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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 27 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 76 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:

   h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   t(t(x)) -> t(c(t(x), x))
   t(x) -> x
   t(x) -> c(0, c(0, c(0, c(0, c(0, x)))))

Q is empty.

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(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:

   H(c(x, y), c(s(z), z), t(w)) -> H(z, c(y, x), t(t(c(x, c(y, t(w))))))
   H(c(x, y), c(s(z), z), t(w)) -> T(t(c(x, c(y, t(w)))))
   H(c(x, y), c(s(z), z), t(w)) -> T(c(x, c(y, t(w))))
   H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w)))
   H(x, c(y, z), t(w)) -> T(c(t(w), w))
   H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   H(c(s(x), c(s(0), y)), z, t(x)) -> T(t(c(x, s(x))))
   H(c(s(x), c(s(0), y)), z, t(x)) -> T(c(x, s(x)))
   T(t(x)) -> T(c(t(x), x))

The TRS R consists of the following rules:

   h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   t(t(x)) -> t(c(t(x), x))
   t(x) -> x
   t(x) -> c(0, c(0, c(0, c(0, c(0, x)))))

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 1 SCC with 6 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w)))
   H(c(x, y), c(s(z), z), t(w)) -> H(z, c(y, x), t(t(c(x, c(y, t(w))))))
   H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))

The TRS R consists of the following rules:

   h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   t(t(x)) -> t(c(t(x), x))
   t(x) -> x
   t(x) -> c(0, c(0, c(0, c(0, c(0, x)))))

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

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


The following pairs can be oriented strictly and are deleted.

   H(c(x, y), c(s(z), z), t(w)) -> H(z, c(y, x), t(t(c(x, c(y, t(w))))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(H(x_1, x_2, x_3)) = x_1 + x_2
   POL(c(x_1, x_2)) = 1 + x_1 + x_2
   POL(s(x_1)) = x_1
   POL(t(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

   H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w)))
   H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))

The TRS R consists of the following rules:

   h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   t(t(x)) -> t(c(t(x), x))
   t(x) -> x
   t(x) -> c(0, c(0, c(0, c(0, c(0, x)))))

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

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


The following pairs can be oriented strictly and are deleted.

   H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = [1/2]
   POL(H(x_1, x_2, x_3)) = [1/2]x_1 + [1/4]x_2
   POL(c(x_1, x_2)) = [1/4]x_1 + x_2
   POL(s(x_1)) = [1/2]x_1
   POL(t(x_1)) = 0
The value of delta used in the strict ordering is 1/64.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

   H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w)))

The TRS R consists of the following rules:

   h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w))))))
   h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w)))
   h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
   t(t(x)) -> t(c(t(x), x))
   t(x) -> x
   t(x) -> c(0, c(0, c(0, c(0, c(0, x)))))

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

(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:
*H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w)))
The graph contains the following edges 2 > 2


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(10)
YES