/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 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:

   app(app(and, true), true) -> true
   app(app(and, x), false) -> false
   app(app(and, false), y) -> false
   app(app(or, true), y) -> true
   app(app(or, x), true) -> true
   app(app(or, false), false) -> false
   app(app(forall, p), nil) -> true
   app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs))
   app(app(forsome, p), nil) -> false
   app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), xs))

Q is empty.

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(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   app(app(and, true), true) -> true
   app(app(and, x), false) -> false
   app(app(and, false), y) -> false
   app(app(or, true), y) -> true
   app(app(or, x), true) -> true
   app(app(or, false), false) -> false
   app(app(forall, p), nil) -> true
   app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs))
   app(app(forsome, p), nil) -> false
   app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

   app(app(and, true), true)
   app(app(and, x0), false)
   app(app(and, false), x0)
   app(app(or, true), x0)
   app(app(or, x0), true)
   app(app(or, false), false)
   app(app(forall, x0), nil)
   app(app(forall, x0), app(app(cons, x1), x2))
   app(app(forsome, x0), nil)
   app(app(forsome, x0), app(app(cons, x1), x2))


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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:

   APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(and, app(p, x)), app(app(forall, p), xs))
   APP(app(forall, p), app(app(cons, x), xs)) -> APP(and, app(p, x))
   APP(app(forall, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(forall, p), xs)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(or, app(p, x)), app(app(forsome, p), xs))
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(or, app(p, x))
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(forsome, p), xs)

The TRS R consists of the following rules:

   app(app(and, true), true) -> true
   app(app(and, x), false) -> false
   app(app(and, false), y) -> false
   app(app(or, true), y) -> true
   app(app(or, x), true) -> true
   app(app(or, false), false) -> false
   app(app(forall, p), nil) -> true
   app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs))
   app(app(forsome, p), nil) -> false
   app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

   app(app(and, true), true)
   app(app(and, x0), false)
   app(app(and, false), x0)
   app(app(or, true), x0)
   app(app(or, x0), true)
   app(app(or, false), false)
   app(app(forall, x0), nil)
   app(app(forall, x0), app(app(cons, x1), x2))
   app(app(forsome, x0), nil)
   app(app(forsome, x0), app(app(cons, x1), x2))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
----------------------------------------

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

   APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(forall, p), xs)
   APP(app(forall, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(forsome, p), xs)

The TRS R consists of the following rules:

   app(app(and, true), true) -> true
   app(app(and, x), false) -> false
   app(app(and, false), y) -> false
   app(app(or, true), y) -> true
   app(app(or, x), true) -> true
   app(app(or, false), false) -> false
   app(app(forall, p), nil) -> true
   app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs))
   app(app(forsome, p), nil) -> false
   app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), xs))

The set Q consists of the following terms:

   app(app(and, true), true)
   app(app(and, x0), false)
   app(app(and, false), x0)
   app(app(or, true), x0)
   app(app(or, x0), true)
   app(app(or, false), false)
   app(app(forall, x0), nil)
   app(app(forall, x0), app(app(cons, x1), x2))
   app(app(forsome, x0), nil)
   app(app(forsome, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
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(7) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(forall, p), xs)
   APP(app(forall, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(p, x)
   APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(forsome, p), xs)

R is empty.
The set Q consists of the following terms:

   app(app(and, true), true)
   app(app(and, x0), false)
   app(app(and, false), x0)
   app(app(or, true), x0)
   app(app(or, x0), true)
   app(app(or, false), false)
   app(app(forall, x0), nil)
   app(app(forall, x0), app(app(cons, x1), x2))
   app(app(forsome, x0), nil)
   app(app(forsome, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
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(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:
*APP(app(forsome, p), app(app(cons, x), xs)) -> APP(p, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(forsome, p), xs)
The graph contains the following edges 1 >= 1, 2 > 2


*APP(app(forall, p), app(app(cons, x), xs)) -> APP(p, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(forall, p), xs)
The graph contains the following edges 1 >= 1, 2 > 2


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(10)
YES