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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 113 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(14) 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:
max/1)YES(
L/1)YES(
N/2(YES,YES)
0/0)
s/1(YES)

Quasi precedence:
N_2 > s_1
0 > s_1


Status:
N_2: multiset status
0: multiset status
s_1: multiset status

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




----------------------------------------

(2)
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.

----------------------------------------

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

(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)))))

The set Q consists of the following terms:

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


----------------------------------------

(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(6)
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.
----------------------------------------

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(8)
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.
----------------------------------------

(9) 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.
----------------------------------------

(10)
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.
----------------------------------------

(11) 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)))


----------------------------------------

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

(13) 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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(14)
YES