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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 68 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 22 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 26 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 11 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(x1))
   #(#(x1)) -> #(x1)
   #(*(x1)) -> *(x1)

Q is empty.

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

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

   #(#(x1)) -> #(x1)
   #(*(x1)) -> *(x1)




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

   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(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:

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

The TRS R consists of the following rules:

   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(x1))

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

(5) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(#(x_1)) = x_1
   POL(#^1(x_1)) = 2 + 3*x_1
   POL($(x_1)) = 2*x_1
   POL(*(x_1)) = x_1
   POL(0(x_1)) = 2 + 2*x_1
   POL(0^1(x_1)) = 3 + 3*x_1
   POL(1(x_1)) = 2 + 2*x_1
   POL(1^1(x_1)) = 3 + 3*x_1


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

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

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

The TRS R consists of the following rules:

   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(x1))

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

(7) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(#(x_1)) = 1 + x_1
   POL($(x_1)) = 2*x_1
   POL(*(x_1)) = 1 + x_1
   POL(0(x_1)) = x_1
   POL(0^1(x_1)) = 1 + x_1
   POL(1(x_1)) = x_1
   POL(1^1(x_1)) = 2 + x_1


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

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

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

The TRS R consists of the following rules:

   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(10)
TRUE