SRS Stand 10685 pair #381717670

details

property	value
status	complete
benchmark	beans2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.97573113441 seconds
cpu usage	8.324958938
max memory	6.92436992E8

stage attributes

key	value
output-size	8601
starexec-result	YES

output

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


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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 21 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 4 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 19 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) TRUE
    (12) QDP
        (13) MNOCProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 20 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

Q is empty.

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

(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:

   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   C(b(x1)) -> B(a(x1))
   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
   L^1(a(a(x1))) -> B(c(x1))
   L^1(a(a(x1))) -> C(x1)
   C(R(x1)) -> B(a(R(x1)))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
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(4)
Complex Obligation (AND)

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

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

   B(a(a(x1))) -> C(x1)

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