/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 35 ms]
(4) QDP
(5) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(6) YES


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

   a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
   a(p(x, y), z) -> p(a(x, z), a(y, z))
   a(a(x, y), z) -> a(x, a(y, z))
   a(id, x) -> x
   a(1, id) -> 1
   a(t, id) -> t
   a(1, p(x, y)) -> x
   a(t, p(x, y)) -> y

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(lambda(x), y) -> A(x, p(1, a(y, t)))
   A(lambda(x), y) -> A(y, t)
   A(p(x, y), z) -> A(x, z)
   A(p(x, y), z) -> A(y, z)
   A(a(x, y), z) -> A(x, a(y, z))
   A(a(x, y), z) -> A(y, z)

The TRS R consists of the following rules:

   a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
   a(p(x, y), z) -> p(a(x, z), a(y, z))
   a(a(x, y), z) -> a(x, a(y, z))
   a(id, x) -> x
   a(1, id) -> 1
   a(t, id) -> t
   a(1, p(x, y)) -> x
   a(t, p(x, y)) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A(lambda(x), y) -> A(x, p(1, a(y, t)))
   A(lambda(x), y) -> A(y, t)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

   POL(1) = 0
   POL(A(x_1, x_2)) = x_1 + x_2
   POL(a(x_1, x_2)) = x_1 + x_2
   POL(id) = 1
   POL(lambda(x_1)) = 1 + x_1
   POL(p(x_1, x_2)) = max(x_1, x_2)
   POL(t) = 0

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

   a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
   a(p(x, y), z) -> p(a(x, z), a(y, z))
   a(a(x, y), z) -> a(x, a(y, z))
   a(id, x) -> x
   a(1, id) -> 1
   a(t, id) -> t
   a(1, p(x, y)) -> x
   a(t, p(x, y)) -> y


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

   A(p(x, y), z) -> A(x, z)
   A(p(x, y), z) -> A(y, z)
   A(a(x, y), z) -> A(x, a(y, z))
   A(a(x, y), z) -> A(y, z)

The TRS R consists of the following rules:

   a(lambda(x), y) -> lambda(a(x, p(1, a(y, t))))
   a(p(x, y), z) -> p(a(x, z), a(y, z))
   a(a(x, y), z) -> a(x, a(y, z))
   a(id, x) -> x
   a(1, id) -> 1
   a(t, id) -> t
   a(1, p(x, y)) -> x
   a(t, p(x, y)) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) 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:
*A(p(x, y), z) -> A(x, z)
The graph contains the following edges 1 > 1, 2 >= 2


*A(p(x, y), z) -> A(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


*A(a(x, y), z) -> A(x, a(y, z))
The graph contains the following edges 1 > 1


*A(a(x, y), z) -> A(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


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(6)
YES