TRS Standard pair #487067108

details

property	value
status	complete
benchmark	pair2hard.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	40.6398 seconds
cpu usage	100.158
user time	97.4236
system time	2.73392
max virtual memory	1.8716956E7
max residence set size	5432600.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesReductionPairsProof [EQUIVALENT, 3 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 20.5 s]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

Q is empty.

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

(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:

   P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2))
   P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2)
   P(a(x0), p(b(a(x1)), x2)) -> A(b(a(x1)))
   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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 2 SCCs with 1 less node.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

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