POL(L(x_1)) = x_1
   POL(N(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(max(x_1)) = x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   max(N(L(0), L(y))) -> y




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

   max(L(x)) -> x
   max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(L(x_1)) = x_1
   POL(N(x_1, x_2)) = x_1 + x_2
   POL(max(x_1)) = x_1
   POL(s(x_1)) = 1 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))




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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   max(L(x)) -> x
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Q is empty.

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(5) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   max(L(x)) -> x
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

   max(L(x0))
   max(N(L(x0), N(x1, x2)))


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAX(N(L(x), N(y, z))) -> MAX(N(L(x), L(max(N(y, z)))))
   MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

The TRS R consists of the following rules:

   max(L(x)) -> x
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

   max(L(x0))
   max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

The TRS R consists of the following rules:

   max(L(x)) -> x
   max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

The set Q consists of the following terms:

   max(L(x0))
   max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

R is empty.
The set Q consists of the following terms:

   max(L(x0))
   max(N(L(x0), N(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   max(L(x0))
   max(N(L(x0), N(x1, x2)))


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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MAX(N(L(x), N(y, z))) -> MAX(N(y, z))
The graph contains the following edges 1 > 1


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(16)
YES