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


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


MAYBE
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 56 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(8) QTRS
(9) DependencyPairsProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QReductionProof [EQUIVALENT, 0 ms]
(14) QDP
(15) NonTerminationLoopProof [COMPLETE, 0 ms]
(16) NO


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

(0)
Obligation:
Conditional term rewrite system:
The TRS R consists of the following rules:

   pin(x) -> pout(g(x))

The conditional TRS C consists of the following conditional rules:

   pin(x) -> pout(f(y)) <= pin(x) -> pout(g(y))


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

(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
----------------------------------------

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

   pin(x) -> U1(pin(x))
   U1(pout(g(y))) -> pout(f(y))
   pin(x) -> pout(g(x))

Q is empty.

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

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

   POL(U1(x_1)) = x_1
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(pin(x_1)) = 2 + 2*x_1
   POL(pout(x_1)) = 1 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   pin(x) -> pout(g(x))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   pin(x) -> U1(pin(x))
   U1(pout(g(y))) -> pout(f(y))

Q is empty.

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

(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
pin/1)YES(
U1/1)YES(
pout/1)YES(
g/1(YES)
f/1(YES)

Quasi precedence:
g_1 > f_1


Status:
g_1: [1]
f_1: [1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   U1(pout(g(y))) -> pout(f(y))




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

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   pin(x) -> U1(pin(x))

Q is empty.

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

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

(8)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   pin(x) -> U1(pin(x))

The set Q consists of the following terms:

   pin(x0)


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

(9) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

   PIN(x) -> PIN(x)

The TRS R consists of the following rules:

   pin(x) -> U1(pin(x))

The set Q consists of the following terms:

   pin(x0)

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

(11) 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.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PIN(x) -> PIN(x)

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

   pin(x0)

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

(13) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   pin(x0)


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PIN(x) -> PIN(x)

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

(15) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = PIN(x) evaluates to  t =PIN(x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from PIN(x) to PIN(x).




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

(16)
NO