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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 39 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 10 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

   1(0(x1)) -> 0(0(0(1(x1))))
   0(1(x1)) -> 1(x1)
   1(1(x1)) -> 0(0(0(0(x1))))
   0(0(x1)) -> 0(x1)

Q is empty.

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

   POL(0(x_1)) = x_1
   POL(1(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   1(1(x1)) -> 0(0(0(0(x1))))




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

   1(0(x1)) -> 0(0(0(1(x1))))
   0(1(x1)) -> 1(x1)
   0(0(x1)) -> 0(x1)

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(0(x1)) -> 0^1(0(0(1(x1))))
   1^1(0(x1)) -> 0^1(0(1(x1)))
   1^1(0(x1)) -> 0^1(1(x1))
   1^1(0(x1)) -> 1^1(x1)

The TRS R consists of the following rules:

   1(0(x1)) -> 0(0(0(1(x1))))
   0(1(x1)) -> 1(x1)
   0(0(x1)) -> 0(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(0(x1)) -> 1^1(x1)

The TRS R consists of the following rules:

   1(0(x1)) -> 0(0(0(1(x1))))
   0(1(x1)) -> 1(x1)
   0(0(x1)) -> 0(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(0(x1)) -> 1^1(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) 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:
*1^1(0(x1)) -> 1^1(x1)
The graph contains the following edges 1 > 1


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(10)
YES