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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 35 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 73 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 28 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

   1(2(1(x1))) -> 2(0(2(x1)))
   0(2(1(x1))) -> 1(0(2(x1)))
   L(2(1(x1))) -> L(1(0(2(x1))))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(0(x1))) -> 1(0(1(x1)))
   L(2(0(x1))) -> L(1(0(1(x1))))
   0(2(R(x1))) -> 1(0(1(R(x1))))

Q is empty.

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

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

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(1(x1))) -> 1^1(0(2(x1)))
   0^1(2(1(x1))) -> 0^1(2(x1))
   L^1(2(1(x1))) -> L^1(1(0(2(x1))))
   L^1(2(1(x1))) -> 1^1(0(2(x1)))
   L^1(2(1(x1))) -> 0^1(2(x1))
   1^1(2(0(x1))) -> 0^1(1(x1))
   1^1(2(0(x1))) -> 1^1(x1)
   1^1(2(R(x1))) -> 0^1(1(R(x1)))
   1^1(2(R(x1))) -> 1^1(R(x1))
   0^1(2(0(x1))) -> 1^1(0(1(x1)))
   0^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 1^1(x1)
   L^1(2(0(x1))) -> L^1(1(0(1(x1))))
   L^1(2(0(x1))) -> 1^1(0(1(x1)))
   L^1(2(0(x1))) -> 0^1(1(x1))
   L^1(2(0(x1))) -> 1^1(x1)
   0^1(2(R(x1))) -> 1^1(0(1(R(x1))))
   0^1(2(R(x1))) -> 0^1(1(R(x1)))
   0^1(2(R(x1))) -> 1^1(R(x1))

The TRS R consists of the following rules:

   1(2(1(x1))) -> 2(0(2(x1)))
   0(2(1(x1))) -> 1(0(2(x1)))
   L(2(1(x1))) -> L(1(0(2(x1))))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(0(x1))) -> 1(0(1(x1)))
   L(2(0(x1))) -> L(1(0(1(x1))))
   0(2(R(x1))) -> 1(0(1(R(x1))))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 10 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   0^1(2(1(x1))) -> 1^1(0(2(x1)))
   1^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(0(x1))) -> 1^1(0(1(x1)))
   1^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 1^1(x1)
   1^1(2(0(x1))) -> 1^1(x1)

The TRS R consists of the following rules:

   1(2(1(x1))) -> 2(0(2(x1)))
   0(2(1(x1))) -> 1(0(2(x1)))
   L(2(1(x1))) -> L(1(0(2(x1))))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(0(x1))) -> 1(0(1(x1)))
   L(2(0(x1))) -> L(1(0(1(x1))))
   0(2(R(x1))) -> 1(0(1(R(x1))))

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

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   0^1(2(1(x1))) -> 1^1(0(2(x1)))
   1^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(0(x1))) -> 1^1(0(1(x1)))
   1^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 1^1(x1)
   1^1(2(0(x1))) -> 1^1(x1)

The TRS R consists of the following rules:

   1(2(1(x1))) -> 2(0(2(x1)))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(1(x1))) -> 1(0(2(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   0(2(R(x1))) -> 1(0(1(R(x1))))

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

(8) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   0^1(2(1(x1))) -> 1^1(0(2(x1)))
   1^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(1(x1))) -> 0^1(2(x1))
   0^1(2(0(x1))) -> 1^1(0(1(x1)))
   1^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 0^1(1(x1))
   0^1(2(0(x1))) -> 1^1(x1)
   1^1(2(0(x1))) -> 1^1(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = x_1
   POL(0^1(x_1)) = 2 + 2*x_1
   POL(1(x_1)) = 1 + x_1
   POL(1^1(x_1)) = 1 + 2*x_1
   POL(2(x_1)) = 2 + x_1
   POL(R(x_1)) = x_1


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

(9)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   1(2(1(x1))) -> 2(0(2(x1)))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(1(x1))) -> 1(0(2(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   0(2(R(x1))) -> 1(0(1(R(x1))))

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

(10) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(11)
YES

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

   L^1(2(0(x1))) -> L^1(1(0(1(x1))))
   L^1(2(1(x1))) -> L^1(1(0(2(x1))))

The TRS R consists of the following rules:

   1(2(1(x1))) -> 2(0(2(x1)))
   0(2(1(x1))) -> 1(0(2(x1)))
   L(2(1(x1))) -> L(1(0(2(x1))))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(0(x1))) -> 1(0(1(x1)))
   L(2(0(x1))) -> L(1(0(1(x1))))
   0(2(R(x1))) -> 1(0(1(R(x1))))

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

(13) 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.
----------------------------------------

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

   L^1(2(0(x1))) -> L^1(1(0(1(x1))))
   L^1(2(1(x1))) -> L^1(1(0(2(x1))))

The TRS R consists of the following rules:

   0(2(1(x1))) -> 1(0(2(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   0(2(R(x1))) -> 1(0(1(R(x1))))
   1(2(1(x1))) -> 2(0(2(x1)))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))

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

(15) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   L^1(2(0(x1))) -> L^1(1(0(1(x1))))
   L^1(2(1(x1))) -> L^1(1(0(2(x1))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 0
   POL(1(x_1)) = x_1
   POL(2(x_1)) = 1
   POL(L^1(x_1)) = x_1
   POL(R(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   1(2(1(x1))) -> 2(0(2(x1)))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))
   0(2(1(x1))) -> 1(0(2(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   0(2(R(x1))) -> 1(0(1(R(x1))))


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

(16)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   0(2(1(x1))) -> 1(0(2(x1)))
   0(2(0(x1))) -> 1(0(1(x1)))
   0(2(R(x1))) -> 1(0(1(R(x1))))
   1(2(1(x1))) -> 2(0(2(x1)))
   1(2(0(x1))) -> 2(0(1(x1)))
   1(2(R(x1))) -> 2(0(1(R(x1))))

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

(17) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(18)
YES