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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 25 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (12) YES


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

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   subst(a, id) -> a
   msubst(a, id) -> a
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))

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:

   CIRC(cons(a, s), t) -> MSUBST(a, t)
   CIRC(cons(a, s), t) -> CIRC(s, t)
   CIRC(cons(lift, s), cons(a, t)) -> CIRC(s, t)
   CIRC(cons(lift, s), cons(lift, t)) -> CIRC(s, t)
   CIRC(circ(s, t), u) -> CIRC(s, circ(t, u))
   CIRC(circ(s, t), u) -> CIRC(t, u)
   CIRC(cons(lift, s), circ(cons(lift, t), u)) -> CIRC(cons(lift, circ(s, t)), u)
   CIRC(cons(lift, s), circ(cons(lift, t), u)) -> CIRC(s, t)
   MSUBST(msubst(a, s), t) -> MSUBST(a, circ(s, t))
   MSUBST(msubst(a, s), t) -> CIRC(s, t)

The TRS R consists of the following rules:

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   subst(a, id) -> a
   msubst(a, id) -> a
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))

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

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


The following pairs can be oriented strictly and are deleted.

   CIRC(cons(a, s), t) -> MSUBST(a, t)
   CIRC(cons(a, s), t) -> CIRC(s, t)
   CIRC(cons(lift, s), cons(a, t)) -> CIRC(s, t)
   CIRC(cons(lift, s), cons(lift, t)) -> CIRC(s, t)
   CIRC(cons(lift, s), circ(cons(lift, t), u)) -> CIRC(cons(lift, circ(s, t)), u)
   CIRC(cons(lift, s), circ(cons(lift, t), u)) -> CIRC(s, t)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(CIRC(x_1, x_2)) = x_1 + x_1*x_2
   POL(MSUBST(x_1, x_2)) = x_1 + x_1*x_2
   POL(circ(x_1, x_2)) = x_1 + x_1*x_2 + x_2
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(id) = 1
   POL(lift) = 0
   POL(msubst(x_1, x_2)) = x_1 + x_1*x_2 + x_2

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

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
   msubst(a, id) -> a


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

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

   CIRC(circ(s, t), u) -> CIRC(s, circ(t, u))
   CIRC(circ(s, t), u) -> CIRC(t, u)
   MSUBST(msubst(a, s), t) -> MSUBST(a, circ(s, t))
   MSUBST(msubst(a, s), t) -> CIRC(s, t)

The TRS R consists of the following rules:

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   subst(a, id) -> a
   msubst(a, id) -> a
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))

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 2 SCCs with 1 less node.
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(6)
Complex Obligation (AND)

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

   CIRC(circ(s, t), u) -> CIRC(t, u)
   CIRC(circ(s, t), u) -> CIRC(s, circ(t, u))

The TRS R consists of the following rules:

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   subst(a, id) -> a
   msubst(a, id) -> a
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(8) 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:
*CIRC(circ(s, t), u) -> CIRC(t, u)
The graph contains the following edges 1 > 1, 2 >= 2


*CIRC(circ(s, t), u) -> CIRC(s, circ(t, u))
The graph contains the following edges 1 > 1


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(9)
YES

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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MSUBST(msubst(a, s), t) -> MSUBST(a, circ(s, t))

The TRS R consists of the following rules:

   circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
   circ(cons(lift, s), cons(a, t)) -> cons(a, circ(s, t))
   circ(cons(lift, s), cons(lift, t)) -> cons(lift, circ(s, t))
   circ(circ(s, t), u) -> circ(s, circ(t, u))
   circ(s, id) -> s
   circ(id, s) -> s
   circ(cons(lift, s), circ(cons(lift, t), u)) -> circ(cons(lift, circ(s, t)), u)
   subst(a, id) -> a
   msubst(a, id) -> a
   msubst(msubst(a, s), t) -> msubst(a, circ(s, t))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) 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:
*MSUBST(msubst(a, s), t) -> MSUBST(a, circ(s, t))
The graph contains the following edges 1 > 1


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(12)
YES